Common root question on quadratics equations to show that $a+b+c=0$

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If $f(x)=ax^2+bx-c$ and $g(x)=ax^2+cx+b$ have a common root , show that $a+b+c=0$.

I tried this by thinking that $alpha$ is the common root and then I got by substituting and solving ,

$(b^2+c^2)(b-c)+a(b+c)^2=0$

how can I proceed from here ? Any ideas ?

algebra-precalculus roots examples-counterexamples quadratics

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edited 11 hours ago

TheSimpliFire

9,18651651

asked 15 hours ago

Angelo Mark

3,98921236

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that $g(1)=0.$
  – prog_SAHIL
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Angelo Mark You can not change the given after answer on your question. Ask new question.
  – Michael Rozenberg
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i missed to mention that f is not identical to g. Really sorry. I'll mark your answer correct and ask this again, :D
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Open new topic, please. Thank you!
  – Michael Rozenberg
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The necessary and sufficient condition for $f$ and $g$ having a common root is that $a(b+c)^2+b(b^2-c^2)-c(b-c)^2=0$. In fact, a common root is a roots of $(f-g)(x)=(b-c)x-(b+c)$. The condition above is equivalent to this root being a root of $f$, but being also a root of $f-g$, it is equivalent to being a root of $g$ too. This condition is not equivalent to $a+b+c=0$, even when $b,cneq 0$.
  – spiralstotheleft
  15 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

If $f(x)=ax^2+bx-c$ and $g(x)=ax^2+cx+b$ have a common root , show that $a+b+c=0$.

I tried this by thinking that $alpha$ is the common root and then I got by substituting and solving ,

$(b^2+c^2)(b-c)+a(b+c)^2=0$

how can I proceed from here ? Any ideas ?

algebra-precalculus roots examples-counterexamples quadratics

share|cite|improve this question

edited 11 hours ago

TheSimpliFire

9,18651651

asked 15 hours ago

Angelo Mark

3,98921236

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that $g(1)=0.$
  – prog_SAHIL
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Angelo Mark You can not change the given after answer on your question. Ask new question.
  – Michael Rozenberg
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i missed to mention that f is not identical to g. Really sorry. I'll mark your answer correct and ask this again, :D
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Open new topic, please. Thank you!
  – Michael Rozenberg
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The necessary and sufficient condition for $f$ and $g$ having a common root is that $a(b+c)^2+b(b^2-c^2)-c(b-c)^2=0$. In fact, a common root is a roots of $(f-g)(x)=(b-c)x-(b+c)$. The condition above is equivalent to this root being a root of $f$, but being also a root of $f-g$, it is equivalent to being a root of $g$ too. This condition is not equivalent to $a+b+c=0$, even when $b,cneq 0$.
  – spiralstotheleft
  15 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

up vote
0
down vote

favorite

If $f(x)=ax^2+bx-c$ and $g(x)=ax^2+cx+b$ have a common root , show that $a+b+c=0$.

I tried this by thinking that $alpha$ is the common root and then I got by substituting and solving ,

$(b^2+c^2)(b-c)+a(b+c)^2=0$

how can I proceed from here ? Any ideas ?

algebra-precalculus roots examples-counterexamples quadratics

share|cite|improve this question

edited 11 hours ago

TheSimpliFire

9,18651651

asked 15 hours ago

Angelo Mark

3,98921236

If $f(x)=ax^2+bx-c$ and $g(x)=ax^2+cx+b$ have a common root , show that $a+b+c=0$.

I tried this by thinking that $alpha$ is the common root and then I got by substituting and solving ,

$(b^2+c^2)(b-c)+a(b+c)^2=0$

how can I proceed from here ? Any ideas ?

algebra-precalculus roots examples-counterexamples quadratics

share|cite|improve this question

edited 11 hours ago

TheSimpliFire

9,18651651

asked 15 hours ago

Angelo Mark

3,98921236

share|cite|improve this question

share|cite|improve this question

edited 11 hours ago

TheSimpliFire

9,18651651

edited 11 hours ago

TheSimpliFire

9,18651651

edited 11 hours ago

TheSimpliFire

9,18651651

9,18651651

asked 15 hours ago

Angelo Mark

3,98921236

asked 15 hours ago

Angelo Mark

3,98921236

asked 15 hours ago

Angelo Mark

3,98921236

3,98921236

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that $g(1)=0.$
  – prog_SAHIL
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Angelo Mark You can not change the given after answer on your question. Ask new question.
  – Michael Rozenberg
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i missed to mention that f is not identical to g. Really sorry. I'll mark your answer correct and ask this again, :D
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Open new topic, please. Thank you!
  – Michael Rozenberg
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The necessary and sufficient condition for $f$ and $g$ having a common root is that $a(b+c)^2+b(b^2-c^2)-c(b-c)^2=0$. In fact, a common root is a roots of $(f-g)(x)=(b-c)x-(b+c)$. The condition above is equivalent to this root being a root of $f$, but being also a root of $f-g$, it is equivalent to being a root of $g$ too. This condition is not equivalent to $a+b+c=0$, even when $b,cneq 0$.
  – spiralstotheleft
  15 hours ago
  
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that $g(1)=0.$
  – prog_SAHIL
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Angelo Mark You can not change the given after answer on your question. Ask new question.
  – Michael Rozenberg
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i missed to mention that f is not identical to g. Really sorry. I'll mark your answer correct and ask this again, :D
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Open new topic, please. Thank you!
  – Michael Rozenberg
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The necessary and sufficient condition for $f$ and $g$ having a common root is that $a(b+c)^2+b(b^2-c^2)-c(b-c)^2=0$. In fact, a common root is a roots of $(f-g)(x)=(b-c)x-(b+c)$. The condition above is equivalent to this root being a root of $f$, but being also a root of $f-g$, it is equivalent to being a root of $g$ too. This condition is not equivalent to $a+b+c=0$, even when $b,cneq 0$.
  – spiralstotheleft
  15 hours ago
  
