# How do you solve sqrt(10–x)+x=8?

Mar 5, 2016

$x = 9 , 6$

#### Explanation:

$\sqrt{10 - x} + x = 8$

$\rightarrow \sqrt{10 - x} = 8 - x$

Square both sides in order to get rid of the radical sign

$\rightarrow {\left(\sqrt{10 - x}\right)}^{2} = {\left(8 - x\right)}^{2}$

Use the formula ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$

$\rightarrow 10 - x = 64 - 16 x + {x}^{2}$

$\rightarrow 10 = 64 - 16 x + {x}^{2} + x$

$\rightarrow 10 = 64 - 15 x + {x}^{2}$

$\rightarrow 0 = 64 - 15 x + {x}^{2} - 10$

$\rightarrow 0 = 54 - 15 x + {x}^{2}$

Rewrite

$\rightarrow {x}^{2} - 15 x + 54 = 0$

Factor the equation

$\rightarrow \left(x - 9\right) \left(x - 6\right) = 0$

If we solve for it we get $x = 9 , 6$

But,

If we take the value of $9$ of $x$ into the equation,

$\sqrt{10 - 9} + 9 = 0$

$\sqrt{1} + 9 = 0$

$1 + 9 \ne 0$

We should take $\sqrt{1}$ as $- \sqrt{1}$