If $c,d$ are the real roots of the equation, $(x-a)(x-b)=f$

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up vote
1
down vote

favorite

EDIT:

Sorry, corrected the Typo

If $c,d$ are the real roots of the equation,
$$(x-a)(x-b)=f$$
Then roots of the equation, $$(x-c)(x-d)+f=0$$
are?

1. $a,b$


2. $fracaf,fracbf$


3. $fracfa,fracfb$


4. $fraccf,fracdf$

I don't have any hints on how to proceed with this. The only thing I was able to establish was, $$a+b=c+d$$

$$ab=cd+f$$

Any hints would be helpful. Thank you.

systems-of-equations quadratics

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edited Aug 2 at 13:06

asked Aug 2 at 12:40

prog_SAHIL

773217

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Did you mean $(x-c)(x-d)+f=0$?
  – lulu
  Aug 2 at 12:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $(x-c)+(x-d)+f=0$ has only a solution. Are you sure this is the correct equation?
  – mfl
  Aug 2 at 12:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have $(x-a)(x-b)-f = (x-c)(x-d)$
  – Hari Shankar
  Aug 2 at 12:47
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

EDIT:

Sorry, corrected the Typo

If $c,d$ are the real roots of the equation,
$$(x-a)(x-b)=f$$
Then roots of the equation, $$(x-c)(x-d)+f=0$$
are?

1. $a,b$


2. $fracaf,fracbf$


3. $fracfa,fracfb$


4. $fraccf,fracdf$

I don't have any hints on how to proceed with this. The only thing I was able to establish was, $$a+b=c+d$$

$$ab=cd+f$$

Any hints would be helpful. Thank you.

systems-of-equations quadratics

share|cite|improve this question

edited Aug 2 at 13:06

asked Aug 2 at 12:40

prog_SAHIL

773217

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Did you mean $(x-c)(x-d)+f=0$?
  – lulu
  Aug 2 at 12:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $(x-c)+(x-d)+f=0$ has only a solution. Are you sure this is the correct equation?
  – mfl
  Aug 2 at 12:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have $(x-a)(x-b)-f = (x-c)(x-d)$
  – Hari Shankar
  Aug 2 at 12:47
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

EDIT:

Sorry, corrected the Typo

If $c,d$ are the real roots of the equation,
$$(x-a)(x-b)=f$$
Then roots of the equation, $$(x-c)(x-d)+f=0$$
are?

1. $a,b$


2. $fracaf,fracbf$


3. $fracfa,fracfb$


4. $fraccf,fracdf$

I don't have any hints on how to proceed with this. The only thing I was able to establish was, $$a+b=c+d$$

$$ab=cd+f$$

Any hints would be helpful. Thank you.

systems-of-equations quadratics

share|cite|improve this question

edited Aug 2 at 13:06

asked Aug 2 at 12:40

prog_SAHIL

773217

EDIT:

Sorry, corrected the Typo

If $c,d$ are the real roots of the equation,
$$(x-a)(x-b)=f$$
Then roots of the equation, $$(x-c)(x-d)+f=0$$
are?

1. $a,b$


2. $fracaf,fracbf$


3. $fracfa,fracfb$


4. $fraccf,fracdf$

I don't have any hints on how to proceed with this. The only thing I was able to establish was, $$a+b=c+d$$

$$ab=cd+f$$

Any hints would be helpful. Thank you.

systems-of-equations quadratics

share|cite|improve this question

edited Aug 2 at 13:06

asked Aug 2 at 12:40

prog_SAHIL

773217

share|cite|improve this question

share|cite|improve this question

edited Aug 2 at 13:06

edited Aug 2 at 13:06

edited Aug 2 at 13:06

asked Aug 2 at 12:40

prog_SAHIL

773217

asked Aug 2 at 12:40

prog_SAHIL

773217

asked Aug 2 at 12:40

prog_SAHIL

773217

773217

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Did you mean $(x-c)(x-d)+f=0$?
  – lulu
  Aug 2 at 12:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $(x-c)+(x-d)+f=0$ has only a solution. Are you sure this is the correct equation?
  – mfl
  Aug 2 at 12:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have $(x-a)(x-b)-f = (x-c)(x-d)$
  – Hari Shankar
  Aug 2 at 12:47
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Did you mean $(x-c)(x-d)+f=0$?
  – lulu
  Aug 2 at 12:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $(x-c)+(x-d)+f=0$ has only a solution. Are you sure this is the correct equation?
  – mfl
  Aug 2 at 12:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have $(x-a)(x-b)-f = (x-c)(x-d)$
  – Hari Shankar
  Aug 2 at 12:47
  