  

  
  
  
  

Note that $g(1)=0.$
– prog_SAHIL
15 hours ago

Note that $g(1)=0.$
– prog_SAHIL
15 hours ago

@Angelo Mark You can not change the given after answer on your question. Ask new question.
– Michael Rozenberg
15 hours ago

@Angelo Mark You can not change the given after answer on your question. Ask new question.
– Michael Rozenberg
15 hours ago

i missed to mention that f is not identical to g. Really sorry. I'll mark your answer correct and ask this again, :D
– Angelo Mark
15 hours ago

i missed to mention that f is not identical to g. Really sorry. I'll mark your answer correct and ask this again, :D
– Angelo Mark
15 hours ago

Open new topic, please. Thank you!
– Michael Rozenberg
15 hours ago

Open new topic, please. Thank you!
– Michael Rozenberg
15 hours ago

The necessary and sufficient condition for $f$ and $g$ having a common root is that $a(b+c)^2+b(b^2-c^2)-c(b-c)^2=0$. In fact, a common root is a roots of $(f-g)(x)=(b-c)x-(b+c)$. The condition above is equivalent to this root being a root of $f$, but being also a root of $f-g$, it is equivalent to being a root of $g$ too. This condition is not equivalent to $a+b+c=0$, even when $b,cneq 0$.
– spiralstotheleft
15 hours ago

The necessary and sufficient condition for $f$ and $g$ having a common root is that $a(b+c)^2+b(b^2-c^2)-c(b-c)^2=0$. In fact, a common root is a roots of $(f-g)(x)=(b-c)x-(b+c)$. The condition above is equivalent to this root being a root of $f$, but being also a root of $f-g$, it is equivalent to being a root of $g$ too. This condition is not equivalent to $a+b+c=0$, even when $b,cneq 0$.
– spiralstotheleft
15 hours ago

 | 
show 1 more comment

9 Answers
9

active

oldest

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up vote
2
down vote



accepted

Your problem is WRONG.

Let $a=2,b=0,c=2$, then $f(x)=2x^2-2, g(x)=2x^2+2x$. It's clear that $-1$ is the common root, but $a+b+c=4neq 0$.

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answered 14 hours ago

Le Anh Dung

701317

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up vote
3
down vote

It's wrong.

Try $b=c=0$ and $a=1$.

share|cite|improve this answer

answered 15 hours ago

Michael Rozenberg

86.9k1575178

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks :) BUT f is not equal to g
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Angelo Mark But $x^2$ and $x^2$ have common root. You can not change the given now. Ask another question.
  – Michael Rozenberg
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay . Do not worry. I'll mark your answer as the correct one. Thanks. And really sorry that I missed to mention that f and g are not identical
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Angelo Mark This happens. All correct.
  – Michael Rozenberg
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why someone down voted?
  – Michael Rozenberg
  14 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

HINT:

Vieta's formulas.

Let the roots of $f$ be $p$ and $q$. Let the roots of $g$ be $p$ and $r$.

Then $$p+q=-frac ba,quad pq=-frac ca$$ and $$p+r=-frac ca,quad pr=frac ba$$

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answered 15 hours ago

TheSimpliFire

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up vote
1
down vote

Counterexample: $a=c, b=0$. That is:
$$f(x)=ax^2+bx-c=ax^2-a=0 Rightarrow x=pm 1;\
g(x)=ax^2+cx+b=ax^2+ax=0 Rightarrow x=-1;0.$$

Backwards checking: if $c=-(a+b)$, then:
$$f(x)= ax^2+bx-c=0 iff ax^2+bx+a+b=0 (1) \
g(x)=ax^2+cx+b=0 iff ax^2-ax-bx+b=0 iff (x-1)(ax+a-b)=0 (2) $$
Then, from $(2)$:
$$x-1=0 stackrel(1)Rightarrow a=-b.\
ax+a-b=0 Rightarrow x=fracb-aa Rightarrow afrac(b-a)^2a^2+bfracb-aa+a+b=0 Rightarrow a^2+b^2=ab.$$

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answered 14 hours ago

farruhota

13.4k2632

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1
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Analysis:

If $f,g$ has one same root, then it is also a root of $f-g$. Since
$$
f(x) - g(x) = (b-c)x -(b+c),
$$
then common root should be
$$
r = frac b+c b-c quad [b neq c].
$$
Plug this into $f(x)$:
$$
a (b+c)^2 + b (b+c)(b-c) - c(b-c)^2 =0,
$$
which is
$$
a(b+c)^2 + b^3 -c^3 +bc^2 -cb^2 =0.
$$
Hence
$$
a + b + c = frac 1 (b+c)^2 (2c^3 + 4b^2c + 2bc^2) = frac 2c (b+c)^2 (c^2 + 2b^2 + bc).
$$
If $c = 0$, and if $bneq 0$, then $a+b + c = 0$. If $c neq 0$ and $b +c neq 0$, then
$$
|a +b +c| = frac 2 (b+c)^2 left( left( c+frac b2right)^2 +frac 74 b^2right) > 0.
$$
Now a counterexample: take $b=0$, then $a = c $. Pick $a =c =1$, then
$$
f(x) = x^2 - 1= (x+1)(x-1),quad g(x)= x^2+x =x(x+1),
$$
clearly they have a common root $-1$ but $a+b+c = g(1)=2neq 0$.

Conclusion: such claim fails.

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answered 14 hours ago

xbh

8806

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Hint: the roots of $f$ are $x = frac-b pm sqrtb^2 + 4ac2a$. (Why? What happened to the minus sign in the quadratic formula?)

What are the roots of $g$?

Now suppose that the first root of $f$ equals the first root of $g$, and three other cases.

