Find the value of $p$ and $q$ of the quadratic equation.

Question

The quadratic equation $x^2+px+q=0$ has roots $-2$ and $6$. Find the value of $p$ and $q$.

Do I have to make two equations?

Something like this?

When $x=-2$, (real and distinct roots) $b^2-4ac>0$

$(-2)^2+p(-2)+q>0$

Making an equation: $q>-4+2p$ ---------------(1)

Then $x=6$....Should it be something like this or any different method?

Answer 1

Use Viete's formulas: the roots of $\;ax^2+bx+c=0\;,\;\;a\neq 0\;$ , are $\;\alpha\,,\,\beta\;$ iff

$$ax^2+bx+c=a(x-\alpha)(x-\beta)$$

and from here

$$\begin{align*}(1)&\;\;\alpha+\beta=-\frac ba\\(2)&\;\;\;\;\;\;\alpha\beta=\frac ca\end{align*}$$

Answer 2

What are the roots of $(x+2)(x-6)=0$? How does that relate to your problem?

Answer 3

In general, it is useful to remember the following: writing your quadratic equation $$x^2-Sx+P=0$$ you have that $S$ is the sum (resp $P$ the product) of the two roots.

Here, you get $p=-(-2+6)=-4$, $q=-2\cdot 6=-12$.