Question

Find the roots of the following quadratic equation

$3y^2+7y+1=0$

• A. $y=\frac{-7\pm \sqrt{37}}{6}$
• B. $y=\frac{7\pm \sqrt{37}}{6}$
• C. $y=\frac{-7\pm \sqrt{-37}}{6}$
• D. $y=\frac{7\pm \sqrt{-37}}{6}$

Answer 1

Answer: (A)

$y=\frac{-7\pm \sqrt{37}}{6}$

Given equation is

$3y^2+7y+1=0$
$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$a=3,b=7,c=1$

$y=\frac{-7\pm \sqrt{49-4\left(1\right)\left(3\right)}}{2\times 3}$

$\therefore y=\frac{-7\pm \sqrt{37}}{6}$

$\therefore$ the roots of the given equation are $\frac{-7\pm \sqrt{37}}{6}$