Let's solve the equation:
[tex]\[
\sqrt{2x + 13} = x - 1
\][/tex]
Step 1: Isolate the square root term.
To remove the square root, we will square both sides of the equation:
[tex]\[
(\sqrt{2x + 13})^2 = (x - 1)^2
\][/tex]
This simplifies to:
[tex]\[
2x + 13 = (x - 1)^2
\][/tex]
Step 2: Expand the right side of the equation.
[tex]\[
2x + 13 = (x - 1)(x - 1)
\][/tex]
Expanding:
[tex]\[
2x + 13 = x^2 - 2x + 1
\][/tex]
Step 3: Move all terms to one side to set the equation to 0.
Subtract [tex]\(2x + 13\)[/tex] from both sides:
[tex]\[
2x + 13 - 2x - 13 = x^2 - 2x + 1 - 2x - 13
\][/tex]
This simplifies to:
[tex]\[
0 = x^2 - 4x - 12
\][/tex]
Or:
[tex]\[
x^2 - 4x - 12 = 0
\][/tex]
Step 4: Solve the quadratic equation.
We use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = -12\)[/tex].
[tex]\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-12)}}{2(1)}
\][/tex]
This simplifies to:
[tex]\[
x = \frac{4 \pm \sqrt{16 + 48}}{2}
\][/tex]
[tex]\[
x = \frac{4 \pm \sqrt{64}}{2}
\][/tex]
[tex]\[
x = \frac{4 \pm 8}{2}
\][/tex]
Step 5: Find the roots.
This results in two possible solutions:
[tex]\[
x = \frac{4 + 8}{2} = \frac{12}{2} = 6
\][/tex]
[tex]\[
x = \frac{4 - 8}{2} = \frac{-4}{2} = -2
\][/tex]
Step 6: Validate the solutions.
We need to check if both potential solutions satisfy the original equation [tex]\( \sqrt{2x + 13} = x - 1 \)[/tex].
For [tex]\( x = 6 \)[/tex]:
[tex]\[
\sqrt{2(6) + 13} = 6 - 1
\][/tex]
[tex]\[
\sqrt{12 + 13} = 5
\][/tex]
[tex]\[
\sqrt{25} = 5
\][/tex]
Which is true. So, [tex]\( x = 6 \)[/tex] is a valid solution.
For [tex]\( x = -2 \)[/tex]:
[tex]\[
\sqrt{2(-2) + 13} = -2 - 1
\][/tex]
[tex]\[
\sqrt{-4 + 13} = -3
\][/tex]
[tex]\[
\sqrt{9} = -3
\][/tex]
[tex]\(\sqrt{9} = 3\)[/tex], not [tex]\(-3\)[/tex], hence [tex]\( x = -2 \)[/tex] is not a valid solution.
Step 7: Conclude the result.
The only valid solution to the equation [tex]\( \sqrt{2x + 13} = x - 1 \)[/tex] is:
[tex]\( x = 6 \)[/tex].
