Question

If the roots of the equation $12x^2+mx+5-0$ are in the ratio 3 : 2, then m equals

• A. $\frac{1}{12}$
• B. $\frac{5}{12}$
• C. Both (A) and (B)
• D. None of these

Answer 1

The correct option is D None of these

Given,

$12x^2+mx+5=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-m\pm \sqrt{m^2-4\cdot 12\cdot 5}}{2\cdot 12}$

$\therefore x=\frac{-m\pm \sqrt{m^2-240}}{24}$

Given,

$\frac{3}{2}=\frac{\frac{-m+\sqrt{m^2-240}}{24}}{\frac{-m-\sqrt{m^2-240}}{24}}$

$3\cdot \frac{-m-\sqrt{m^2-240}}{24}=2\cdot \frac{-m+\sqrt{m^2-240}}{24}$

$3\cdot \frac{-m-\sqrt{m^2-240}}{24}-2\cdot \frac{-m+\sqrt{m^2-240}}{24}=2\cdot \frac{-m+\sqrt{m^2-240}}{24}-2\cdot \frac{-m+\sqrt{m^2-240}}{24}$

$\Rightarrow \frac{-m-5\sqrt{m^2-240}}{24}=0$

$-m-5\sqrt{m^2-240}=0$

$-5\sqrt{m^2-240}=m$

${(-5\sqrt{m^2-240})}^2=m^2$

$25m^2-6000=m^2$

$m=5\sqrt{10},m=-5\sqrt{10}$

$\therefore m=-5\sqrt{10}$ is true solution