How do you solve $\sqrt{x + 1} + 2 = 4$ $sqrt(x+1)+2=4$ and check the solution?

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Answer 1 · 2017-06-23T17:35:14

x = 3

1. Subtract 2 from both sides to isolate the square root.

$\sqrt{x + 1} + 2 - 2 = 4 - 2$ $sqrt(x+1) + 2 - 2 = 4 - 2$
$\sqrt{x + 1} = 2$ $sqrt(x+1) = 2$

2. To get rid of the square root, square both sides.

${(\sqrt{x + 1})}^{2} = 2^{2}$ $(sqrt(x+1))^2 = 2^2$
$x + 1 = 4$ $x + 1 = 4$

3. Subtract 1 from both sides to find x.

$x + 1 - 1 = 4 - 1$ $x + 1 - 1 = 4 - 1$
$x = 3$ $x = 3$

Now that we know that x = 3, check the solution by plugging its value back into the original equation.

$\sqrt{3 + 1} + 2 = 4$ $sqrt(3 + 1) + 2 = 4$
$\sqrt{4} + 2 = 4$ $sqrt(4) + 2 = 4$
$2 + 2 = 4$ $2 + 2 = 4$
$4 = 4$ $4 = 4$

The solution, x = 3, works!

How do you solve sqrt(x+1)+2=4√x+1+2=4sqrt(x+1)+2=4 and check the solution?