parheliubdr

2023-02-24

What are the zeros of the function $f\left(x\right)={x}^{2}+5x+5$ written in simplest radical form?

Bruno Stanley

We have:
$f\left(x\right)={x}^{2}+5x+5$
Method 1 - Completing the square
Solve:
$0=4f\left(x\right)$
$0=4\left({x}^{2}+5x+5\right)$
$0=4{x}^{2}+20x+20$
$0={\left(2x\right)}^{2}+2\left(2x\right)\left(5\right)+25-5$
$0={\left(2x+5\right)}^{2}-{\left(\sqrt{5}\right)}^{2}$
$0=\left(\left(2x+5\right)-\sqrt{5}\right)\left(\left(2x+5\right)+\sqrt{5}\right)$
$0=\left(2x+5-\sqrt{5}\right)\left(2x+5+\sqrt{5}\right)$
Then:
$2x=-5±\sqrt{5}$
Dividing both sides by 2, we find:
$x=-\frac{5}{2}±\frac{\sqrt{5}}{2}$

Note that f(x) is in standard quadratic form:
$f\left(x\right)=a{x}^{2}+bx+c$
with a=1, b=5 and c=5.
According to the quadratic formula, this has zeros:
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
$x=\frac{-\left(5\right)±\sqrt{{\left(5\right)}^{2}-4\left(1\right)\left(5\right)}}{2\left(1\right)}$
$x=\frac{-5±\sqrt{25-20}}{2}$
$x=\frac{-5±\sqrt{5}}{2}$
$x=-\frac{5}{2}±\frac{\sqrt{5}}{2}$

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