(Yes, that's a very pedestrian way to approach this problem, but it'll get you there.)

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edited 15 hours ago

Ángel Mario Gallegos

18.1k11229

answered 15 hours ago

John Hughes

59.3k23685

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tried this way , but its hard to get the answer
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's why it's an exercise. If it were easy, it'd just be a remark in the textbook. Why don't you show us your work on this approach (use the "edit" button below your question to fill in what you've done), and we can help you proceed. Yeah, it'll mean typing a lot of math...but that's what you're asking us to do, and you're the one getting the benefit, so perhaps you should be doing more of the work.
  – John Hughes
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did this whole day
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good for you. But we don't know what you've done, because you haven't bothered to show us. Sigh. We're really not here to do your homework for you.
  – John Hughes
  15 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

Hint 2: An alternative approach is to say that

$$
f(x) = a (x - u) (x - v)
$$
for some numbers $u$ and $v$ (the roots of $f$), and
$$
g(x) = a(x-u) (x - w)
$$

(the $u$ is repeated because $f$ and $g$ share a root; it's also possible that $v = w$, but not required by any means.)

Now if you expand out $f$, you'll find a relation between $b$ and $u, v$, and between $c$ and $u, v$; similarly for $g$. See where that leads you.

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answered 15 hours ago

John Hughes

59.3k23685

• 
  
  
  

  
  

  
  
  
  
  
  

  
  did it but leads to the same place where I got stuck
  – Angelo Mark
  15 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

$$f(x)=ax^2+bx-c$$
$$g(x)=ax^2+cx+b$$
for f(x), the roots are given by:
$$(1),x=frac-b+sqrtb^2+4ac2a textand, x=frac-b-sqrtb^2+4ac2a$$
for g(x), the roots are given by:
$$(2),x=frac-c+sqrtc^2-4ab2a textand,x=frac-c-sqrtc^2-4ab2a$$
since $2a$ is common on the bottom you could remove this then make the two sides equal to each other?

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answered 15 hours ago

Henry Lee

48210

add a comment | 

up vote
0
down vote

If $f(x)=ax^2+bx-c$ and $g(x)=ax^2+cx+b$ have a common root , show that $a+b+c=0$.

The statement is false, as pointed out in several answers already.

The problem most likely has a typo, however, and the following statement is in fact true:

If $b ne c$, then $f(x)=ax^2+bx colorred+c$ and $g(x)=ax^2+cx+b$ have a common root iff $a+b+c=0$.

The proof follows immediately by subtracting the two equations, which gives $,(b-c)(x-1)=0,$, therefore the common root must be $,x=1,$.

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answered 9 hours ago

dxiv

53.6k64696

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9 Answers
9

active

oldest

votes

9 Answers
9

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Your problem is WRONG.

Let $a=2,b=0,c=2$, then $f(x)=2x^2-2, g(x)=2x^2+2x$. It's clear that $-1$ is the common root, but $a+b+c=4neq 0$.

share|cite|improve this answer

answered 14 hours ago

Le Anh Dung

701317

add a comment | 

up vote
2
down vote



accepted

Your problem is WRONG.

Let $a=2,b=0,c=2$, then $f(x)=2x^2-2, g(x)=2x^2+2x$. It's clear that $-1$ is the common root, but $a+b+c=4neq 0$.

share|cite|improve this answer

answered 14 hours ago

Le Anh Dung

701317

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Your problem is WRONG.

Let $a=2,b=0,c=2$, then $f(x)=2x^2-2, g(x)=2x^2+2x$. It's clear that $-1$ is the common root, but $a+b+c=4neq 0$.

share|cite|improve this answer

answered 14 hours ago

Le Anh Dung

701317

Your problem is WRONG.

Let $a=2,b=0,c=2$, then $f(x)=2x^2-2, g(x)=2x^2+2x$. It's clear that $-1$ is the common root, but $a+b+c=4neq 0$.

share|cite|improve this answer

answered 14 hours ago

Le Anh Dung

701317

share|cite|improve this answer

share|cite|improve this answer

answered 14 hours ago

Le Anh Dung

701317

answered 14 hours ago

Le Anh Dung

701317

answered 14 hours ago

Le Anh Dung

701317

701317

add a comment | 

add a comment | 

up vote
3
down vote

It's wrong.

Try $b=c=0$ and $a=1$.

share|cite|improve this answer

answered 15 hours ago

Michael Rozenberg

86.9k1575178

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks :) BUT f is not equal to g
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Angelo Mark But $x^2$ and $x^2$ have common root. You can not change the given now. Ask another question.
  – Michael Rozenberg
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay . Do not worry. I'll mark your answer as the correct one. Thanks. And really sorry that I missed to mention that f and g are not identical
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Angelo Mark This happens. All correct.
  – Michael Rozenberg
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why someone down voted?
  – Michael Rozenberg
  14 hours ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

It's wrong.

Try $b=c=0$ and $a=1$.

share|cite|improve this answer

answered 15 hours ago

Michael Rozenberg

86.9k1575178

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks :) BUT f is not equal to g
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Angelo Mark But $x^2$ and $x^2$ have common root. You can not change the given now. Ask another question.
  – Michael Rozenberg
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay . Do not worry. I'll mark your answer as the correct one. Thanks. And really sorry that I missed to mention that f and g are not identical
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Angelo Mark This happens. All correct.
  – Michael Rozenberg
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why someone down voted?
  – Michael Rozenberg
  14 hours ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

It's wrong.

Try $b=c=0$ and $a=1$.

share|cite|improve this answer

answered 15 hours ago

Michael Rozenberg

86.9k1575178

It's wrong.

