Transforming a quadratic from its zeroes

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In below problem I've been trying to figure out how simply letting $y=dfracxx+10$ gives the quadratic with roots scaled by $dfrac1alpha+10$. I'm a bit clueless why it works.

My thoughts :

- To get a quadratic $g(x)$ whose roots are $k$ times bigger than $f(x)$, we can simply replace $x$ by $x/k$. This works because $f(alpha) = 0 implies g(kalpha) = 0$, where $g(x) = f(x/k).$

- To get a quadratic $g(x)$ whose roots are $h$ bigger than $f(x)$, we can simply replace $x$ by $x-h$. This works because $f(alpha) = 0 implies g(alpha+h) = 0$, where $g(x) = f(x-h).$

quadratics

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asked Jul 22 at 9:25

rsadhvika

1,4891026

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It's certainly odd that they give an answer with a common factor of $2$, but the solution given seems perfectly lucid. What's your objection to it?
  – Lord Shark the Unknown
  Jul 22 at 9:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown I've no objection as such, I just want to understand how it works. I know how to form a quadratic whose roots are scaled by a constant factor. But here they're scaling by a variable, which I feel is throwing me off..
  – rsadhvika
  Jul 22 at 9:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Btw thanks for responding @LordSharktheUnknown :) I guess I'm looking for some reasoning/explanation behind letting $y=x/(x+10)$
  – rsadhvika
  Jul 22 at 9:42
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You have $y=phi(x)$ for some function. Then $x=phi^-1(y)$ and if $Q(x)=0$ then $Q(phi^-1(y))=0$. All the examples here are of this form.
  – Lord Shark the Unknown
  Jul 22 at 9:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If the desired equation reads $x^2+px+q=0$ you know that $q=fracaa+10fracbb+10$. Now exploit that $ab=-2$ and $a+b=-13$.
  – Michael Hoppe
  Jul 22 at 10:14
  

  
  
  
  

 | 
show 6 more comments

up vote
2
down vote

favorite

In below problem I've been trying to figure out how simply letting $y=dfracxx+10$ gives the quadratic with roots scaled by $dfrac1alpha+10$. I'm a bit clueless why it works.

My thoughts :

- To get a quadratic $g(x)$ whose roots are $k$ times bigger than $f(x)$, we can simply replace $x$ by $x/k$. This works because $f(alpha) = 0 implies g(kalpha) = 0$, where $g(x) = f(x/k).$

- To get a quadratic $g(x)$ whose roots are $h$ bigger than $f(x)$, we can simply replace $x$ by $x-h$. This works because $f(alpha) = 0 implies g(alpha+h) = 0$, where $g(x) = f(x-h).$

quadratics

share|cite|improve this question

asked Jul 22 at 9:25

rsadhvika

1,4891026

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It's certainly odd that they give an answer with a common factor of $2$, but the solution given seems perfectly lucid. What's your objection to it?
  – Lord Shark the Unknown
  Jul 22 at 9:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown I've no objection as such, I just want to understand how it works. I know how to form a quadratic whose roots are scaled by a constant factor. But here they're scaling by a variable, which I feel is throwing me off..
  – rsadhvika
  Jul 22 at 9:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Btw thanks for responding @LordSharktheUnknown :) I guess I'm looking for some reasoning/explanation behind letting $y=x/(x+10)$
  – rsadhvika
  Jul 22 at 9:42
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You have $y=phi(x)$ for some function. Then $x=phi^-1(y)$ and if $Q(x)=0$ then $Q(phi^-1(y))=0$. All the examples here are of this form.
  – Lord Shark the Unknown
  Jul 22 at 9:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If the desired equation reads $x^2+px+q=0$ you know that $q=fracaa+10fracbb+10$. Now exploit that $ab=-2$ and $a+b=-13$.
  – Michael Hoppe
  Jul 22 at 10:14
  

  
  
  
  

 | 
show 6 more comments

up vote
2
down vote

favorite

up vote
2
down vote

favorite

In below problem I've been trying to figure out how simply letting $y=dfracxx+10$ gives the quadratic with roots scaled by $dfrac1alpha+10$. I'm a bit clueless why it works.

