# How do you solve x^2-10x+16=0 using the quadratic formula?

Mar 20, 2018

$x = 8 \mathmr{and} x = 2$

#### Explanation:

If $\textcolor{red}{a {x}^{2} + b x + c = 0} ,$

then, $\textcolor{red}{x = \frac{- b \pm \sqrt{\triangle}}{2 a} \mathmr{and} \triangle = {b}^{2} - 4 a c}$

Comparing , color(blue)(x^2-10x+16=0, with quad.equn ,we get

$a = 1 , b = - 10 , c = 16$

First we find $\triangle$

$\triangle = {\left(- 10\right)}^{2} - 4 \left(1\right) \left(16\right) = 100 - 64 = 36$

$\implies \sqrt{\triangle} = \sqrt{36} = 6$

So,

$x = \frac{- b \pm \sqrt{\triangle}}{2 a} = \frac{- \left(- 10\right) \pm 6}{2 \left(1\right)} = \frac{10 \pm 6}{2}$

$i . e . x = \frac{10 + 6}{2} \mathmr{and} x = \frac{10 - 6}{2}$

$\implies x = \frac{16}{2} \mathmr{and} x = \frac{4}{2}$

$\implies x = 8 \mathmr{and} x = 2$

Mar 20, 2018

$x = 8 , 2$

#### Explanation:

$a {x}^{2} + b x + c = 0$

${x}^{2} - 10 x + 16 = 0$

What we are trying to do is find the values of $a , b$ and $c$. They are located in the same spot in your equation as they are in the quadratic form.

$a {x}^{2} + b x + c = 0$

$\left(1\right) {x}^{2} + \left(- 10\right) x + \left(16\right) = 0$
$a = 1$
$b = - 10$
$c = 16$

Now plug these in to the quadratic formula:
$x = \frac{- b \setminus \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(- 10\right) \setminus \pm \sqrt{- {10}^{2} - 4 \left(1\right) \left(16\right)}}{2 \left(1\right)}$

$x = \frac{\left(\textcolor{red}{+} 10\right) \setminus \pm \sqrt{\textcolor{red}{100} - \left(\textcolor{b l u e}{64}\right)}}{2}$

$x = \frac{10 \setminus \pm \sqrt{100 - 64}}{2}$

$x = \frac{10 \setminus \pm \sqrt{36}}{2}$

$x = \frac{10 \setminus \pm 6}{2}$

Now solve by doing addition alone and subtraction alone to get your two values:

$x = \frac{10 \textcolor{red}{+} 6}{2} \rightarrow x = \frac{16}{2} \rightarrow x = 8$

$x = \frac{10 \textcolor{red}{-} 6}{2} \rightarrow x = \frac{4}{2} \rightarrow x = 2$

So the solution is $x = 8 , 2$