# How do you solve 5sqrt(a-3)+4=14 and check your solution?

Apr 11, 2017

$a = 7$

#### Explanation:

$5 \sqrt{a - 3} + 4 = 14$

reduce $4$ to both sides.
$5 \sqrt{a - 3} + 4 - 4 = 14 - 4$
$5 \sqrt{a - 3} = 10$

divide $5$ to both sides
$\frac{5 \sqrt{a - 3}}{5} = \frac{10}{5}$
$\sqrt{a - 3} = 2$

square to both sides
${\left(\sqrt{a - 3}\right)}^{2} = {2}^{2}$
$a - 3 = 4$

add $3$ to both sides
$a - 3 + 3 = 4 + 3$
$a = 7$

we check Left hand side to prove right hand side by plug in $a = 4$.
$5 \sqrt{7 - 3} + 4 = 5 \sqrt{4} + 4 = 5 \cdot 2 + 4 = 14 \to$proved

Apr 11, 2017

$a = 7$

#### Explanation:

$\textcolor{b l u e}{\text{Isolate}}$ the root on the left side and place numeric values on the right side.

subtract 4 from both sides.

$5 \sqrt{a - 3} \cancel{+ 4} \cancel{- 4} = 14 - 4$

$\Rightarrow 5 \sqrt{a - 3} = 10$

divide both sides by 5

$\frac{{\cancel{5}}^{1} \sqrt{a - 3}}{\cancel{5}} ^ 1 = \frac{10}{5}$

$\rightarrow \sqrt{a - 3} = 2 \leftarrow \textcolor{red}{\text{ root isolated on left side}}$

$\text{to'undo' the root "color(blue)"square both sides}$

${\left(\sqrt{a - 3}\right)}^{2} = {2}^{2}$

$\Rightarrow a - 3 = 4$

add 3 to both sides.

$a \cancel{- 3} \cancel{+ 3} = 4 + 3$

$\Rightarrow a = 7$

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

Substitute this value into the left side and if equal to the right side then it is the solution.

$5 \sqrt{7 - 3} + 4 = 5 \sqrt{4} + 4 = \left(5 \times 2\right) + 4 = 14$

$\Rightarrow a = 7 \text{ is the solution}$