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

Beginner2023-02-25Added 7 answers

We have:

$f\left(x\right)={x}^{2}+5x+5$

Method 1 - Completing the square

Solve:

$0=4f\left(x\right)$

${0}=4({x}^{2}+5x+5)$

${0}=4{x}^{2}+20x+20$

${0}={\left(2x\right)}^{2}+2\left(2x\right)\left(5\right)+25-5$

$0}={(2x+5)}^{2}-{\left(\sqrt{5}\right)}^{2$

${0}=((2x+5)-\sqrt{5})((2x+5)+\sqrt{5})$

${0}=(2x+5-\sqrt{5})(2x+5+\sqrt{5})$

Then:

$2x=-5\pm \sqrt{5}$

Dividing both sides by 2, we find:

$x=-\frac{5}{2}\pm \frac{\sqrt{5}}{2}$

$f\left(x\right)={x}^{2}+5x+5$

Method 1 - Completing the square

Solve:

$0=4f\left(x\right)$

${0}=4({x}^{2}+5x+5)$

${0}=4{x}^{2}+20x+20$

${0}={\left(2x\right)}^{2}+2\left(2x\right)\left(5\right)+25-5$

$0}={(2x+5)}^{2}-{\left(\sqrt{5}\right)}^{2$

${0}=((2x+5)-\sqrt{5})((2x+5)+\sqrt{5})$

${0}=(2x+5-\sqrt{5})(2x+5+\sqrt{5})$

Then:

$2x=-5\pm \sqrt{5}$

Dividing both sides by 2, we find:

$x=-\frac{5}{2}\pm \frac{\sqrt{5}}{2}$

Addison Stuart

Beginner2023-02-26Added 5 answers

Method 2 - Quadratic formula

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:

$x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

$x}=\frac{-\left({5}\right)\pm \sqrt{{\left({5}\right)}^{2}-4\left({1}\right)\left({5}\right)}}{2\left({1}\right)$

$x}=\frac{-5\pm \sqrt{25-20}}{2$

$x}=\frac{-5\pm \sqrt{5}}{2$

$x}=-\frac{5}{2}\pm \frac{\sqrt{5}}{2$

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:

$x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

$x}=\frac{-\left({5}\right)\pm \sqrt{{\left({5}\right)}^{2}-4\left({1}\right)\left({5}\right)}}{2\left({1}\right)$

$x}=\frac{-5\pm \sqrt{25-20}}{2$

$x}=\frac{-5\pm \sqrt{5}}{2$

$x}=-\frac{5}{2}\pm \frac{\sqrt{5}}{2$