Chapter 9: Q.458. (page 984)

Solve the given equation.
${(2 y + 3)}^{2} + 3 (2 y + 3) - 28 = 0$

Short Answer

Expert verified

The solutions are $y = - 5$ and $y = \frac{1}{2}$ .

Step by step solution

Step 1. Given information.

The given equation is ${(2 y + 3)}^{2} + 3 (2 y + 3) - 28 = 0$ .

Step 2. Simplify the equation.

$\begin{array}{rcl} {(2 y + 3)}^{2} + 3 (2 y + 3) - 28 & = & 0 \\ {(2 y + 3)}^{2} + 7 (2 y + 3) - 4 (2 y + 3) - 28 & = & 0 \\ (2 y + 3) (2 y + 3 + 7) - 4 (2 y + 3 + 7) & = & 0 \\ (2 y + 3 + 7) (2 y + 3 - 4) & = & 0 \end{array}$

Step 3. Solve the equation for y.

$\begin{array}{rcl} 2 y + 3 + 7 & = & 0 \\ 2 y + 10 & = & 0 \\ 2 y & = & - 10 \\ y & = & - \frac{10}{2} \\ y & = & - 5 \end{array}$ and $\begin{array}{rcl} 2 y + 3 - 4 & = & 0 \\ 2 y - 1 & = & 0 \\ 2 y & = & 1 \\ y & = & \frac{1}{2} \end{array}$

Step 4. Simplified answer.

Hence, the required solutions are $y = - 5$ and $y = \frac{1}{2}$ .

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Solve the given equation.
${(2 y + 3)}^{2} + 3 (2 y + 3) - 28 = 0$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Simplify the equation.

Step 3. Solve the equation for y.

Step 4. Simplified answer.

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Step 2. Simplify the equation.

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