# How do you solve \sqrt { 5v - 9} = v - 1?

Sep 28, 2017

Eliminate the radical by squaring both sides of the equation
Combine like terms

#### Explanation:

Eliminate the radical by squaring both sides of the equation

$5 v - 9 = {v}^{2} - 2 v + 1$

Combine like terms:

${v}^{2} - 7 v + 10 = 0$

$\left(v - 2\right) \left(v - 5\right) = 0$

$v = 2 \mathmr{and} v = 5$

$\sqrt{5 \left(2\right) - 9} = 2 - 1$

$\sqrt{1} = 1$

$1 = 1$

$\sqrt{5 \left(5\right) - 9} = 5 - 1$

$\sqrt{16} = 4$

$4 = 4$

Sep 28, 2017

$v = 2$ or $v = 5$

#### Explanation:

$\sqrt{5 v - 9} = v - 1$ This type of equation is called 'Radical equation'.

Step 1) Square both sides of the equation to eliminate square root.
${\left[\sqrt{5 v - 9}\right]}^{2} = {\left(v - 1\right)}^{2}$

Step 2) Use FOIL and simplify
$| 5 v - 9 | = {v}^{2} - 2 v + 1$

Step 3) Consider when $| 5 v - 9 |$ is positive or negative, then simplify.

i) $5 v - 9 > 0$, $v > \frac{9}{5}$

$5 v - 9 = {v}^{2} - 2 v + 1$
−v^2+7v−10=0
${v}^{2} - 7 v + 10 = 0$

Factor left side of the equation
$\left(v - 5\right) \left(v - 2\right) = 0$
$v - 5 = 0$ or $v - 2 = 0$

∴v=2 or $v = 5$
Check if it satisfies range $v > \frac{9}{5}$ (Yes)
∴v=2 or v=5(∵v>9/5) ------------- 1

ii) $5 v - 9 < 0$, $v < \frac{9}{5}$
$- \left(5 v - 9\right) = {v}^{2} - 2 v + 1$
${v}^{2} + 3 v + 8 = 0$

v=-(3±sqrt41)/2
Check if it satisfies range $v < \frac{9}{5}$ (Yes)
But if you substitute v=-(3±sqrt41)/2 to $\sqrt{5 v - 9} = v - 1$, It won't satisfy the equation.
∴No Solution (∵Doesn't satisfies when you substitute the value.) ------------- 2
∴v=2 or $v = 5$