# Thread: Sum & Product of the Roots of a Quadratic Equation

1. ## Sum & Product of the Roots of a Quadratic Equation

Having a bit of trouble with this one. Can anyone help?

Many thanks.

Q.: (i)
The roots of the equation $2x^2+6x+3=0$ are $\alpha$ & $\beta$. Show that $\alpha^2+\beta^2=6$. (ii) The roots of $2x^2+px+q=0$ are $2\alpha+\beta$ & $\alpha+2\beta$. Find the value of p & the value of q.

Attempt: (i) a = 2, b = 6, c = 3

$\alpha+\beta=\frac{-b}{a}$ => $\frac{-6}{2}$ => -3

$\alpha\beta=\frac{c}{a}$ => $\frac{3}{2}$

Sum of roots: $\alpha^2+\beta^2$ => $(\alpha+\beta)^2-2\alpha\beta$ => $(-3)^2-2(\frac{3}{2})$ => $9-\frac{6}{2}$ => 9 - 3 => 6

(ii) a = 2, b = p, c = q

Sum of roots: $2\alpha+\beta+\alpha+2\beta$ => $3\alpha+3\beta$ => $3(\alpha+\beta)$ => $\frac{-b}{a}$ => $\frac{-p}{2}$ => $3(\alpha+\beta)$ => $3(\frac{-p}{2})$ => $\frac{-3p}{2}$

Product of roots: $(2\alpha+\beta)(\alpha+2\beta)$ => $2\alpha^2+5\alpha\beta+2\beta^2$ => $2(\alpha^2+\beta^2)+5\alpha\beta$ => $2((\alpha+\beta)^2-2\alpha\beta)+5\alpha\beta$ => $2(\alpha+\beta)^2+\alpha\beta$ => $2(\frac{-p}{2})^2+(\frac{c}{a})$ => $\frac{2p^2}{4}+\frac{q}{2}$ => $\frac{p^2}{2}+\frac{q}{2}$ => $\frac{p^2+q}{2}$

$x^2-$(sum of roots)x + (product of roots) => $x^2-(\frac{-3p}{2})x+\frac{p^2+q}{2}$ => $2x^2+3px+p^2+q$...

Ans.: (From text book): p = 18, q = 39

2. ## Re: Sum & Product of the Roots of a Quadratic Equation

It seems that more informations are needed. Since if $\alpha=0,\beta=1$ then the roots will be $1,2$. And hence $p=-6,q=4$.
Also, if
$\alpha=1,\beta=-1$ then the roots will be $1,5$. And hence $p=0,q=-2$.

3. ## Re: Sum & Product of the Roots of a Quadratic Equation

I've made the changes to the question I think you're referring to. There was an earlier piece to the question, but I had already solved it and wasn't sure it was relevant.

4. ## Re: Sum & Product of the Roots of a Quadratic Equation

Yes, you do have to use the $\alpha$ and $\beta$ from part a).

Sum of roots:
$3 \alpha + 3 \beta = -\frac{p}{2}$. From part a), $\alpha + \beta = -3$, substituting yields $-9 = -\frac{p}{2} \Rightarrow p = 18$

Product of roots:
$2 \alpha^2 + 5 \alpha \beta + 2 \beta^2 = \frac{q}{2} \Rightarrow 2(\alpha^2 + \beta^2) + 5 \alpha \beta = \frac{q}{2}$. From part a), $\alpha^2 + \beta^2 = 6$, and $\alpha \beta = \frac{3}{2}$. Therefore,

$2(6) + 5(\frac{3}{2}) = \frac{q}{2} \Rightarrow \frac{39}{2} = \frac{q}{2} \Rightarrow q = 39$.

5. ## Re: Sum & Product of the Roots of a Quadratic Equation

So, from the first part you have
$\alpha^2+\beta^2=6$ and also $\alpha+\beta=-3$.
We also need $\alpha\beta$ which is from the first equation
$\alpha\beta=\frac{3}{2}.$
Now,
$p=-((\alpha+2\beta)+(2\alpha+\beta))=-6(\alpha+\beta)=-6*-3=18.$

$q=(2\alpha+\beta)(\alpha+2\beta)=4(\alpha^2+\beta^ 2)+10\alpha\beta=4*6+10\frac{3}{2}=39$

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