Question

Solve for $y : 2y^{2}+6y+\frac{17}{2}=0$.

• A. $\frac{3}{2} \pm i \sqrt{2}$
• B. $-\frac{3}{2} \pm i \sqrt{2}$
• C. $\frac{1}{2} \pm i \sqrt{2}$
• D. $-\frac{1}{2} \pm i \sqrt{2}$

Answer 1

Answer: (B)

$-\frac{3}{2} \pm i \sqrt{2}$

Answer 2

We have, $2y^{2}+6y+\frac{17}{2}=0$ $\Rightarrow\, y^{2}+3y+\frac{17}{4}=0\quad$ (dividing both sides by 2) On adding and subtracting $\left(\frac{3}{2}\right)^{2}$ , we get $y^{2}+2\left(y\right)\left(\frac{3}{2}\right)+\left(\frac{3}{2}\right)^{2} -\left(\frac{3}{2}\right)^{2} +\frac{17}{4}=0$ $\Rightarrow\,\left(y+\frac{3}{2}\right)^{2} +2=0$ $\left(\because\, a^{2}+2ab+b^{2}=\left(a+b\right)^2\right)$ $\Rightarrow\, \left(y+\frac{3}{2}\right)^{2}-\left(i\sqrt{2}\right)^{2}=0$ $\Rightarrow\, \left(y+\frac{3}{2}+i\sqrt{2}\right)\left(y+\frac{3}{2}-i \sqrt{2}\right)=0$ $\left(\because\, a^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\right)$ Either $y+\frac{3}{2}+i \sqrt{2}=0$ or $y+\frac{3}{2}-i\sqrt{2}=0$ $\Rightarrow\, y=-\frac{3}{2}-i\sqrt{2}$ or $y=-\frac{3}{2}+i\sqrt{2}$ Hence, roots of the given equation are $-\frac{3}{2}-i\sqrt{2}$ and $-\frac{3}{2}+i \sqrt{2}$.