Try $b=c=0$ and $a=1$.

share|cite|improve this answer

answered 15 hours ago

Michael Rozenberg

86.9k1575178

share|cite|improve this answer

share|cite|improve this answer

answered 15 hours ago

Michael Rozenberg

86.9k1575178

answered 15 hours ago

Michael Rozenberg

86.9k1575178

answered 15 hours ago

Michael Rozenberg

86.9k1575178

86.9k1575178

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks :) BUT f is not equal to g
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Angelo Mark But $x^2$ and $x^2$ have common root. You can not change the given now. Ask another question.
  – Michael Rozenberg
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay . Do not worry. I'll mark your answer as the correct one. Thanks. And really sorry that I missed to mention that f and g are not identical
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Angelo Mark This happens. All correct.
  – Michael Rozenberg
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why someone down voted?
  – Michael Rozenberg
  14 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks :) BUT f is not equal to g
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Angelo Mark But $x^2$ and $x^2$ have common root. You can not change the given now. Ask another question.
  – Michael Rozenberg
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay . Do not worry. I'll mark your answer as the correct one. Thanks. And really sorry that I missed to mention that f and g are not identical
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Angelo Mark This happens. All correct.
  – Michael Rozenberg
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why someone down voted?
  – Michael Rozenberg
  14 hours ago
  

  
  
  
  

1

1

Thanks :) BUT f is not equal to g
– Angelo Mark
15 hours ago

Thanks :) BUT f is not equal to g
– Angelo Mark
15 hours ago

@Angelo Mark But $x^2$ and $x^2$ have common root. You can not change the given now. Ask another question.
– Michael Rozenberg
15 hours ago

@Angelo Mark But $x^2$ and $x^2$ have common root. You can not change the given now. Ask another question.
– Michael Rozenberg
15 hours ago

Okay . Do not worry. I'll mark your answer as the correct one. Thanks. And really sorry that I missed to mention that f and g are not identical
– Angelo Mark
15 hours ago

Okay . Do not worry. I'll mark your answer as the correct one. Thanks. And really sorry that I missed to mention that f and g are not identical
– Angelo Mark
15 hours ago

@Angelo Mark This happens. All correct.
– Michael Rozenberg
15 hours ago

@Angelo Mark This happens. All correct.
– Michael Rozenberg
15 hours ago

Why someone down voted?
– Michael Rozenberg
14 hours ago

Why someone down voted?
– Michael Rozenberg
14 hours ago

add a comment | 

up vote
2
down vote

HINT:

Vieta's formulas.

Let the roots of $f$ be $p$ and $q$. Let the roots of $g$ be $p$ and $r$.

Then $$p+q=-frac ba,quad pq=-frac ca$$ and $$p+r=-frac ca,quad pr=frac ba$$

share|cite|improve this answer

answered 15 hours ago

TheSimpliFire

9,18651651

add a comment | 

up vote
2
down vote

HINT:

Vieta's formulas.

Let the roots of $f$ be $p$ and $q$. Let the roots of $g$ be $p$ and $r$.

Then $$p+q=-frac ba,quad pq=-frac ca$$ and $$p+r=-frac ca,quad pr=frac ba$$

share|cite|improve this answer

answered 15 hours ago

TheSimpliFire

9,18651651

add a comment | 

up vote
2
down vote

up vote
2
down vote

HINT:

Vieta's formulas.

Let the roots of $f$ be $p$ and $q$. Let the roots of $g$ be $p$ and $r$.

Then $$p+q=-frac ba,quad pq=-frac ca$$ and $$p+r=-frac ca,quad pr=frac ba$$

share|cite|improve this answer

answered 15 hours ago

TheSimpliFire

9,18651651

HINT:

Vieta's formulas.

Let the roots of $f$ be $p$ and $q$. Let the roots of $g$ be $p$ and $r$.

Then $$p+q=-frac ba,quad pq=-frac ca$$ and $$p+r=-frac ca,quad pr=frac ba$$

share|cite|improve this answer

answered 15 hours ago

TheSimpliFire

9,18651651

share|cite|improve this answer

share|cite|improve this answer

answered 15 hours ago

TheSimpliFire

9,18651651

answered 15 hours ago

TheSimpliFire

9,18651651

answered 15 hours ago

TheSimpliFire

9,18651651

9,18651651

add a comment | 

add a comment | 

up vote
1
down vote

Counterexample: $a=c, b=0$. That is:
$$f(x)=ax^2+bx-c=ax^2-a=0 Rightarrow x=pm 1;\
g(x)=ax^2+cx+b=ax^2+ax=0 Rightarrow x=-1;0.$$

Backwards checking: if $c=-(a+b)$, then:
$$f(x)= ax^2+bx-c=0 iff ax^2+bx+a+b=0 (1) \
g(x)=ax^2+cx+b=0 iff ax^2-ax-bx+b=0 iff (x-1)(ax+a-b)=0 (2) $$
Then, from $(2)$:
$$x-1=0 stackrel(1)Rightarrow a=-b.\
ax+a-b=0 Rightarrow x=fracb-aa Rightarrow afrac(b-a)^2a^2+bfracb-aa+a+b=0 Rightarrow a^2+b^2=ab.$$

share|cite|improve this answer

answered 14 hours ago

farruhota

13.4k2632

add a comment | 

up vote
1
down vote

Counterexample: $a=c, b=0$. That is:
$$f(x)=ax^2+bx-c=ax^2-a=0 Rightarrow x=pm 1;\
g(x)=ax^2+cx+b=ax^2+ax=0 Rightarrow x=-1;0.$$

