Question

If one root of the equation $a{{x}^{2}}+bx+c=0$ is three times the other, then
a. $3{{b}^{2}}=4ac$
b. $9{{b}^{2}}=25ac$
c. $3{{b}^{2}}=16ac$
d. None of these

Answer 1

It is given in the question that one root of the equation $a{{x}^{2}}+bx+c=0$ is three times the other, so we will assume the two roots as $\alpha$ and $3\alpha$. We then use the concept of the sum and product of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ as $\dfrac{-b}{a}$ and $\dfrac{c}{a}$ for the assumed roots. We then make use of these two conditions and make the necessary calculations to get the required result.

Complete step-by-step solution
We have been given the question that one root of the equation $a{{x}^{2}}+bx+c=0$ is three times the other. It means that one root is thrice the other root. Then we have to determine the condition and choose the most appropriate option from the given options.
Let us assume the first root of the equation as $\alpha$, then we can write the other root of the equation as $3\alpha$. Now, we know that the sum of roots is equal to $-\dfrac{b}{a}$. And here, we have the roots as $\alpha$ and $3\alpha$, so their sum will be,
$\Rightarrow \alpha +3\alpha =-\dfrac{b}{a}$.
$\Rightarrow 4\alpha =-\dfrac{b}{a}.........\left( i \right)$.
And we know that the product of roots is equal to $\dfrac{c}{a}$. So, we can write the product of the roots, $\alpha$ and $3\alpha$ as,
$\Rightarrow \alpha \left( 3\alpha \right)=\dfrac{c}{a}$.
$\Rightarrow 3{{\alpha }^{2}}=\dfrac{c}{a}.........\left( ii \right)$.
From equation (i), we can write $\alpha =-\dfrac{b}{4a}$, so on putting the value of $\alpha$ as $-\dfrac{b}{4a}$ in equation (ii), we get
$\Rightarrow 3{{\left( -\dfrac{b}{4a} \right)}^{2}}=\dfrac{c}{a}$.
$\Rightarrow \dfrac{3{{b}^{2}}}{16{{a}^{2}}}=\dfrac{c}{a}$.
On cross multiplying, we get,
$\Rightarrow 3a{{b}^{2}}=16{{a}^{2}}c$.
Canceling the similar terms on both the sides, we get,
$\Rightarrow 3{{b}^{2}}=16ac$.
Thus option (c) is the correct answer.

Note: We should not confuse between sum and product of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$. We can solve the problem by finding the roots of the given quadratic equation using $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and then taking one of the roots equal to the three times of the other root. Here we consider the roots as $\alpha$ and $3\alpha$ in order to reduce the calculation time. We can also take the roots as $\alpha$ and $\beta$ to solve the problem in which we need an extra step to get the required result.