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

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Answer 1 · 2018-04-04T21:31:19

$x = 3$ $x = 3$

$\sqrt{2 x + 3} = 6 - x$ $sqrt (2x + 3) = 6 - x$

Square both sides:

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

Notice that $2 x + 3 \geq 0$ $2x + 3 >= 0$ and $6 - x \geq 0$ $6 - x >= 0$
$\Rightarrow - \frac{3}{2} \leq x \leq 6$ $=> -3/2 <= x <= 6$

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

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

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

$x = 3, 11$ $x = 3, 11$

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

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