How do you solve $\sqrt{x - 3} + 4 = 6$ $sqrt(x-3)+4=6$ ?

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Answer 1 · 2015-07-20T15:15:52

$x = 7$ $color(blue)(x =7$

$\sqrt{x - 3} + 4 = 6$ $sqrt(x-3) +4 =6$

$\sqrt{x - 3} = 6 - 4$ $sqrt(x-3) =6-4$

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

Squaring both sides:
${(\sqrt{x - 3})}^{2} = 2^{2}$ $(sqrt(x-3))^2 =2^2$
$x - 3 = 4$ $x-3 =4$

$x = 7$ $color(blue)(x =7$

How do you solve √x−3+4=6sqrt(x-3)+4=6?