Solve the equation
-$16 y^{2}=(2 y-3)^{2}$

A) $\left\{-\frac{3}{2}\right\}$

B) $\left\{-\frac{1}{2}, 4\right\}$

C) $\left\{\frac{3}{2},-\frac{3}{2}\right\}$

D) $\left\{-\frac{3}{2}, \frac{1}{2}\right\}$

Question

Solve the equation
-$16 y^{2}=(2 y-3)^{2}$

A) $\left\{-\frac{3}{2}\right\}$

B) $\left\{-\frac{1}{2}, 4\right\}$

C) $\left\{\frac{3}{2},-\frac{3}{2}\right\}$

D) $\left\{-\frac{3}{2}, \frac{1}{2}\right\}$

Quizplus · Accepted Answer

Expanding and simplifying both sides of the equation, we get $16y^2 = 4y^2 - 12y + 9$. Rearranging gives $12y^2 - 12y - 9 = 0$, which simplifies to $4y^2 - 4y - 3 = 0$. Solving this quadratic equation using the quadratic formula yields the solutions $y = -\frac{3}{2}$ and $y = \frac{1}{2}$.

Solve the Equation - 16y2=(2y−3)216 y^{2}=(2 y-3)^{2}16y2=(2y−3)2 A) {−32}\left\{-\frac{3}{2}\right\}{−23​} B) {−12,4}\left\{-\frac{1}{2}, 4\right\}{−21​,4} C) {32,−32}\left\{\frac{3}{2},-\frac{3}{2}\right\}{23​,−23​}

Solve the Equation
- $16 y^{2}=(2 y-3)^{2}$ A) $\left\{-\frac{3}{2}\right\}$ B) $\left\{-\frac{1}{2}, 4\right\}$ C) $\left\{\frac{3}{2},-\frac{3}{2}\right\}$