Backwards checking: if $c=-(a+b)$, then:
$$f(x)= ax^2+bx-c=0 iff ax^2+bx+a+b=0 (1) \
g(x)=ax^2+cx+b=0 iff ax^2-ax-bx+b=0 iff (x-1)(ax+a-b)=0 (2) $$
Then, from $(2)$:
$$x-1=0 stackrel(1)Rightarrow a=-b.\
ax+a-b=0 Rightarrow x=fracb-aa Rightarrow afrac(b-a)^2a^2+bfracb-aa+a+b=0 Rightarrow a^2+b^2=ab.$$

share|cite|improve this answer

answered 14 hours ago

farruhota

13.4k2632

add a comment | 

up vote
1
down vote

up vote
1
down vote

Counterexample: $a=c, b=0$. That is:
$$f(x)=ax^2+bx-c=ax^2-a=0 Rightarrow x=pm 1;\
g(x)=ax^2+cx+b=ax^2+ax=0 Rightarrow x=-1;0.$$

Backwards checking: if $c=-(a+b)$, then:
$$f(x)= ax^2+bx-c=0 iff ax^2+bx+a+b=0 (1) \
g(x)=ax^2+cx+b=0 iff ax^2-ax-bx+b=0 iff (x-1)(ax+a-b)=0 (2) $$
Then, from $(2)$:
$$x-1=0 stackrel(1)Rightarrow a=-b.\
ax+a-b=0 Rightarrow x=fracb-aa Rightarrow afrac(b-a)^2a^2+bfracb-aa+a+b=0 Rightarrow a^2+b^2=ab.$$

share|cite|improve this answer

answered 14 hours ago

farruhota

13.4k2632

Counterexample: $a=c, b=0$. That is:
$$f(x)=ax^2+bx-c=ax^2-a=0 Rightarrow x=pm 1;\
g(x)=ax^2+cx+b=ax^2+ax=0 Rightarrow x=-1;0.$$

Backwards checking: if $c=-(a+b)$, then:
$$f(x)= ax^2+bx-c=0 iff ax^2+bx+a+b=0 (1) \
g(x)=ax^2+cx+b=0 iff ax^2-ax-bx+b=0 iff (x-1)(ax+a-b)=0 (2) $$
Then, from $(2)$:
$$x-1=0 stackrel(1)Rightarrow a=-b.\
ax+a-b=0 Rightarrow x=fracb-aa Rightarrow afrac(b-a)^2a^2+bfracb-aa+a+b=0 Rightarrow a^2+b^2=ab.$$

share|cite|improve this answer

answered 14 hours ago

farruhota

13.4k2632

share|cite|improve this answer

share|cite|improve this answer

answered 14 hours ago

farruhota

13.4k2632

answered 14 hours ago

farruhota

13.4k2632

answered 14 hours ago

farruhota

13.4k2632

13.4k2632

add a comment | 

add a comment | 

up vote
1
down vote

Analysis:

If $f,g$ has one same root, then it is also a root of $f-g$. Since
$$
f(x) - g(x) = (b-c)x -(b+c),
$$
then common root should be
$$
r = frac b+c b-c quad [b neq c].
$$
Plug this into $f(x)$:
$$
a (b+c)^2 + b (b+c)(b-c) - c(b-c)^2 =0,
$$
which is
$$
a(b+c)^2 + b^3 -c^3 +bc^2 -cb^2 =0.
$$
Hence
$$
a + b + c = frac 1 (b+c)^2 (2c^3 + 4b^2c + 2bc^2) = frac 2c (b+c)^2 (c^2 + 2b^2 + bc).
$$
If $c = 0$, and if $bneq 0$, then $a+b + c = 0$. If $c neq 0$ and $b +c neq 0$, then
$$
|a +b +c| = frac 2 (b+c)^2 left( left( c+frac b2right)^2 +frac 74 b^2right) > 0.
$$
Now a counterexample: take $b=0$, then $a = c $. Pick $a =c =1$, then
$$
f(x) = x^2 - 1= (x+1)(x-1),quad g(x)= x^2+x =x(x+1),
$$
clearly they have a common root $-1$ but $a+b+c = g(1)=2neq 0$.

Conclusion: such claim fails.

share|cite|improve this answer

answered 14 hours ago

xbh

8806

add a comment | 

up vote
1
down vote

Analysis:

If $f,g$ has one same root, then it is also a root of $f-g$. Since
$$
f(x) - g(x) = (b-c)x -(b+c),
$$
then common root should be
$$
r = frac b+c b-c quad [b neq c].
$$
Plug this into $f(x)$:
$$
a (b+c)^2 + b (b+c)(b-c) - c(b-c)^2 =0,
$$
which is
$$
a(b+c)^2 + b^3 -c^3 +bc^2 -cb^2 =0.
$$
Hence
$$
a + b + c = frac 1 (b+c)^2 (2c^3 + 4b^2c + 2bc^2) = frac 2c (b+c)^2 (c^2 + 2b^2 + bc).
$$
If $c = 0$, and if $bneq 0$, then $a+b + c = 0$. If $c neq 0$ and $b +c neq 0$, then
$$
|a +b +c| = frac 2 (b+c)^2 left( left( c+frac b2right)^2 +frac 74 b^2right) > 0.
$$
Now a counterexample: take $b=0$, then $a = c $. Pick $a =c =1$, then
$$
f(x) = x^2 - 1= (x+1)(x-1),quad g(x)= x^2+x =x(x+1),
$$
clearly they have a common root $-1$ but $a+b+c = g(1)=2neq 0$.

