Question

$3a^2{x}^2+8abx+4b^2=0,a\ne 0$

• A. No, Roots are imaginary
• B. .$\frac{-2b}{a},\frac{-2b}{3a}$, real
• C. Cannot be determined
  
• D. Incorrect question
  

Answer 1

Answer: (B)

.$\frac{-2b}{a},\frac{-2b}{3a}$, real

In $3a^2{x}^2+8abx+4b^2=0$

Roots are $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Roots are $=\frac{-8ab\pm \sqrt{(8ab{)}^2-4\times 3a^2\times 4b^2}}{2\times 3a^2}$

Therefore, $\frac{-8ab\pm \sqrt{64a^2{b}^2-48a^2{b}^2}}{6a^2}\Rightarrow \frac{-8ab\pm 4ab}{6a^2}$

If $\frac{-8ab-4ab}{6a^2}=\frac{-12ab}{6a^2}=\frac{-2b}{a}$

Or $\frac{-8ab+4ab}{6a^2}=\frac{-4ab}{6a^2}=\frac{-2b}{3a}$

Then roots are $\frac{-2b}{a}$ and $\frac{-2b}{3a}$.