# If 8sqrt(4z^2 - 43) = 40, what is the value of z?

Aug 27, 2016

Assumption: the question is: $\text{ } 8 \sqrt{4 {z}^{2}} - 43 = 40$

$z = + \frac{83}{16} \to 5 \frac{3}{16}$

#### Explanation:

$\textcolor{b r o w n}{\text{The objective is to manipulate the equation such that you have a}}$$\textcolor{b r o w n}{\text{single z. This is to be on one side of the equals sign and everything}}$ $\textcolor{b r o w n}{\text{else on the other side.}}$

This is done in stages. First you have all the terms with z on the LHS of = and all the terms without z on the other side.
Then you manipulate the LHS side until the only thing left is the single z.

color(blue)("Step 1 - Isolate "sqrt(4z^2)

Add 43 to both sides giving

$8 \sqrt{4 {z}^{2}} + 0 = 40 + 43$

Divide both sides by 8. Same as multiply by $\frac{1}{8}$ to get rif og the 8 from $8 \sqrt{4 {z}^{2}}$

$\frac{8}{8} \times \sqrt{4 {z}^{2}} = \frac{83}{8}$

But $\frac{8}{8} = 1$

$\sqrt{4 {z}^{2}} = \frac{83}{8}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 2 - Isolate z}}$

But $4 = {2}^{2} \to 4 {z}^{2} = {2}^{2} {z}^{2}$

Write as:$\text{ } \sqrt{{2}^{2} {z}^{2}} = \frac{83}{8}$

Taking the root

$\implies \pm 2 z = \frac{83}{8}$

divide both side by 2 to get rid of the 2 from $2 z$

$\pm z = \frac{83}{16} \to 5 \frac{3}{16}$

After testing $z = + \frac{83}{16}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check: Consider the left hand side only for $z = + \frac{83}{16}$

$8 \sqrt{4 {z}^{2}} - 43$

$8 \times \left(2 z\right) - 43$

$8 \times 2 \times \frac{83}{16} - 43$

But $2 \times 8 = 16$

$\cancel{16} \times \frac{83}{\cancel{16}} - 43$

$83 - 43$

$40$

Thus LHS=RHS so

$\textcolor{red}{\text{ The equation will not works for } z = - \frac{83}{16}}$

Aug 27, 2016

$z = \pm \sqrt{17}$

#### Explanation:

$8 \sqrt{4 {z}^{2} - 43} = 40$

Isolate the square-root.

$\sqrt{4 {z}^{2} - 43} = 5$

${\left(\sqrt{4 {z}^{2} - 43}\right)}^{2} = {5}^{2}$

$4 {z}^{2} - 43 = 25$

$4 {z}^{2} = 25 + 43$

$4 {z}^{2} = 68$

${z}^{2} = 17$

z = +-sqrt(17

Checking in the original equation, both solutions work.