How do you solve $\sqrt{8 x} + 1 = 65$ $sqrt(8x)+1=65$ and check the solution?

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Answer 1 · 2017-05-23T18:09:36

See a solution process below:

First, subtract $1$ $color(red)(1)$ from each side of the equation to isolate the radical while keeping the equation balanced:

$\sqrt{8 x} + 1 - 1 = 65 - 1$ $sqrt(8x) + 1 - color(red)(1) = 65 - color(red)(1)$

$\sqrt{8 x} + 0 = 64$ $sqrt(8x) + 0 = 64$

$\sqrt{8 x} = 64$ $sqrt(8x) = 64$

Next, square each side of the equation to remove the radical while keeping the equation balanced:

${(\sqrt{8 x})}^{2} = 64^{2}$ $(sqrt(8x))^2 = 64^2$

$8 x = 4096$ $8x = 4096$

Now, divide each side of the equation by $8$ $color(red)(8)$ to solve for $x$ $x$ while keeping the equation balanced:

$\frac{8 x}{8} = \frac{4096}{8}$ $(8x)/color(red)(8) = 4096/color(red)(8)$

$(color(red)(cancel(color(black)(8)))x)/cancel(color(red)(8)) = 512$

$x = 512$

To check the solution substitute $color(red)(512)$ for $color(red)(x)$ and calculate each side of the equation to ensure both sides are equal:

$sqrt(8color(red)(x)) + 1 = 65$ becomes:

$sqrt(8 * color(red)(512)) + 1 = 65$

$sqrt(4096) + 1 = 65$

$64 + 1 = 65$

$65 + 65$

Both sides of the equation are equal therefore the solution is valid.

How do you solve sqrt(8x)+1=65√8x+1=65sqrt(8x)+1=65 and check the solution?