How do you solve $sqrt(2x +3) = 6-x$ ?

Question

2749 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-04T21:31:19

$x = 3$

$sqrt (2x + 3) = 6 - x$

Square both sides:

$sqrt (2x + 3)^2 = (6 - x)^2$

Notice that $2x + 3 >= 0$ and $6 - x >= 0$
$=> -3/2 <= x <= 6$

$2x + 3 = 36 - 12x + x^2$

$x^2 - 14x + 33 = 0$

$(x - 11)(x - 3) = 0$

$x = 3, 11$

Since $-3/2 <= x <= 6$ , $x = 11$ will not work in the original eqaution and the answer is $x = 3$ .

How do you solve sqrt(2x +3) = 6-xsqrt(2x +3) = 6-x?