# 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: 20th Mar 2023

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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

$\begin{align}

& \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}}} \\

\end{align}$

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

$3{{b}^{2}}=16ac$

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;

\[\begin{align}

& 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 \\

\end{align}\]

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;

$\begin{align}

& 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 \\

\end{align}$

As, we have suppose $t\ \text{as}\ \dfrac{\beta }{\alpha }$ , Hence we get;

$\begin{align}

& \dfrac{\beta }{\alpha }=\dfrac{1}{3}\ \And \dfrac{\beta }{\alpha }=3 \\

& or \\

& 3\beta =\alpha ,\beta =3\alpha \\

\end{align}$

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

$\begin{align}

& 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 \\

\end{align}$

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

$\begin{align}

& \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}}}} \\

\end{align}$

Which is wrong. Hence, we need to apply the above relations very carefully. Correct relation is given as;

$\begin{align}

& \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}}} \\

\end{align}$

${{x}_{1}}+{{x}_{2}}=\dfrac{-B}{A}\And {{x}_{1}}{{x}_{2}}=\dfrac{C}{A}$

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

$\begin{align}

& \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}}} \\

\end{align}$

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

$3{{b}^{2}}=16ac$

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;

\[\begin{align}

& 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 \\

\end{align}\]

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;

$\begin{align}

& 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 \\

\end{align}$

As, we have suppose $t\ \text{as}\ \dfrac{\beta }{\alpha }$ , Hence we get;

$\begin{align}

& \dfrac{\beta }{\alpha }=\dfrac{1}{3}\ \And \dfrac{\beta }{\alpha }=3 \\

& or \\

& 3\beta =\alpha ,\beta =3\alpha \\

\end{align}$

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

$\begin{align}

& 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 \\

\end{align}$

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

$\begin{align}

& \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}}}} \\

\end{align}$

Which is wrong. Hence, we need to apply the above relations very carefully. Correct relation is given as;

$\begin{align}

& \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}}} \\

\end{align}$

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