How do you solve $\sqrt{x - 2} + 5 = 6$ $sqrt(x-2) + 5=6$ ?

Question

1548 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-15T12:29:58

See a solution process below:

First, subtract $5$ $color(red)(5)$ from each side of the equation to isolate the radical while keeping the equation balanced:

$\sqrt{x - 2} + 5 - 5 = 6 - 5$ $sqrt(x - 2) + 5 - color(red)(5) = 6 - color(red)(5)$

$\sqrt{x - 2} + 0 = 1$ $sqrt(x - 2) + 0 = 1$

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

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

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

$x - 2 = 1$ $x - 2 = 1$

Now, add $2$ $color(red)(2)$ to each side of the equation to solve for $x$ $x$ while keeping the equation balanced:

$x - 2 + 2 = 1 + 2$ $x - 2 + color(red)(2) = 1 + color(red)(2)$

$x - 0 = 3$ $x - 0 = 3$

$x = 3$ $x = 3$

How do you solve √x−2+5=6sqrt(x-2) + 5=6?