How do you solve $\sqrt{4 x + 4} - 8 = 0$ $sqrt(4x+4)-8=0$ ?

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Answer 1 · 2016-01-30T07:34:58

$x = 15$ $x=15$

$\sqrt{4 x - 4} - 8 = 0$ $sqrt(4x-4)-8=0$

$\to \sqrt{4 x - 4} = 8$ $rarrsqrt(4x-4)=8$

Square both sides

$\to {(\sqrt{4 x - 4})}^{2} = 8^{2}$ $rarr(sqrt(4x-4))^2=8^2$

$\to 4 x - 4 = 64$ $rarr4x-4=64$

$\to 4 x = 64 - 4$ $rarr4x=64-4$

$\to 4 x = 60$ $rarr4x=60$

$\to x = \frac{60}{4} = 15$ $rarrx=60/4=15$

How do you solve √4x+4−8=0sqrt(4x+4)-8=0?