Conclusion: such claim fails.

share|cite|improve this answer

answered 14 hours ago

xbh

8806

add a comment | 

up vote
1
down vote

up vote
1
down vote

Analysis:

If $f,g$ has one same root, then it is also a root of $f-g$. Since
$$
f(x) - g(x) = (b-c)x -(b+c),
$$
then common root should be
$$
r = frac b+c b-c quad [b neq c].
$$
Plug this into $f(x)$:
$$
a (b+c)^2 + b (b+c)(b-c) - c(b-c)^2 =0,
$$
which is
$$
a(b+c)^2 + b^3 -c^3 +bc^2 -cb^2 =0.
$$
Hence
$$
a + b + c = frac 1 (b+c)^2 (2c^3 + 4b^2c + 2bc^2) = frac 2c (b+c)^2 (c^2 + 2b^2 + bc).
$$
If $c = 0$, and if $bneq 0$, then $a+b + c = 0$. If $c neq 0$ and $b +c neq 0$, then
$$
|a +b +c| = frac 2 (b+c)^2 left( left( c+frac b2right)^2 +frac 74 b^2right) > 0.
$$
Now a counterexample: take $b=0$, then $a = c $. Pick $a =c =1$, then
$$
f(x) = x^2 - 1= (x+1)(x-1),quad g(x)= x^2+x =x(x+1),
$$
clearly they have a common root $-1$ but $a+b+c = g(1)=2neq 0$.

Conclusion: such claim fails.

share|cite|improve this answer

answered 14 hours ago

xbh

8806

Analysis:

If $f,g$ has one same root, then it is also a root of $f-g$. Since
$$
f(x) - g(x) = (b-c)x -(b+c),
$$
then common root should be
$$
r = frac b+c b-c quad [b neq c].
$$
Plug this into $f(x)$:
$$
a (b+c)^2 + b (b+c)(b-c) - c(b-c)^2 =0,
$$
which is
$$
a(b+c)^2 + b^3 -c^3 +bc^2 -cb^2 =0.
$$
Hence
$$
a + b + c = frac 1 (b+c)^2 (2c^3 + 4b^2c + 2bc^2) = frac 2c (b+c)^2 (c^2 + 2b^2 + bc).
$$
If $c = 0$, and if $bneq 0$, then $a+b + c = 0$. If $c neq 0$ and $b +c neq 0$, then
$$
|a +b +c| = frac 2 (b+c)^2 left( left( c+frac b2right)^2 +frac 74 b^2right) > 0.
$$
Now a counterexample: take $b=0$, then $a = c $. Pick $a =c =1$, then
$$
f(x) = x^2 - 1= (x+1)(x-1),quad g(x)= x^2+x =x(x+1),
$$
clearly they have a common root $-1$ but $a+b+c = g(1)=2neq 0$.

Conclusion: such claim fails.

share|cite|improve this answer

answered 14 hours ago

xbh

8806

share|cite|improve this answer

share|cite|improve this answer

answered 14 hours ago

xbh

8806

answered 14 hours ago

xbh

8806

answered 14 hours ago

xbh

8806

8806

add a comment | 

add a comment | 

up vote
0
down vote

Hint: the roots of $f$ are $x = frac-b pm sqrtb^2 + 4ac2a$. (Why? What happened to the minus sign in the quadratic formula?)

What are the roots of $g$?

Now suppose that the first root of $f$ equals the first root of $g$, and three other cases.

(Yes, that's a very pedestrian way to approach this problem, but it'll get you there.)

share|cite|improve this answer

edited 15 hours ago

Ángel Mario Gallegos

18.1k11229

answered 15 hours ago

John Hughes

59.3k23685

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tried this way , but its hard to get the answer
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's why it's an exercise. If it were easy, it'd just be a remark in the textbook. Why don't you show us your work on this approach (use the "edit" button below your question to fill in what you've done), and we can help you proceed. Yeah, it'll mean typing a lot of math...but that's what you're asking us to do, and you're the one getting the benefit, so perhaps you should be doing more of the work.
  – John Hughes
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did this whole day
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good for you. But we don't know what you've done, because you haven't bothered to show us. Sigh. We're really not here to do your homework for you.
  – John Hughes
  15 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

Hint: the roots of $f$ are $x = frac-b pm sqrtb^2 + 4ac2a$. (Why? What happened to the minus sign in the quadratic formula?)

What are the roots of $g$?

Now suppose that the first root of $f$ equals the first root of $g$, and three other cases.

(Yes, that's a very pedestrian way to approach this problem, but it'll get you there.)

share|cite|improve this answer

edited 15 hours ago

Ángel Mario Gallegos

18.1k11229

answered 15 hours ago

John Hughes

59.3k23685

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tried this way , but its hard to get the answer
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's why it's an exercise. If it were easy, it'd just be a remark in the textbook. Why don't you show us your work on this approach (use the "edit" button below your question to fill in what you've done), and we can help you proceed. Yeah, it'll mean typing a lot of math...but that's what you're asking us to do, and you're the one getting the benefit, so perhaps you should be doing more of the work.
  – John Hughes
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did this whole day
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good for you. But we don't know what you've done, because you haven't bothered to show us. Sigh. We're really not here to do your homework for you.
  – John Hughes
  15 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Hint: the roots of $f$ are $x = frac-b pm sqrtb^2 + 4ac2a$. (Why? What happened to the minus sign in the quadratic formula?)

What are the roots of $g$?

Now suppose that the first root of $f$ equals the first root of $g$, and three other cases.

(Yes, that's a very pedestrian way to approach this problem, but it'll get you there.)

share|cite|improve this answer

edited 15 hours ago

Ángel Mario Gallegos

18.1k11229

answered 15 hours ago

John Hughes

59.3k23685

Hint: the roots of $f$ are $x = frac-b pm sqrtb^2 + 4ac2a$. (Why? What happened to the minus sign in the quadratic formula?)

What are the roots of $g$?

Now suppose that the first root of $f$ equals the first root of $g$, and three other cases.