  
  
  
  

2

2

Did you mean $(x-c)(x-d)+f=0$?
– lulu
Aug 2 at 12:42

Did you mean $(x-c)(x-d)+f=0$?
– lulu
Aug 2 at 12:42

1

1

$(x-c)+(x-d)+f=0$ has only a solution. Are you sure this is the correct equation?
– mfl
Aug 2 at 12:42

$(x-c)+(x-d)+f=0$ has only a solution. Are you sure this is the correct equation?
– mfl
Aug 2 at 12:42

You have $(x-a)(x-b)-f = (x-c)(x-d)$
– Hari Shankar
Aug 2 at 12:47

You have $(x-a)(x-b)-f = (x-c)(x-d)$
– Hari Shankar
Aug 2 at 12:47

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

The answer can be achieved by using the same technique again.

That is, consider the roots of the equation
$$(x-a)(x-b)-f = x^2 - (a+b)x +(ab-f) = 0$$

As the OP has already realised, the sum of the roots of this quadratic are $(a+b)$, and the product is $(ab-f)$.

Given that the roots are $c,d$, we can thus conclude that
$$ c+d = a+b quad cd = ab-f $$

We now simply apply this same logic to the second equation.
$$ (x-c)(x-d) + f = 0 implies x^2-(c+d)x +(cd-f) = 0$$

So the sum of the roots of this equation are $c+d$ and the product of the roots are ($cd-f)$.

But we have already determined that $$ c+d = a+b quad cd = ab-f $$

Thus, the sum roots of the second equation are $a+b$, and the product is $ab$.

Thus it is clear that the answer is option 1: the roots are $a,b$

share|cite|improve this answer

edited Aug 2 at 21:35

answered Aug 2 at 12:54

Martin Roberts

1,204318

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It was a typo in the textbook. That's what had me stuck. Thank you.
  – prog_SAHIL
  Aug 2 at 13:07
  

  
  
  
  

add a comment | 

up vote
2
down vote

Trusting that $(x-c)(x-d)+f=0$ was intended, just remark that $$(x-a)(x-b)-f=(x-c)(x-d)implies (x-c)(x-d)+f=(x-a)(x-b)$$

Easy to see that this generalizes to higher degree.

share|cite|improve this answer

answered Aug 2 at 12:57

lulu

34.7k14071

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

The answer can be achieved by using the same technique again.

That is, consider the roots of the equation
$$(x-a)(x-b)-f = x^2 - (a+b)x +(ab-f) = 0$$

As the OP has already realised, the sum of the roots of this quadratic are $(a+b)$, and the product is $(ab-f)$.

Given that the roots are $c,d$, we can thus conclude that
$$ c+d = a+b quad cd = ab-f $$

We now simply apply this same logic to the second equation.
$$ (x-c)(x-d) + f = 0 implies x^2-(c+d)x +(cd-f) = 0$$

So the sum of the roots of this equation are $c+d$ and the product of the roots are ($cd-f)$.

But we have already determined that $$ c+d = a+b quad cd = ab-f $$

Thus, the sum roots of the second equation are $a+b$, and the product is $ab$.

Thus it is clear that the answer is option 1: the roots are $a,b$

share|cite|improve this answer

edited Aug 2 at 21:35

answered Aug 2 at 12:54

Martin Roberts

1,204318

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It was a typo in the textbook. That's what had me stuck. Thank you.
  – prog_SAHIL
  Aug 2 at 13:07
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

The answer can be achieved by using the same technique again.

That is, consider the roots of the equation
$$(x-a)(x-b)-f = x^2 - (a+b)x +(ab-f) = 0$$

As the OP has already realised, the sum of the roots of this quadratic are $(a+b)$, and the product is $(ab-f)$.

Given that the roots are $c,d$, we can thus conclude that
$$ c+d = a+b quad cd = ab-f $$

We now simply apply this same logic to the second equation.
$$ (x-c)(x-d) + f = 0 implies x^2-(c+d)x +(cd-f) = 0$$

So the sum of the roots of this equation are $c+d$ and the product of the roots are ($cd-f)$.

But we have already determined that $$ c+d = a+b quad cd = ab-f $$

Thus, the sum roots of the second equation are $a+b$, and the product is $ab$.

Thus it is clear that the answer is option 1: the roots are $a,b$

share|cite|improve this answer

edited Aug 2 at 21:35

answered Aug 2 at 12:54

Martin Roberts

1,204318

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It was a typo in the textbook. That's what had me stuck. Thank you.
  – prog_SAHIL
  Aug 2 at 13:07
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

The answer can be achieved by using the same technique again.