My thoughts :

- To get a quadratic $g(x)$ whose roots are $k$ times bigger than $f(x)$, we can simply replace $x$ by $x/k$. This works because $f(alpha) = 0 implies g(kalpha) = 0$, where $g(x) = f(x/k).$

- To get a quadratic $g(x)$ whose roots are $h$ bigger than $f(x)$, we can simply replace $x$ by $x-h$. This works because $f(alpha) = 0 implies g(alpha+h) = 0$, where $g(x) = f(x-h).$

quadratics

share|cite|improve this question

asked Jul 22 at 9:25

rsadhvika

1,4891026

In below problem I've been trying to figure out how simply letting $y=dfracxx+10$ gives the quadratic with roots scaled by $dfrac1alpha+10$. I'm a bit clueless why it works.

My thoughts :

- To get a quadratic $g(x)$ whose roots are $k$ times bigger than $f(x)$, we can simply replace $x$ by $x/k$. This works because $f(alpha) = 0 implies g(kalpha) = 0$, where $g(x) = f(x/k).$

- To get a quadratic $g(x)$ whose roots are $h$ bigger than $f(x)$, we can simply replace $x$ by $x-h$. This works because $f(alpha) = 0 implies g(alpha+h) = 0$, where $g(x) = f(x-h).$

quadratics

share|cite|improve this question

asked Jul 22 at 9:25

rsadhvika

1,4891026

share|cite|improve this question

share|cite|improve this question

asked Jul 22 at 9:25

rsadhvika

1,4891026

asked Jul 22 at 9:25

rsadhvika

1,4891026

asked Jul 22 at 9:25

rsadhvika

1,4891026

1,4891026

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It's certainly odd that they give an answer with a common factor of $2$, but the solution given seems perfectly lucid. What's your objection to it?
  – Lord Shark the Unknown
  Jul 22 at 9:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown I've no objection as such, I just want to understand how it works. I know how to form a quadratic whose roots are scaled by a constant factor. But here they're scaling by a variable, which I feel is throwing me off..
  – rsadhvika
  Jul 22 at 9:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Btw thanks for responding @LordSharktheUnknown :) I guess I'm looking for some reasoning/explanation behind letting $y=x/(x+10)$
  – rsadhvika
  Jul 22 at 9:42
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You have $y=phi(x)$ for some function. Then $x=phi^-1(y)$ and if $Q(x)=0$ then $Q(phi^-1(y))=0$. All the examples here are of this form.
  – Lord Shark the Unknown
  Jul 22 at 9:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If the desired equation reads $x^2+px+q=0$ you know that $q=fracaa+10fracbb+10$. Now exploit that $ab=-2$ and $a+b=-13$.
  – Michael Hoppe
  Jul 22 at 10:14
  

  
  
  
  

 | 
show 6 more comments

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It's certainly odd that they give an answer with a common factor of $2$, but the solution given seems perfectly lucid. What's your objection to it?
  – Lord Shark the Unknown
  Jul 22 at 9:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown I've no objection as such, I just want to understand how it works. I know how to form a quadratic whose roots are scaled by a constant factor. But here they're scaling by a variable, which I feel is throwing me off..
  – rsadhvika
  Jul 22 at 9:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Btw thanks for responding @LordSharktheUnknown :) I guess I'm looking for some reasoning/explanation behind letting $y=x/(x+10)$
  – rsadhvika
  Jul 22 at 9:42
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  You have $y=phi(x)$ for some function. Then $x=phi^-1(y)$ and if $Q(x)=0$ then $Q(phi^-1(y))=0$. All the examples here are of this form.
  – Lord Shark the Unknown
  Jul 22 at 9:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If the desired equation reads $x^2+px+q=0$ you know that $q=fracaa+10fracbb+10$. Now exploit that $ab=-2$ and $a+b=-13$.
  – Michael Hoppe
  Jul 22 at 10:14
  

  
  
  
  

2

2

It's certainly odd that they give an answer with a common factor of $2$, but the solution given seems perfectly lucid. What's your objection to it?
– Lord Shark the Unknown
Jul 22 at 9:38

It's certainly odd that they give an answer with a common factor of $2$, but the solution given seems perfectly lucid. What's your objection to it?
– Lord Shark the Unknown
Jul 22 at 9:38

@LordSharktheUnknown I've no objection as such, I just want to understand how it works. I know how to form a quadratic whose roots are scaled by a constant factor. But here they're scaling by a variable, which I feel is throwing me off..
– rsadhvika
Jul 22 at 9:41

@LordSharktheUnknown I've no objection as such, I just want to understand how it works. I know how to form a quadratic whose roots are scaled by a constant factor. But here they're scaling by a variable, which I feel is throwing me off..
– rsadhvika
Jul 22 at 9:41

