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Which is the solution of the quadratic equation [tex](4y-3)^2 = 72[/tex]?

A. [tex]y = \frac{3 + 6\sqrt{2}}{4}[/tex] and [tex]y = \frac{3 - 6\sqrt{2}}{4}[/tex]

B. [tex]y = \frac{3 + 6\sqrt{2}}{4}[/tex] and [tex]y = \frac{-3 - 6\sqrt{2}}{4}[/tex]

C. [tex]y = \frac{2\sqrt{2}}{4}[/tex] and [tex]y = \frac{4\sqrt{2}}{4}[/tex]

D. [tex]y = \frac{9\sqrt{2}}{4}[/tex] and [tex]y = \frac{3\sqrt{2}}{4}[/tex]


Sagot :

To determine the solutions of the given quadratic equation [tex]\((4y - 3)^2 = 72\)[/tex], we can follow these steps:

1. Write down the given equation:
[tex]\[(4y - 3)^2 = 72\][/tex]

2. Take the square root of both sides of the equation to eliminate the square:
[tex]\[ 4y - 3 = \pm \sqrt{72} \][/tex]

3. Simplify [tex]\(\sqrt{72}\)[/tex]:
[tex]\[ \sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2} \][/tex]
Therefore, we can write:
[tex]\[ 4y - 3 = \pm 6\sqrt{2} \][/tex]

4. Separate this into two distinct equations:
[tex]\[ 4y - 3 = 6\sqrt{2} \][/tex]
and
[tex]\[ 4y - 3 = -6\sqrt{2} \][/tex]

5. Solve each equation for [tex]\(y\)[/tex]:

- For [tex]\(4y - 3 = 6\sqrt{2}\)[/tex]:
[tex]\[ 4y = 6\sqrt{2} + 3 \][/tex]
[tex]\[ y = \frac{6\sqrt{2} + 3}{4} \][/tex]

- For [tex]\(4y - 3 = -6\sqrt{2}\)[/tex]:
[tex]\[ 4y = -6\sqrt{2} + 3 \][/tex]
[tex]\[ y = \frac{3 - 6\sqrt{2}}{4} \][/tex]

6. Hence, the solutions to the quadratic equation are:
[tex]\[ y = \frac{3 + 6\sqrt{2}}{4} \quad \text{and} \quad y = \frac{3 - 6\sqrt{2}}{4} \][/tex]

Given the results, the correct option is:
[tex]\[ \boxed{y = \frac{3+6\sqrt{2}}{4} \text{ and } y = \frac{3-6\sqrt{2}}{4}} \][/tex]