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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 $x^2 - (\alpha + \beta)x + (\alpha\beta)$.]

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5 Answers

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 $(2a+1)(2b+1)=4ab+2(a+b)+1$?

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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).

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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 $x^2-x-2=0$,

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

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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: $x^2-S'x+P'$.

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

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