Btw thanks for responding @LordSharktheUnknown :) I guess I'm looking for some reasoning/explanation behind letting $y=x/(x+10)$
– rsadhvika
Jul 22 at 9:42

Btw thanks for responding @LordSharktheUnknown :) I guess I'm looking for some reasoning/explanation behind letting $y=x/(x+10)$
– rsadhvika
Jul 22 at 9:42

3

3

You have $y=phi(x)$ for some function. Then $x=phi^-1(y)$ and if $Q(x)=0$ then $Q(phi^-1(y))=0$. All the examples here are of this form.
– Lord Shark the Unknown
Jul 22 at 9:45

You have $y=phi(x)$ for some function. Then $x=phi^-1(y)$ and if $Q(x)=0$ then $Q(phi^-1(y))=0$. All the examples here are of this form.
– Lord Shark the Unknown
Jul 22 at 9:45

1

1

If the desired equation reads $x^2+px+q=0$ you know that $q=fracaa+10fracbb+10$. Now exploit that $ab=-2$ and $a+b=-13$.
– Michael Hoppe
Jul 22 at 10:14

If the desired equation reads $x^2+px+q=0$ you know that $q=fracaa+10fracbb+10$. Now exploit that $ab=-2$ and $a+b=-13$.
– Michael Hoppe
Jul 22 at 10:14

 | 
show 6 more comments

1 Answer
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The roots of $$x^2-13x-2=0$$ satisfy $$a+b =13, ab=-2$$

The transformed equation is $$(x-frac aa-10)(x-frac bb-10) $$

Upon simplification we get $$(ab-10(a+b)+100)x^2 -(2ab-10(a+b))x+ab=0$$

Substitution for a+b and ab gives us $$-32x^2+134x-2=0$$

Thus the correct answer is $$-32x^2+134x-2=0$$

share|cite|improve this answer

answered Jul 22 at 10:17

Mohammad Riazi-Kermani

27.5k41852

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The roots of $$x^2-13x-2=0$$ satisfy $$a+b =13, ab=-2$$

The transformed equation is $$(x-frac aa-10)(x-frac bb-10) $$

Upon simplification we get $$(ab-10(a+b)+100)x^2 -(2ab-10(a+b))x+ab=0$$

Substitution for a+b and ab gives us $$-32x^2+134x-2=0$$

Thus the correct answer is $$-32x^2+134x-2=0$$

share|cite|improve this answer

answered Jul 22 at 10:17

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
1
down vote

The roots of $$x^2-13x-2=0$$ satisfy $$a+b =13, ab=-2$$

The transformed equation is $$(x-frac aa-10)(x-frac bb-10) $$

Upon simplification we get $$(ab-10(a+b)+100)x^2 -(2ab-10(a+b))x+ab=0$$

Substitution for a+b and ab gives us $$-32x^2+134x-2=0$$

Thus the correct answer is $$-32x^2+134x-2=0$$

share|cite|improve this answer

answered Jul 22 at 10:17

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
1
down vote

up vote
1
down vote

The roots of $$x^2-13x-2=0$$ satisfy $$a+b =13, ab=-2$$

The transformed equation is $$(x-frac aa-10)(x-frac bb-10) $$

Upon simplification we get $$(ab-10(a+b)+100)x^2 -(2ab-10(a+b))x+ab=0$$

Substitution for a+b and ab gives us $$-32x^2+134x-2=0$$

Thus the correct answer is $$-32x^2+134x-2=0$$

share|cite|improve this answer

answered Jul 22 at 10:17

Mohammad Riazi-Kermani

27.5k41852

The roots of $$x^2-13x-2=0$$ satisfy $$a+b =13, ab=-2$$

The transformed equation is $$(x-frac aa-10)(x-frac bb-10) $$

Upon simplification we get $$(ab-10(a+b)+100)x^2 -(2ab-10(a+b))x+ab=0$$

Substitution for a+b and ab gives us $$-32x^2+134x-2=0$$

Thus the correct answer is $$-32x^2+134x-2=0$$

share|cite|improve this answer

answered Jul 22 at 10:17

Mohammad Riazi-Kermani

27.5k41852

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 10:17

Mohammad Riazi-Kermani

27.5k41852

answered Jul 22 at 10:17

Mohammad Riazi-Kermani

27.5k41852

answered Jul 22 at 10:17

Mohammad Riazi-Kermani

27.5k41852

27.5k41852

add a comment | 

add a comment | 

 

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