Question

If $\alpha ,\beta $are the roots of the quadratic equation $a{x^2} + bx + c = 0$ and $3{b^2} = 16ac$ then:
A.$\alpha = 4\beta {\text{ or }}\beta = 4\alpha $
B.$\alpha = - 4\beta {\text{ or }}\beta = - \alpha $
C.$\alpha = 3\beta {\text{ or }}\beta = 3\alpha $
D.$\alpha = - 3\beta {\text{ or }}\beta = - 3\alpha $

Answer 1

Hint: Use the information that roots always satisfy the quadratic equation. Also, if quadratic equation is in the standard from, which is $l{x^2} + mx + n = 0$then sum of the roots $ = \dfrac{{ - m}}{l}$and multiplication of the roots $ = \dfrac{n}{l}$.

According to the given information in the hint, $\alpha + \beta = \dfrac{{ - b}}{a}$ and $\alpha \beta = \dfrac{c}{a}$. Then ${(\alpha + \beta )^2} = \dfrac{{{b^2}}}{{{a^2}}}$ . we have given, $3{b^2} = 16ac$. On putting the value of ${b^2}$ we’ll get,
\[
  {(\alpha + \beta )^2} = \dfrac{{{b^2}}}{{{a^2}}} \Rightarrow {b^2} = {a^2}{(\alpha + \beta )^2} \\
  3{b^2} = 16ac \\
   \Rightarrow 3{a^2}{(\alpha + \beta )^2} = 16a \times a(\alpha \beta ){\text{ }}[{\text{Using, }}\alpha \beta = \dfrac{c}{a}] \\
   \Rightarrow 3{a^2}({\alpha ^2} + {\beta ^2} + 2\alpha \beta ) = 16{a^2}\alpha \beta \\
   \Rightarrow 3({\alpha ^2} + {\beta ^2} + 2\alpha \beta ) = 16\alpha \beta \\
   \Rightarrow 3{\alpha ^2} + 3{\beta ^2} + 6\alpha \beta = 16\alpha \beta \\
   \Rightarrow 3{\alpha ^2} - 10\alpha \beta + 3{\beta ^2} = 0 \\
   \Rightarrow 3{\alpha ^2} - (9\beta + \beta )\alpha + 3{\beta ^2} = 0 \\
   \Rightarrow 3{\alpha ^2} - 9\beta \alpha - \beta \alpha + 3{\beta ^2} = 0 \\
   \Rightarrow 3\alpha (\alpha - 3\beta ) - \beta (\alpha - 3\beta ) = 0 \\
   \Rightarrow (\alpha - 3\beta )(3\alpha - \beta ) = 0 \\
   \Rightarrow \alpha = 3\beta ,\dfrac{\beta }{3} \\
\]

Hence, $\alpha = 3\beta $ and $\beta = 3\alpha $ are the correct set of equations. So, option C will be the correct option.

Note: While solving the question, be careful with the calculation part. Then only you’ll be able to solve the problem properly.