(Yes, that's a very pedestrian way to approach this problem, but it'll get you there.)

share|cite|improve this answer

edited 15 hours ago

Ángel Mario Gallegos

18.1k11229

answered 15 hours ago

John Hughes

59.3k23685

share|cite|improve this answer

share|cite|improve this answer

edited 15 hours ago

Ángel Mario Gallegos

18.1k11229

edited 15 hours ago

Ángel Mario Gallegos

18.1k11229

edited 15 hours ago

Ángel Mario Gallegos

18.1k11229

18.1k11229

answered 15 hours ago

John Hughes

59.3k23685

answered 15 hours ago

John Hughes

59.3k23685

answered 15 hours ago

John Hughes

59.3k23685

59.3k23685

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tried this way , but its hard to get the answer
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's why it's an exercise. If it were easy, it'd just be a remark in the textbook. Why don't you show us your work on this approach (use the "edit" button below your question to fill in what you've done), and we can help you proceed. Yeah, it'll mean typing a lot of math...but that's what you're asking us to do, and you're the one getting the benefit, so perhaps you should be doing more of the work.
  – John Hughes
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did this whole day
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good for you. But we don't know what you've done, because you haven't bothered to show us. Sigh. We're really not here to do your homework for you.
  – John Hughes
  15 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tried this way , but its hard to get the answer
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's why it's an exercise. If it were easy, it'd just be a remark in the textbook. Why don't you show us your work on this approach (use the "edit" button below your question to fill in what you've done), and we can help you proceed. Yeah, it'll mean typing a lot of math...but that's what you're asking us to do, and you're the one getting the benefit, so perhaps you should be doing more of the work.
  – John Hughes
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I did this whole day
  – Angelo Mark
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good for you. But we don't know what you've done, because you haven't bothered to show us. Sigh. We're really not here to do your homework for you.
  – John Hughes
  15 hours ago
  

  
  
  
  

Tried this way , but its hard to get the answer
– Angelo Mark
15 hours ago

Tried this way , but its hard to get the answer
– Angelo Mark
15 hours ago

That's why it's an exercise. If it were easy, it'd just be a remark in the textbook. Why don't you show us your work on this approach (use the "edit" button below your question to fill in what you've done), and we can help you proceed. Yeah, it'll mean typing a lot of math...but that's what you're asking us to do, and you're the one getting the benefit, so perhaps you should be doing more of the work.
– John Hughes
15 hours ago

That's why it's an exercise. If it were easy, it'd just be a remark in the textbook. Why don't you show us your work on this approach (use the "edit" button below your question to fill in what you've done), and we can help you proceed. Yeah, it'll mean typing a lot of math...but that's what you're asking us to do, and you're the one getting the benefit, so perhaps you should be doing more of the work.
– John Hughes
15 hours ago

I did this whole day
– Angelo Mark
15 hours ago

I did this whole day
– Angelo Mark
15 hours ago

Good for you. But we don't know what you've done, because you haven't bothered to show us. Sigh. We're really not here to do your homework for you.
– John Hughes
15 hours ago

Good for you. But we don't know what you've done, because you haven't bothered to show us. Sigh. We're really not here to do your homework for you.
– John Hughes
15 hours ago

add a comment | 

up vote
0
down vote

Hint 2: An alternative approach is to say that

$$
f(x) = a (x - u) (x - v)
$$
for some numbers $u$ and $v$ (the roots of $f$), and
$$
g(x) = a(x-u) (x - w)
$$

(the $u$ is repeated because $f$ and $g$ share a root; it's also possible that $v = w$, but not required by any means.)

Now if you expand out $f$, you'll find a relation between $b$ and $u, v$, and between $c$ and $u, v$; similarly for $g$. See where that leads you.

share|cite|improve this answer

answered 15 hours ago

John Hughes

59.3k23685

• 
  
  
  

  
  

  
  
  
  
  
  

  
  did it but leads to the same place where I got stuck
  – Angelo Mark
  15 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

Hint 2: An alternative approach is to say that

$$
f(x) = a (x - u) (x - v)
$$
for some numbers $u$ and $v$ (the roots of $f$), and
$$
g(x) = a(x-u) (x - w)
$$

(the $u$ is repeated because $f$ and $g$ share a root; it's also possible that $v = w$, but not required by any means.)

Now if you expand out $f$, you'll find a relation between $b$ and $u, v$, and between $c$ and $u, v$; similarly for $g$. See where that leads you.

share|cite|improve this answer

answered 15 hours ago

John Hughes

59.3k23685

• 
  
  
  

  
  

  
  
  
  
  
  

  
  did it but leads to the same place where I got stuck
  – Angelo Mark
  15 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Hint 2: An alternative approach is to say that

$$
f(x) = a (x - u) (x - v)
$$
for some numbers $u$ and $v$ (the roots of $f$), and
$$
g(x) = a(x-u) (x - w)
$$

(the $u$ is repeated because $f$ and $g$ share a root; it's also possible that $v = w$, but not required by any means.)

Now if you expand out $f$, you'll find a relation between $b$ and $u, v$, and between $c$ and $u, v$; similarly for $g$. See where that leads you.

share|cite|improve this answer

answered 15 hours ago

John Hughes

59.3k23685

Hint 2: An alternative approach is to say that

$$
f(x) = a (x - u) (x - v)
$$
for some numbers $u$ and $v$ (the roots of $f$), and
$$
g(x) = a(x-u) (x - w)
$$

(the $u$ is repeated because $f$ and $g$ share a root; it's also possible that $v = w$, but not required by any means.)

