# How to find x in sqrt(x+3)+sqrt(2-x)=3? Please show complete solution. Thanks so much.

Oct 10, 2017

$x = 1 , \setminus q \quad x = - 2$

#### Explanation:

$\setminus \sqrt{x + 3} + \setminus \sqrt{2 - x} = 3$

$\setminus R i g h t a r r o w \setminus \sqrt{x + 3} = 3 - \setminus \sqrt{2 x - 2}$

$\setminus R i g h t a r r o w {\left(\setminus \sqrt{x + 3}\right)}^{2} = {\left(2 - \setminus \sqrt{2 - x}\right)}^{2}$

$\setminus R i g h t a r r o w x + 3 = 11 - 6 \setminus \sqrt{2 x} - x$

$\setminus R i g h t a r r o w 2 x - 8 = - 6 \setminus \sqrt{2 - x}$

$\setminus R i g h t a r r o w {\left(2 x - 8\right)}^{2} = {\left(- 6 \setminus \sqrt{2 - x}\right)}^{2}$

$\setminus R i g h t a r r o w 4 x - 32 x + 64 = 72 - 36 x$

$\setminus R i g h t a r r o w 4 {x}^{2} + 4 x - 9 = 0$

$x = \setminus \frac{- 4 + \setminus \sqrt{{4}^{2} - 4 \left(4\right) \left(- 8\right)}}{2 \setminus \cdot 4}$

$= 1$

$x = \setminus \frac{- 4 - \setminus \sqrt{{4}^{2} - 4 \left(4\right) \left(- 8\right)}}{2 \setminus \cdot 4}$
$= - 2$
$\setminus \therefore \setminus \quad x = 1 , \setminus q \quad x = - 2$