# How do you solve sqrt(20-X) + 8= sqrt( 9-X) +11?

May 19, 2016

$x = \frac{80}{9}$

#### Explanation:

Given,

$\sqrt{20 - x} + 8 = \sqrt{9 - x} + 11$

Subtract $8$ from both sides.

$\sqrt{20 - x} + 8 \textcolor{w h i t e}{i} \textcolor{red}{- 8} = \sqrt{9 - x} + 11 \textcolor{w h i t e}{i} \textcolor{red}{- 8}$

$\sqrt{20 - x} = \sqrt{9 - x} + 3$

Square both sides to get rid of the radical signs.

${\left(\sqrt{20 - x}\right)}^{2} = {\left(\sqrt{9 - x} + 3\right)}^{2}$

$\left(\sqrt{20 - x}\right) \left(\sqrt{20 - x}\right) = \left(\sqrt{9 - x} + 3\right) \left(\sqrt{9 - x} + 3\right)$

Simplify.

$20 - x = \left(9 - x\right) + 6 \sqrt{9 - x} + 9$

$20 - x = 9 - x + 6 \sqrt{9 - x} + 9$

$2 = 6 \sqrt{9 - x}$

Solve for $x$.

$\frac{1}{3} = \sqrt{9 - x}$

${\left(\frac{1}{3}\right)}^{2} = {\left(\sqrt{9 - x}\right)}^{2}$

$\frac{1}{9} = 9 - x$

$x = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{80}{9}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

May 19, 2016

$x = \frac{80}{9}$

#### Explanation:

Regrouping and powering
${\left(\sqrt{20 - x} - \sqrt{9 - x}\right)}^{2} = {\left(11 - 8\right)}^{2}$
giving
$29 - 2 \sqrt{9 - x} \sqrt{20 - x} - 2 x = 9$
Regrouping and powering
${\left(2 \sqrt{9 - x} \sqrt{20 - x}\right)}^{2} = {\left(29 - 9 - 2 x\right)}^{2}$
giving
$9 x = 80$
Solving for $x$ we get $x = \frac{80}{9}$