# (t - 9)^(1/2) - t^(1/2) = 3? solve the radical equations, if possible.

Apr 13, 2017

No solution

#### Explanation:

Given: ${\left(t - 9\right)}^{\frac{1}{2}} - {t}^{\frac{1}{2}} = 3 \text{ or } \sqrt{t - 9} - \sqrt{t} = 3$

Add the $\sqrt{t}$ to both sides of the equation:

$\sqrt{t - 9} - \sqrt{t} + \sqrt{t} = 3 + \sqrt{t}$

Simplify: $\sqrt{t - 9} = 3 + \sqrt{t}$

Square both sides of the equation:

${\left(\sqrt{t - 9}\right)}^{2} = {\left(3 + \sqrt{t}\right)}^{2}$

$t - 9 = \left(3 + \sqrt{t}\right) \left(3 + \sqrt{t}\right)$

Distribute the right side of the equation:

$t - 9 = 9 + 3 \sqrt{t} + 3 \sqrt{t} + \sqrt{t} \sqrt{t}$

Simplify by adding like terms and using $\sqrt{m} \sqrt{m} = \sqrt{m \cdot m} = \sqrt{{m}^{2}} = m$:

$t - 9 = 9 + 6 \sqrt{t} + t$

Subtract $t$ from both sides:

$- 9 = 9 + 6 \sqrt{t}$

Subtract $- 9$ from both sides:

$- 18 = 6 \sqrt{t}$

Divide both sides by $6$:

$- 3 = \sqrt{t}$

Square both sides:

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

$t = 9$

Check:
Always check your answer for radical problems by putting it back into the original equation to see if it works:

$\sqrt{9 - 9} - \sqrt{9} = 0 - 3 = - 3 \ne 3$

No solution