# What is x^2-8x-20=0 solving by completing the square?

Oct 9, 2017

$x = 10$

#### Explanation:

${x}^{2} - 8 x - 20 = 0$

${x}^{2} - 8 x = 20$

When completed we should have a function of the form ${\left(x + a\right)}^{2}$. This function expanded would be ${x}^{2} + 2 a x + {a}^{2}$. If $2 a x = - 8 x$, then $a = - 4$, meaning our term will be ${\left(x - 4\right)}^{2}$. Expanded this would give us ${x}^{2} - 8 x + 16$, so to complete the square we have to add 16 to both sides...

${x}^{2} - 8 x + 16 = 20 + 16$

Now change it into our ${\left(x + a\right)}^{2}$ form...

${\left(x - 4\right)}^{2} = 36$

Square root both sides:

$x - 4 = 6$

And finally add 4 to both sides to isolate x.

$x = 10$

Oct 9, 2017

$x = 10 , \setminus q \quad \setminus q \quad x = - 2$

#### Explanation:

First, move the $c$ value to the RHS:

${x}^{2} - 8 x = 20$

Add ${\left(\setminus \frac{b}{2}\right)}^{2}$ to both sides:

${x}^{2} - 8 x + {\left(\setminus \frac{- 8}{2}\right)}^{2} = 20 + {\left(\setminus \frac{- 8}{2}\right)}^{2}$

Simplifying the fractions:

${x}^{2} - 8 x + 16 = 20 + 16$

Now that the LHS is a perfect square, we can factor it as ${\left(x - \setminus \frac{b}{2}\right)}^{2}$

${\left(x - 4\right)}^{2} = 36$

Taking the real (non-principal) square root:

$\setminus \sqrt{{\left(x - 4\right)}^{2}} = \setminus \sqrt{36}$

Simplifying:

$x - 4 = \setminus \pm 6$

Isolating for $x$:

$x = \setminus \pm 6 + 4$

$\setminus \quad x = - 6 + 4 , \setminus q \quad x = 6 + 4$

$\setminus \therefore x = - 2 , \setminus q \quad \setminus q \quad x = 10$