To solve the equation [tex]\( x = \sqrt{2x + 15} \)[/tex], follow these steps:
1. Square both sides to eliminate the square root:
[tex]\[
x^2 = (\sqrt{2x + 15})^2
\][/tex]
This simplifies to:
[tex]\[
x^2 = 2x + 15
\][/tex]
2. Rearrange the equation to bring all terms to one side:
[tex]\[
x^2 - 2x - 15 = 0
\][/tex]
3. Factor the quadratic equation:
Find two numbers that multiply to [tex]\(-15\)[/tex] and add to [tex]\(-2\)[/tex]. Those numbers are [tex]\(-5\)[/tex] and [tex]\(3\)[/tex]. Therefore, the equation can be factored as:
[tex]\[
(x - 5)(x + 3) = 0
\][/tex]
4. Solve for [tex]\(x\)[/tex] by setting each factor to zero:
[tex]\[
x - 5 = 0 \quad \text{or} \quad x + 3 = 0
\][/tex]
Solving these gives:
[tex]\[
x = 5 \quad \text{or} \quad x = -3
\][/tex]
5. Check for extraneous solutions by substituting back into the original equation [tex]\(x = \sqrt{2x + 15}\)[/tex]:
- For [tex]\(x = 5\)[/tex]:
[tex]\[
5 = \sqrt{2(5) + 15} \Rightarrow 5 = \sqrt{10 + 15} \Rightarrow 5 = \sqrt{25} \Rightarrow 5 = 5
\][/tex]
This is true, so [tex]\(x = 5\)[/tex] is a valid solution.
- For [tex]\(x = -3\)[/tex]:
[tex]\[
-3 = \sqrt{2(-3) + 15} \Rightarrow -3 = \sqrt{-6 + 15} \Rightarrow -3 = \sqrt{9} \Rightarrow -3 = 3
\][/tex]
This is not true, so [tex]\(x = -3\)[/tex] is not a valid solution.
6. Conclusion:
The only solution is:
[tex]\[
x = 5
\][/tex]
