which is the solution of the quadratic equation (4y - 3)^2 = 72?\no y = \\frac{3 + 6\\sqrt{2}}{4} and y =…

which is the solution of the quadratic equation (4y - 3)^2 = 72?\no y = \\frac{3 + 6\\sqrt{2}}{4} and y = \\frac{3 - 6\\sqrt{2}}{4}\no y = \\frac{3 + 6\\sqrt{2}}{4} and y = \\frac{-3 - 6\\sqrt{2}}{4}\no y = \\frac{9\\sqrt{2}}{4} and y = \\frac{-3\\sqrt{2}}{4}\no y = \\frac{9\\sqrt{2}}{4} and y = \\frac{3\\sqrt{2}}{4}

which is the solution of the quadratic equation (4y - 3)^2 = 72?\no y = \\frac{3 + 6\\sqrt{2}}{4} and y = \\frac{3 - 6\\sqrt{2}}{4}\no y = \\frac{3 + 6\\sqrt{2}}{4} and y = \\frac{-3 - 6\\sqrt{2}}{4}\no y = \\frac{9\\sqrt{2}}{4} and y = \\frac{-3\\sqrt{2}}{4}\no y = \\frac{9\\sqrt{2}}{4} and y = \\frac{3\\sqrt{2}}{4}

Answer

Explanation:

Step1: Take square - root of both sides

$$4y - 3=\pm\sqrt{72}$$ Since $\sqrt{72}=\sqrt{36\times2}=6\sqrt{2}$, we have $4y - 3=\pm6\sqrt{2}$.

Step2: Solve for $y$ when $4y - 3 = 6\sqrt{2}$

Add 3 to both sides: $4y=3 + 6\sqrt{2}$. Then divide by 4: $y=\frac{3 + 6\sqrt{2}}{4}$.

Step3: Solve for $y$ when $4y - 3=-6\sqrt{2}$

Add 3 to both sides: $4y=3-6\sqrt{2}$. Then divide by 4: $y=\frac{3 - 6\sqrt{2}}{4}$.

Answer:

A. $y=\frac{3 + 6\sqrt{2}}{4}$ and $y=\frac{3 - 6\sqrt{2}}{4}$