# How do you solve 1/2sqrt(2x-5)-1/2sqrt(3x+4)=-1?

Sep 30, 2017

$x = 15 \mathmr{and} x = 7$

#### Explanation:

$\frac{1}{2} \sqrt{2 x - 5} - \frac{1}{2} \sqrt{3 x + 4} = - 1 \text{ } \leftarrow$ multiply through by $2$

$\sqrt{2 x - 5} - \sqrt{3 x + 4} = - 2$

${\left(\sqrt{2 x - 5} - \sqrt{3 x + 4}\right)}^{2} = {\left(- 2\right)}^{2} \text{ } \leftarrow$ square both sides

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

$2 x - 5 - 2 \sqrt{2 x - 5} \times \sqrt{3 x + 4} + 3 x + 4 - 4 = 0$

$5 x - 5 = 2 \sqrt{2 x - 5} \times \sqrt{3 x + 4}$

${\left(5 x - 5\right)}^{2} = {\left(2 \sqrt{2 x - 5} \times \sqrt{3 x + 4}\right)}^{2} \text{ } \leftarrow$ square both sides

$25 {x}^{2} - 50 x + 25 = 4 \left(2 x - 5\right) \left(3 x + 4\right)$

$25 {x}^{2} - 50 x + 25 = 4 \left(6 {x}^{2} + 8 x - 15 x - 20\right)$

$25 {x}^{2} - 50 x + 25 = 24 {x}^{2} + 32 x - 60 x - 80$

${x}^{2} - 22 x + 105 = 0 \text{ } \leftarrow$ factorise

$\left(x - 15\right) \left(x - 7\right) = 0$

$x = 15 \mathmr{and} x = 7$