Question

If the roots of the equation $2{{x}^{2}}-7x+8=0$ are $\alpha \text{ and }\beta $, then the equation whose roots are $\left( 3\alpha -4\beta \right)\text{ and }\left( 3\beta -4\alpha \right)$ is
(a) $2{{x}^{2}}+7x+98=0$
(b) ${{x}^{2}}+7x+98=0$
(c) $2{{x}^{2}}-7x-98=0$
(d) $2{{x}^{2}}-7x+98=0$

Answer 1

Hint: Find the value of the product of roots and the sum of the roots of the equation, $2{{x}^{2}}-7x+8=0$. Now, write the general form of the quadratic equation if its two roots are known and substitute the value of the roots to get the equation.

Complete step-by-step answer:
We have been provided with the quadratic equation given by: $2{{x}^{2}}-7x+8=0$ and its roots are $\alpha \text{ and }\beta $. We know that,
Sum of the roots of a quadratic equation $=\dfrac{-\text{coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}}$
Therefore, $\alpha +\beta =\dfrac{-(-7)}{2}=\dfrac{7}{2}$
Also, product of the roots of a quadratic equation $=\dfrac{\text{constant term}}{\text{coefficient of }{{\text{x}}^{\text{2}}}}$
Therefore, $\alpha \beta =\dfrac{8}{2}=4$
Now, we have to find the quadratic equation whose roots are $\left( 3\alpha -4\beta \right)\text{ and }\left( 3\beta -4\alpha \right)$. We know that a quadratic equation having roots $a\text{ and }b$ is given as: k[${{x}^{2}}-\left( a+b \right)x+ab=0$] where k is any integer. Therefore, the quadratic equation having roots $\left( 3\alpha -4\beta \right)\text{ and }\left( 3\beta -4\alpha \right)$ can be written as (First let k=1):
${{x}^{2}}-\left[ \left( 3\alpha -4\beta \right)+\left( 3\beta -4\alpha \right) \right]x+\left( 3\alpha -4\beta \right)\times \left( 3\beta -4\alpha \right)=0$
Simplifying the above equation we get,
$\begin{align}
  & {{x}^{2}}-\left( -\alpha -\beta \right)x+9\alpha \beta -12{{\alpha }^{2}}-12{{\beta }^{2}}+16\alpha \beta =0 \\
 & \Rightarrow {{x}^{2}}+\left( \alpha +\beta \right)x+25\alpha \beta -12{{\alpha }^{2}}-12{{\beta }^{2}}=0 \\
\end{align}$
Now, $-12{{\alpha }^{2}}-12{{\beta }^{2}}=-12\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)=-12\left[ {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta \right]$
Therefore, the quadratic equation becomes,
$\begin{align}
  & {{x}^{2}}-\left( \alpha +\beta \right)x+25\alpha \beta -12{{\left( \alpha +\beta \right)}^{2}}+24\alpha \beta =0 \\
 & {{x}^{2}}+\left( \alpha +\beta \right)x+49\alpha \beta -12{{\left( \alpha +\beta \right)}^{2}}=0 \\
\end{align}$
Now, substituting the value of $\left( \alpha +\beta \right)\text{ and }\alpha \beta $ in the above equation we get,
$\begin{align}
  & {{x}^{2}}+\dfrac{7}{2}x+49\times 4-12\times {{\left( \dfrac{7}{2} \right)}^{2}}=0 \\
 & \Rightarrow {{x}^{2}}+\dfrac{7}{2}x+196-12\times \dfrac{49}{4}=0 \\
 & \Rightarrow {{x}^{2}}+\dfrac{7}{2}x+196-147=0 \\
 & \Rightarrow {{x}^{2}}+\dfrac{7}{2}x+49=0 \\
\end{align}$
Now, multiplying the equation by 2 on both sides, we get,
$\Rightarrow 2{{x}^{2}}+7x+98=0$
Hence, option (a) is the correct answer.

Note: Always remember the general form of a quadratic equation when its roots are given. If any term of the quadratic equation comes in fraction then multiply both sides of the equation by the denominator to get rid of the fraction. One can also remember the sum and product of roots of the quadratic equation, $a{{x}^{2}}+bx+c$ as $\dfrac{-b}{a}\text{ and }\dfrac{c}{a}$ respectively, in short form.