# How do you solve for sqrt(4s+17)-s-3=0?

May 3, 2018

$s = 2$

#### Explanation:

$\sqrt{4 s + 17} - s - 3 = 0$
$\sqrt{4 s + 17} = s + 3$
${\left(\sqrt{4 s + 17}\right)}^{2} = {\left(s + 2\right)}^{2}$
$4 s + 17 = {s}^{2} + 6 s + 9$
${s}^{2} + 6 s + 9 - 4 s - 17 = 0$
${s}^{2} + 2 s - 8 = 0$

$D = {2}^{2} - 4 \cdot 1 \cdot \left(- 8\right) = 4 + 32 = 36$

${s}_{1} = \frac{- 2 + 6}{2} = 2$
${s}_{2} = \frac{- 2 - 6}{2} = - 4$

-4 is not the solution of the equation. Because square root cannot be negative. $\left(s + 3\right) = \left(- 4 + 3\right) = - 1$

$\sqrt{4 \cdot 2 + 17} = 2 + 3$
$\sqrt{25} = 5$
$5 = 5$

May 3, 2018

$s = 2$

#### Explanation:

$\text{subtract "-s-3" from both sides of the equation}$

$\Rightarrow \sqrt{4 s + 17} = s + 3$

$\textcolor{b l u e}{\text{square both sides}}$

${\left(\sqrt{4 s + 17}\right)}^{2} = {\left(s + 3\right)}^{2} \leftarrow \textcolor{b l u e}{\text{distribute}}$

$\Rightarrow 4 s + 17 = {s}^{2} + 6 s + 9$

$\text{collect like terms and equate to zero}$

${s}^{2} + 2 s - 8 = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{the factors of - 8 which sum to + 2 are + 4 and - 2}$

$\Rightarrow \left(s + 4\right) \left(s - 2\right) = 0$

$\text{equate each factor to zero and solve for s}$

$s + 4 = 0 \Rightarrow s = - 4$

$s - 2 = 0 \Rightarrow s = 2$

$\textcolor{b l u e}{\text{As a check}}$

Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.

$s = - 4 \to \sqrt{- 16 + 17} + 4 - 3 = 1 + 4 - 3 = 2$

$2 \ne 0 \Rightarrow s = - 4 \text{ is extraneous}$

$s = 2 \to \sqrt{8 + 17} - 2 - 3 = \sqrt{25} - 5 = 5 - 5 = 0$

$\Rightarrow s = 2 \text{ is the solution}$