# How do you solve \sqrt { 2a + 5} - 2\sqrt { 2a } = 1?

Aug 5, 2018

$a = \frac{2}{9}$

#### Explanation:

Here ,

$\sqrt{2 a + 5} - 2 \sqrt{2 a} = 1. . . \to \left(1\right)$

$\therefore \sqrt{2 a + 5} = 1 + 2 \sqrt{2 a}$

Squaring both sides of eqn.

${\left(\sqrt{2 a + 5}\right)}^{2} = {\left(1 + 2 \sqrt{2 a}\right)}^{2}$

$\therefore 2 a + 5 = 1 + 4 \sqrt{2 a} + 8 a$

$\therefore 2 a - 8 a + 5 - 1 = 4 \sqrt{2 a}$

$\therefore 4 - 6 a = 4 \sqrt{2 a}$

$\therefore 2 - 3 a = 2 \sqrt{2 a}$

Again squaring both sides of eqn.

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

$\therefore 4 - 12 a + 9 {a}^{2} = 8 a$

$\therefore 9 {a}^{2} - 20 a + 4 = 0$

Now ,

$\left(- 18 a\right) + \left(- 2 a\right) = - 20 a \mathmr{and} \left(- 18 a\right) \times \left(- 2 a\right) = 36 {a}^{2}$

$9 {a}^{2} - 18 a - 2 a + 4 = 0$

$\therefore 9 a \left(a - 2\right) - 2 \left(a - 2\right) = 0$

$\therefore \left(a - 2\right) \left(9 a - 2\right) = 0$

$\therefore a - 2 = 0 \mathmr{and} 9 a - 2 = 0$

$\therefore a = 2 \mathmr{and} a = \frac{2}{9}$

Check :

$\left(i\right) a = 2$

$L H S = \sqrt{2 a + 5} - 2 \sqrt{2 a}$

$L H S = \sqrt{2 \left(2\right) + 5} - 2 \sqrt{2 \left(2\right)}$

$\therefore L H S = \sqrt{9} - 2 \sqrt{4} = 3 - 4 = - 1 \ne R H S$

$\therefore a \ne 2$

$\left(i i\right) a = \frac{2}{9}$

$L H S = \sqrt{2 a + 5} - 2 \sqrt{2 a}$

$L H S = \sqrt{2 \left(\frac{2}{9}\right) + 5} - 2 \sqrt{2 \left(\frac{2}{9}\right)}$

$\therefore L H S = \sqrt{\frac{4}{9} + 5} - 2 \sqrt{\frac{4}{9}} = \frac{7}{3} - \frac{4}{3} = \frac{3}{3} = 1 = R H S$

$\therefore a = \frac{2}{9}$