3
$\begingroup$

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

$\endgroup$
9
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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