Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $a$ and $b$ are the zeros of a polynomial $p(x) = x^2 - x-2$, find a polynomial whose zeroes are $2a+1$ and $2b+1$.

[Hint: I know that if we have the value of $\alpha + \beta$ and $\alpha\beta$, where $\alpha$ and $\beta$ are the roots of the required polynomial, then we can apply the formula $x^2 - (\alpha + \beta)x + (\alpha\beta)$.]

share|cite|improve this question
up vote 9 down vote accepted

Hint: First remember that a quadratic equation is basically $$x^2-(\text{Sum of roots})x+\text{Product of roots}=0$$

From the given condition $a+b=1$ and $ab=-2$.

Now what would be the values of $(2a+1)+(2b+1)=2(a+b)+2$ and $(2a+1)(2b+1)=4ab+2(a+b)+1$?

share|cite|improve this answer

By looking at it and factoring, $p(x)=(x-2)(x+1)$ so $a=2$ and $b=-1$. Therefore you want zeros at $2a+1=5$ and $2b+1=-1$ for the new polynomial, call it $q(x)$. This polynomial will give the desired roots if $q(x)=(x-5)(x+1)=x^2-4x-5$.

This also works your way of using the product-sum formula for roots as you've mentioned, but just factoring it is simpler (note factoring does not always produce "nice" results like this).

share|cite|improve this answer

If x is a root of $x^2-x-2=0,$ we need to find an equation in $y$ such that $y=2x+1\implies x=\frac{y-1}2$

As x is a root of $x^2-x-2=0$,

$$ \left(\frac{y-1}2\right)^2-\frac{y-1}2-2=0\implies (y-1)^2-2(y-1)-8=0\implies y^2-4y-5=0$$

share|cite|improve this answer

For a polynomial function of degree $2$ denoted $p(x)$, one can rewrite it $p(x)=x^2-Sx+P$ where $S = a+b$ and $P=ab$ with $a,b$ the zeros of $p(x)$. In your case, $S=1$ and $P=-2$. You have two equations: $a+b=1$ and $ab=-2$. Your new polynomial should satisfy the equations: $(2a+1)+(2b+1)=S'$ and $(2a+1)(2b+1)=P'$. Just expand these equations and regroup terms that you know, i.e. $ab$ and $a+b$. You'll find $S'$ and $P'$ and then your new polynomial will be: $x^2-S'x+P'$.

share|cite|improve this answer

We know $\rm\:\color{#0A0}{a\!+\!b},\ \color{#C00}{ab},\:$ we seek $\rm\:2a\!+\!1+2b\!+\!1 = 2(\color{#0A0}{a\!+\!b}\!+\!1),\ (2a\!+\!1)(2b\!+\!1) = 4\color{#C00}{ab}+2(\color{#0A0}{a\!+\!b})+1$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.