How do you solve $\sqrt{x + 1} = 8$ $sqrt(x+1) = 8$ ?

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How do you solve $\sqrt{x + 1} = 8$ $sqrt(x+1) = 8$ ?

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Answer 1 · 2016-04-10T18:35:06

x=63

Explanation:

$\sqrt{x + 1} = 8$ $sqrt(x+1)=8$
Now you have to square both sides to cencle out root.

${(\sqrt{x + 1})}^{2} = 8^{2}$ $(sqrt(x+1))^2=8^2$
$x + 1 = 64$ $x+1=64$
$x = 64 - 1$ $x=64-1$
$x = 63$ $x=63$

In order to verify: $\sqrt{63 + 1} = 8$ $sqrt(63+1)=8$
$\sqrt{64} = 8$ $sqrt64=8$
$\pm 8 = 8$ $+-8=8$
In the verification $\pm$ $+-$ is just because of the rule any number squared gives a positive value. Accordingly ${(- 8)}^{2} = 64$ $(-8)^2=64$ as well as $8^{2} = 64$ $8^2=64$ .

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How do you solve $\sqrt{x + 1} = 8$ $sqrt(x+1) = 8$ ?

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