To find all real solutions for the equation [tex]\((6z + 4)^2 - 22 = 0\)[/tex], we will follow these steps:
1. Simplify the Equation:
Start with the given equation:
[tex]\[
(6z + 4)^2 - 22 = 0
\][/tex]
2. Isolate the Square Term:
Add 22 to both sides to isolate the square term:
[tex]\[
(6z + 4)^2 = 22
\][/tex]
3. Take the Square Root:
Take the square root of both sides of the equation:
[tex]\[
6z + 4 = \pm \sqrt{22}
\][/tex]
This results in two separate equations:
[tex]\[
6z + 4 = \sqrt{22} \quad \text{and} \quad 6z + 4 = -\sqrt{22}
\][/tex]
4. Solve for [tex]\(z\)[/tex]:
For the first equation:
[tex]\[
6z + 4 = \sqrt{22}
\][/tex]
Subtract 4 from both sides:
[tex]\[
6z = \sqrt{22} - 4
\][/tex]
Divide both sides by 6:
[tex]\[
z = \frac{\sqrt{22} - 4}{6}
\][/tex]
Simplify the expression:
[tex]\[
z = -\frac{2}{3} + \frac{\sqrt{22}}{6}
\][/tex]
For the second equation:
[tex]\[
6z + 4 = -\sqrt{22}
\][/tex]
Subtract 4 from both sides:
[tex]\[
6z = -\sqrt{22} - 4
\][/tex]
Divide both sides by 6:
[tex]\[
z = \frac{-\sqrt{22} - 4}{6}
\][/tex]
Simplify the expression:
[tex]\[
z = -\frac{\sqrt{22}}{6} - \frac{2}{3}
\][/tex]
5. List All Real Solutions:
Therefore, the real solutions to the equation [tex]\((6z + 4)^2 - 22 = 0\)[/tex] are:
[tex]\[
z = -\frac{2}{3} + \frac{\sqrt{22}}{6}
\][/tex]
and
[tex]\[
z = -\frac{\sqrt{22}}{6} - \frac{2}{3}
\][/tex]
These are the solutions to the given quadratic equation.
