How do you solve $x - 9 = \sqrt{x - 3}$ $x - 9 = sqrt(x - 3)$ ?

Question

1565 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-01T10:15:29

$x = 12$ $x=12$

First you fix the domain of the solutions:

$x - 3 \geq 0 and x - 9 \geq 0$ $x-3>=0 and x-9>=0$

that's $x \geq 9$ $x>=9$

then you square both members:

${(x - 9)}^{2} = {(\sqrt{x - 3})}^{2}$ $(x-9)^2=(sqrt(x-3))^2$

that's

$x^{2} - 18 x + 81 = x - 3$ $x^2-18x+81=x-3$

$x^{2} - 18 x + 81 - x + 3 = 0$ $x^2-18x+81-x+3=0$

$x^{2} - 19 x + 84 = 0$ $x^2-19x+84=0$

$x = \frac{19 \pm \sqrt{19^{2} - 4 \cdot 84}}{2}$ $x=(19+-sqrt(19^2-4*84))/2$

$x = \frac{19 \pm 5}{2}$ $x=(19+-5)/2$

$x = 7 and x = 12$ $x=7 and x=12$

but only x=12 belongs to the calculated domain

How do you solve x−9=√x−3x - 9 = sqrt(x - 3)?