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

Oct 16, 2016

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

#### Explanation:

Given:$\text{ } \sqrt{20 - n} + 8 = \sqrt{9 - n} + 11$

$\textcolor{b r o w n}{\text{When ever you have roots try and get rid of them. This may not always work}}$

Subtract 8 from both sides isolating one of the roots.

$\text{ } \sqrt{20 - n} = \sqrt{9 - n} + 3$

Square both sides

$20 \cancel{- n} = \left[9 \cancel{- n}\right] + 6 \sqrt{9 - n} + 9$

Subtract 18 from both sides (9+9=18)

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

Divide both sides by 6

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

Square both sides

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

$n = \frac{80}{9}$
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$\textcolor{b l u e}{\text{Check}}$ comparing left to right

$\sqrt{20 - n} + 8 \to \sqrt{9 - n} + 11$

$\sqrt{20 - \frac{80}{9}} + 8 \to \sqrt{9 - \frac{80}{9}} + 11$

$\sqrt{\frac{100}{9}} + 8 \to \sqrt{\frac{1}{9}} + 11$

$\frac{10}{3} + 8 \to \frac{1}{3} + 11$

$\frac{34}{3} \to \frac{34}{3}$ Thus LHS = RHS $\to \textcolor{red}{\text{ True}}$

Dec 7, 2016

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

$\implies \sqrt{20 - n} - \sqrt{9 - n} = 11 - 8$

=>sqrt(20-n)-sqrt(9-n)=3.... .[1]

$\implies \frac{1}{\sqrt{20 - n} - \sqrt{9 - n}} = \frac{1}{3}$

$\implies \frac{\sqrt{20 - n} + \sqrt{9 - n}}{20 - n - 9 + n} = \frac{1}{3}$

$\implies \frac{\sqrt{20 - n} + \sqrt{9 - n}}{11} = \frac{1}{3}$

$\implies \left(\sqrt{20 - n} + \sqrt{9 - n}\right) = \frac{11}{3.} \ldots . . \left[2\right]$

Adding [1] and [2] we get

$2 \sqrt{20 - n} = 3 + \frac{11}{3} = \frac{20}{3}$

$\implies \sqrt{20 - n} = \frac{10}{3}$

$\implies \left(20 - n\right) = \frac{100}{9}$

$\implies n = 20 - \frac{100}{9} = \frac{80}{9}$