How do you solve ${(\sqrt{x + 5} - 6)}^{2} + 4 = 20$ $(sqrt(x+5)-6)^2+4=20$ ?

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Answer 1 · 2018-03-14T09:45:28

The solution is $S = {- 1, 95}$ $S={-1, 95}$

First, what's under the $\sqrt{}$ $sqrt$ $sign$ $"sign"$ is $\geq 0$ $>=0$

Therefore,

$x + 5 \geq 0$ $x+5>=0$

$x \geq - 5$ $x>=-5$

${(\sqrt{x + 5} - 6)}^{2} + 4 = 20$ $(sqrt(x +5) - 6)^2 + 4 = 20$

${(\sqrt{x + 5} - 6)}^{2} = 16$ $(sqrt(x +5) - 6)^2 = 16$ //subtract both sides by 4

$\sqrt{x + 5} - 6 = \pm 4$ $sqrt(x +5) - 6 = ± 4$ //take the square root of both sides

$\sqrt{x + 5} = 6 \pm 4$ $sqrt(x +5) = 6± 4$ //add 6 to both sides

$x + 5 = {(6 \pm 4)}^{2}$ $x +5 = (6± 4)^2$ //square both sides

$x = {(6 \pm 4)}^{2} - 5$ $x = (6± 4)^2-5$ // subtract both sides by 5

so...

$x = {(6 \pm 4)}^{2} - 5$ $x = (6± 4)^2-5$

$x_{1} = {(6 + 4)}^{2} - 5$ $x_1 = (6+4)^2-5$
$x_{1} = {(10)}^{2} - 5$ $x_1 = (10)^2-5$
$x_{1} = 95$ $x_1=95$

$x_{2} = {(6 - 4)}^{2} - 5$ $x_2 = (6-4)^2-5$
$x_{2} = {(2)}^{2} - 5$ $x_2 = (2)^2-5$
$x_{2} = - 1$ $x_2 = -1$

The solution is $S = {- 1, 95}$ $S={-1, 95}$

graph{(sqrt(x+5)-6)^2-16 [-84.1, 126.9, -46.6, 58.8]}

How do you solve (√x+5−6)2+4=20(sqrt(x+5)-6)^2+4=20?