To solve the equation [tex]\(\sqrt{2x + 9} + x = 13\)[/tex], follow these steps:
1. Isolate the Square Root Term:
First, isolate the square root term by subtracting [tex]\(x\)[/tex] from both sides of the equation:
[tex]\[
\sqrt{2x + 9} = 13 - x
\][/tex]
2. Square Both Sides:
To eliminate the square root, square both sides of the equation:
[tex]\[
(\sqrt{2x + 9})^2 = (13 - x)^2
\][/tex]
Simplify both sides:
[tex]\[
2x + 9 = (13 - x)(13 - x)
\][/tex]
[tex]\[
2x + 9 = 169 - 26x + x^2
\][/tex]
3. Form a Quadratic Equation:
Rearrange the equation to form a standard quadratic equation:
[tex]\[
x^2 - 28x + 160 = 0
\][/tex]
4. Solve the Quadratic Equation:
Now solve the quadratic equation [tex]\(x^2 - 28x + 160 = 0\)[/tex]. Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], we have:
[tex]\[
a = 1, \quad b = -28, \quad c = 160
\][/tex]
Substitute these values into the quadratic formula:
[tex]\[
x = \frac{28 \pm \sqrt{(-28)^2 - 4 \cdot 1 \cdot 160}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{28 \pm \sqrt{784 - 640}}{2}
\][/tex]
[tex]\[
x = \frac{28 \pm \sqrt{144}}{2}
\][/tex]
[tex]\[
x = \frac{28 \pm 12}{2}
\][/tex]
5. Find the Possible Solutions:
Calculate the two possible values for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{28 + 12}{2} = \frac{40}{2} = 20
\][/tex]
[tex]\[
x = \frac{28 - 12}{2} = \frac{16}{2} = 8
\][/tex]
6. Check the Solutions:
Substitute each solution back into the original equation to verify which values satisfy it. Let's check [tex]\(x = 20\)[/tex]:
[tex]\[
\sqrt{2 \cdot 20 + 9} + 20 = 13
\][/tex]
[tex]\[
\sqrt{49} + 20 = 13
\][/tex]
[tex]\[
7 + 20 \neq 13
\][/tex]
Thus, [tex]\(x = 20\)[/tex] does not satisfy the equation.
Now let's check [tex]\(x = 8\)[/tex]:
[tex]\[
\sqrt{2 \cdot 8 + 9} + 8 = 13
\][/tex]
[tex]\[
\sqrt{25} + 8 = 13
\][/tex]
[tex]\[
5 + 8 = 13
\][/tex]
Thus, [tex]\(x = 8\)[/tex] satisfies the equation.
Therefore, the solution to the equation [tex]\(\sqrt{2x + 9} + x = 13\)[/tex] is:
[tex]\[
x = 8
\][/tex]
