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

Last updated date: 17th Jul 2024
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Hint: If we have any quadratic $A{{x}^{2}}+Bx+C=0$ with roots ${{x}_{1}}\And {{x}_{2}}$ , then we have
${{x}_{1}}+{{x}_{2}}=\dfrac{-B}{A}\And {{x}_{1}}{{x}_{2}}=\dfrac{C}{A}$
Use the given relation to solve the given equation.

We have given that $\left( \alpha ,\beta \right)$ are roots of the quadratic $a{{x}^{2}}+bx+c=0\text{ }$
Another information given here is
$3{{b}^{2}}=16ac...........\left( 1 \right)$
Now, $\left( \alpha ,\beta \right)$ are roots of quadratic $a{{x}^{2}}+bx+c=0\text{ }$
Hence, relation between roots and coefficients of any quadratic is given as
  & \text{ sum of roots = }\dfrac{\text{-}\left( \text{coefficient of }x \right)}{\text{coefficient of }{{x}^{\text{2}}}} \\
 & \text{Product of roots = }\dfrac{\text{constant term}}{\text{coefficient of }{{x}^{2}}} \\
Therefore, we have quadratic equation as;
$a{{x}^{2}}+bx+c=0\text{ }$
Sum of roots $=\alpha +\beta =\dfrac{-b}{a}...................\left( 2 \right)$
Product of roots $=\alpha \beta =\dfrac{c}{a}.....................\left( 3 \right)$
Now, from equation (1), we have
Putting value of ‘b’ from equation (2) i.e. $-a\left( \alpha +\beta \right)$ in the above equation, we get;
$3{{a}^{2}}{{\left( \alpha +\beta \right)}^{2}}=16ac$
Transferring ${{a}^{2}}$ to another side, we get;
$3{{\left( \alpha +\beta \right)}^{2}}=\dfrac{16ac}{{{a}^{2}}}=16\dfrac{c}{a}$
From equation (3), we can replace $'\dfrac{c}{a}'$ by $\alpha \text{ }\beta $ from above equation;
$3{{\left( \alpha +\beta \right)}^{2}}=16\alpha \beta $
We have identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab,$ applying it with the above equation we get;
  & 3\left( {{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta \right)=16\alpha \beta \\
 & 3{{\alpha }^{2}}+3{{\beta }^{2}}+6\alpha \beta -16\alpha \beta =0 \\
 & 3{{\alpha }^{2}}+3{{\beta }^{2}}-10\alpha \beta =0 \\
Dividing whole equation by ‘${{\alpha }^{2}}$’, we get;
$3+3{{\left( \dfrac{\beta }{\alpha } \right)}^{2}}-10\left( \dfrac{\beta }{\alpha } \right)=0$
Let $\dfrac{\beta }{\alpha }='t'$ we can write above equation as;
$3{{t}^{2}}-10t+3=0$ …………………(4)
Now, splitting the middle term to get summation of 10 and product ‘9’, we get;
  & 3{{t}^{2}}-9t-t+3=0 \\
 & 3t\left( t-3 \right)-1\left( t-3 \right)=0 \\
 & \left( t-3 \right)\left( t-\dfrac{1}{3} \right)=0 \\
 & t=\dfrac{1}{3},t=3 \\
As, we have suppose $t\ \text{as}\ \dfrac{\beta }{\alpha }$ , Hence we get;
  & \dfrac{\beta }{\alpha }=\dfrac{1}{3}\ \And \dfrac{\beta }{\alpha }=3 \\
 & or \\
 & 3\beta =\alpha ,\beta =3\alpha \\
Therefore, Option (C) is the correct answer.

Note: One can go wrong while factoring $3{{\alpha }^{2}}+3{{\beta }^{2}}-10\alpha \beta =0$. To minimize the confusion, divide the whole equation by ${{\alpha }^{2}}\ \And \ {{\beta }^{2}}$.
We can factorize $3{{\alpha }^{2}}-10\alpha \beta +3{{\beta }^{2}}=0$ as splitting middle term to $-9\alpha \beta \ and\ -\alpha \beta $ as
  & 3{{\alpha }^{2}}-9\alpha \beta -\alpha \beta +3{{\beta }^{2}}=0 \\
 & 3\alpha \left( \alpha -3\beta \right)-\beta \left( \alpha -3\beta \right)=0 \\
 & \left( \alpha -3\beta \right)\left( 3\alpha -\beta \right)=0 \\
Hence, we get $\alpha =3\beta \ or\ \beta =3\alpha $.
One can go wrong while writing the sum of roots and product of roots. He/she may write
  & \text{ sum of roots = }\dfrac{\text{constant term}}{\text{coefficient of }{{x}^{2}}} \\
 & \text{Product of roots = }\dfrac{\text{- coefficient of }x}{\text{coefficient of }{{x}^{\text{2}}}} \\
Which is wrong. Hence, we need to apply the above relations very carefully. Correct relation is given as;
  & \text{ sum of roots = }\dfrac{\text{- coefficient of }x}{\text{coefficient of }{{x}^{\text{2}}}} \\
 & \text{Product of roots = }\dfrac{\text{constant term}}{\text{coefficient of }{{x}^{2}}} \\