tripplek7i

2022-01-30

Find the value of the product of roots of this quadratic equation
It is given that one of the roots of the quadratic equation :
${x}^{2}+\left(p+3\right)x-{p}^{2}=0$

coolbananas03ok

Step 1
Without applying the factorization of a "difference of two squares" or Viete's relations, we can still use the information stated in the problem. If we call the two roots of the quadratic equation r and -r , then we have
${r}^{2}+\left(p+3\right)×r-{p}^{2}=0$
and
${\left[-r\right]}^{2}+\left(p+3\right)×\left[-r\right]-{p}^{2}$
$={r}^{2}-\left(p+3\right)×r-{p}^{2}=0$
This means that
${r}^{2}=-\left(p+3\right)×r+{p}^{2}$
$=\left(p+3\right)×r+{p}^{2}⇒2×\left(p+3\right)×r=0$
So either $r=0$ or $p=-3$
But if $r=0=-r$, then
${0}^{2}+\left(p+3\right)×0-{p}^{2}=0$ would reqiure $p=0$,
which would then make the quadratic equation
${x}^{3}+3×x=0$
But that polynomial factors as
$x×\left(x+3\right)=0$,
so we couldn't have both roots equal to zero.
Instead, it must be that $p=-3$, making the equation
${x}^{2}+0×x-{\left(-3\right)}^{2}={x}^{2}-9=0$,
for which the roots are given by

the product of the roots is thus -9
Step 2
Another way to arrive at this conclusion is that
$y={x}^{2}+\left(p+3\right)×x-{p}^{2}$
is the equation of an "upward-opening" parabola, for which we want the x-intercepts to be $x=-r$ and $x=r$. Its axis of symmetry is located midway between these intercepts, so we have $h=0$ in the "vertex form" of the parabola's equation,
$y={\left(x-0\right)}^{2}-{p}^{2}$
(The vertex is definitely "below" the x-axis at , so we know these x-intercepts exist.) The equation of the parabola is therefore $y={x}^{2}-{p}^{2}$, making $p+3=0$ and the rest of the argument above follows.

lirwerwammete9t

Step 1
We know that this equation has at most two roots in the set of reals.
Let's denote them with a and -a.
${x}^{2}-{a}^{2}=0$.
Therefore p must be equal to -3.
Hence the equation is
${x}^{2}-9=0$
and the roots are 3 and -3. So the desired product is -9.

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