Question

Roots of quadratic equation $3y^2+7y+4=0$ are

• A. $y=-1,\frac{-4}{3}$
• B. $y=2,\frac{-4}{3}$
• C. $y=-2,\frac{-4}{3}$
• D. $y=-1,\frac{-4}{5}$

Answer 1

Answer: (A)

$y=-1,\frac{-4}{3}$

Given equation is

$3y^2+7y+4=0$
$3y^2+3y+4y+4=0$
$3y(y+1)+4(y+1)=0$
$(3y+4)(y+1)=0$

$\therefore y=\frac{-4}{3},-1$

$\therefore$ the roots of the given equation are $\frac{-4}{3},-1$