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Find all real solutions of the equation:

[tex]\[
(6z + 4)^2 - 22 = 0
\][/tex]

Sagot :

GinnyAnswer
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.
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