That is, consider the roots of the equation
$$(x-a)(x-b)-f = x^2 - (a+b)x +(ab-f) = 0$$

As the OP has already realised, the sum of the roots of this quadratic are $(a+b)$, and the product is $(ab-f)$.

Given that the roots are $c,d$, we can thus conclude that
$$ c+d = a+b quad cd = ab-f $$

We now simply apply this same logic to the second equation.
$$ (x-c)(x-d) + f = 0 implies x^2-(c+d)x +(cd-f) = 0$$

So the sum of the roots of this equation are $c+d$ and the product of the roots are ($cd-f)$.

But we have already determined that $$ c+d = a+b quad cd = ab-f $$

Thus, the sum roots of the second equation are $a+b$, and the product is $ab$.

Thus it is clear that the answer is option 1: the roots are $a,b$

share|cite|improve this answer

edited Aug 2 at 21:35

answered Aug 2 at 12:54

Martin Roberts

1,204318

The answer can be achieved by using the same technique again.

That is, consider the roots of the equation
$$(x-a)(x-b)-f = x^2 - (a+b)x +(ab-f) = 0$$

As the OP has already realised, the sum of the roots of this quadratic are $(a+b)$, and the product is $(ab-f)$.

Given that the roots are $c,d$, we can thus conclude that
$$ c+d = a+b quad cd = ab-f $$

We now simply apply this same logic to the second equation.
$$ (x-c)(x-d) + f = 0 implies x^2-(c+d)x +(cd-f) = 0$$

So the sum of the roots of this equation are $c+d$ and the product of the roots are ($cd-f)$.

But we have already determined that $$ c+d = a+b quad cd = ab-f $$

Thus, the sum roots of the second equation are $a+b$, and the product is $ab$.

Thus it is clear that the answer is option 1: the roots are $a,b$

share|cite|improve this answer

edited Aug 2 at 21:35

answered Aug 2 at 12:54

Martin Roberts

1,204318

share|cite|improve this answer

share|cite|improve this answer

edited Aug 2 at 21:35

edited Aug 2 at 21:35

edited Aug 2 at 21:35

answered Aug 2 at 12:54

Martin Roberts

1,204318

answered Aug 2 at 12:54

Martin Roberts

1,204318

answered Aug 2 at 12:54

Martin Roberts

1,204318

1,204318

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It was a typo in the textbook. That's what had me stuck. Thank you.
  – prog_SAHIL
  Aug 2 at 13:07
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It was a typo in the textbook. That's what had me stuck. Thank you.
  – prog_SAHIL
  Aug 2 at 13:07
  

  
  
  
  

It was a typo in the textbook. That's what had me stuck. Thank you.
– prog_SAHIL
Aug 2 at 13:07

It was a typo in the textbook. That's what had me stuck. Thank you.
– prog_SAHIL
Aug 2 at 13:07

add a comment | 

up vote
2
down vote

Trusting that $(x-c)(x-d)+f=0$ was intended, just remark that $$(x-a)(x-b)-f=(x-c)(x-d)implies (x-c)(x-d)+f=(x-a)(x-b)$$

Easy to see that this generalizes to higher degree.

share|cite|improve this answer

answered Aug 2 at 12:57

lulu

34.7k14071

add a comment | 

up vote
2
down vote

Trusting that $(x-c)(x-d)+f=0$ was intended, just remark that $$(x-a)(x-b)-f=(x-c)(x-d)implies (x-c)(x-d)+f=(x-a)(x-b)$$

Easy to see that this generalizes to higher degree.

share|cite|improve this answer

answered Aug 2 at 12:57

lulu

34.7k14071

add a comment | 

up vote
2
down vote

up vote
2
down vote

Trusting that $(x-c)(x-d)+f=0$ was intended, just remark that $$(x-a)(x-b)-f=(x-c)(x-d)implies (x-c)(x-d)+f=(x-a)(x-b)$$

Easy to see that this generalizes to higher degree.

share|cite|improve this answer

answered Aug 2 at 12:57

lulu

34.7k14071

Trusting that $(x-c)(x-d)+f=0$ was intended, just remark that $$(x-a)(x-b)-f=(x-c)(x-d)implies (x-c)(x-d)+f=(x-a)(x-b)$$

Easy to see that this generalizes to higher degree.

share|cite|improve this answer

answered Aug 2 at 12:57

lulu

34.7k14071

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 12:57

lulu

34.7k14071

answered Aug 2 at 12:57

lulu

34.7k14071

answered Aug 2 at 12:57

lulu

34.7k14071

34.7k14071

add a comment | 

add a comment | 

 

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