Now if you expand out $f$, you'll find a relation between $b$ and $u, v$, and between $c$ and $u, v$; similarly for $g$. See where that leads you.

share|cite|improve this answer

answered 15 hours ago

John Hughes

59.3k23685

share|cite|improve this answer

share|cite|improve this answer

answered 15 hours ago

John Hughes

59.3k23685

answered 15 hours ago

John Hughes

59.3k23685

answered 15 hours ago

John Hughes

59.3k23685

59.3k23685

• 
  
  
  

  
  

  
  
  
  
  
  

  
  did it but leads to the same place where I got stuck
  – Angelo Mark
  15 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  did it but leads to the same place where I got stuck
  – Angelo Mark
  15 hours ago
  

  
  
  
  

did it but leads to the same place where I got stuck
– Angelo Mark
15 hours ago

did it but leads to the same place where I got stuck
– Angelo Mark
15 hours ago

add a comment | 

up vote
0
down vote

$$f(x)=ax^2+bx-c$$
$$g(x)=ax^2+cx+b$$
for f(x), the roots are given by:
$$(1),x=frac-b+sqrtb^2+4ac2a textand, x=frac-b-sqrtb^2+4ac2a$$
for g(x), the roots are given by:
$$(2),x=frac-c+sqrtc^2-4ab2a textand,x=frac-c-sqrtc^2-4ab2a$$
since $2a$ is common on the bottom you could remove this then make the two sides equal to each other?

share|cite|improve this answer

answered 15 hours ago

Henry Lee

48210

add a comment | 

up vote
0
down vote

$$f(x)=ax^2+bx-c$$
$$g(x)=ax^2+cx+b$$
for f(x), the roots are given by:
$$(1),x=frac-b+sqrtb^2+4ac2a textand, x=frac-b-sqrtb^2+4ac2a$$
for g(x), the roots are given by:
$$(2),x=frac-c+sqrtc^2-4ab2a textand,x=frac-c-sqrtc^2-4ab2a$$
since $2a$ is common on the bottom you could remove this then make the two sides equal to each other?

share|cite|improve this answer

answered 15 hours ago

Henry Lee

48210

add a comment | 

up vote
0
down vote

up vote
0
down vote

$$f(x)=ax^2+bx-c$$
$$g(x)=ax^2+cx+b$$
for f(x), the roots are given by:
$$(1),x=frac-b+sqrtb^2+4ac2a textand, x=frac-b-sqrtb^2+4ac2a$$
for g(x), the roots are given by:
$$(2),x=frac-c+sqrtc^2-4ab2a textand,x=frac-c-sqrtc^2-4ab2a$$
since $2a$ is common on the bottom you could remove this then make the two sides equal to each other?

share|cite|improve this answer

answered 15 hours ago

Henry Lee

48210

$$f(x)=ax^2+bx-c$$
$$g(x)=ax^2+cx+b$$
for f(x), the roots are given by:
$$(1),x=frac-b+sqrtb^2+4ac2a textand, x=frac-b-sqrtb^2+4ac2a$$
for g(x), the roots are given by:
$$(2),x=frac-c+sqrtc^2-4ab2a textand,x=frac-c-sqrtc^2-4ab2a$$
since $2a$ is common on the bottom you could remove this then make the two sides equal to each other?

share|cite|improve this answer

answered 15 hours ago

Henry Lee

48210

share|cite|improve this answer

share|cite|improve this answer

answered 15 hours ago

Henry Lee

48210

answered 15 hours ago

Henry Lee

48210

answered 15 hours ago

Henry Lee

48210

48210

add a comment | 

add a comment | 

up vote
0
down vote

If $f(x)=ax^2+bx-c$ and $g(x)=ax^2+cx+b$ have a common root , show that $a+b+c=0$.

The statement is false, as pointed out in several answers already.

The problem most likely has a typo, however, and the following statement is in fact true:

If $b ne c$, then $f(x)=ax^2+bx colorred+c$ and $g(x)=ax^2+cx+b$ have a common root iff $a+b+c=0$.

The proof follows immediately by subtracting the two equations, which gives $,(b-c)(x-1)=0,$, therefore the common root must be $,x=1,$.

share|cite|improve this answer

answered 9 hours ago

dxiv

53.6k64696

add a comment | 

up vote
0
down vote

If $f(x)=ax^2+bx-c$ and $g(x)=ax^2+cx+b$ have a common root , show that $a+b+c=0$.

The statement is false, as pointed out in several answers already.

The problem most likely has a typo, however, and the following statement is in fact true:

If $b ne c$, then $f(x)=ax^2+bx colorred+c$ and $g(x)=ax^2+cx+b$ have a common root iff $a+b+c=0$.

The proof follows immediately by subtracting the two equations, which gives $,(b-c)(x-1)=0,$, therefore the common root must be $,x=1,$.

share|cite|improve this answer

answered 9 hours ago

dxiv

53.6k64696

add a comment | 

up vote
0
down vote

up vote
0
down vote

If $f(x)=ax^2+bx-c$ and $g(x)=ax^2+cx+b$ have a common root , show that $a+b+c=0$.

The statement is false, as pointed out in several answers already.

The problem most likely has a typo, however, and the following statement is in fact true:

If $b ne c$, then $f(x)=ax^2+bx colorred+c$ and $g(x)=ax^2+cx+b$ have a common root iff $a+b+c=0$.

The proof follows immediately by subtracting the two equations, which gives $,(b-c)(x-1)=0,$, therefore the common root must be $,x=1,$.

share|cite|improve this answer

answered 9 hours ago

dxiv

53.6k64696

If $f(x)=ax^2+bx-c$ and $g(x)=ax^2+cx+b$ have a common root , show that $a+b+c=0$.

The statement is false, as pointed out in several answers already.

The problem most likely has a typo, however, and the following statement is in fact true:

If $b ne c$, then $f(x)=ax^2+bx colorred+c$ and $g(x)=ax^2+cx+b$ have a common root iff $a+b+c=0$.

The proof follows immediately by subtracting the two equations, which gives $,(b-c)(x-1)=0,$, therefore the common root must be $,x=1,$.

share|cite|improve this answer

answered 9 hours ago

dxiv

53.6k64696

share|cite|improve this answer

share|cite|improve this answer

answered 9 hours ago

dxiv

53.6k64696

answered 9 hours ago

dxiv

53.6k64696

answered 9 hours ago

dxiv

53.6k64696

53.6k64696

add a comment | 

add a comment | 

 

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