Question

If one root of the equation ${{x}^{2}}-30x+p=0$ is square of the other, then p is equal to.
a. Only 125
b. 125, -216
c. 125, 215
d. Only 216

Answer 1

Hint: We have been given information that one root is square of other, so we can consider one root as $\alpha $, so the other root will be ${{\alpha }^{2}}$. We will use the concept of the sum of roots and the product of roots to solve this question. For general equation, $a{{x}^{2}}+bx+c=0$, we get the sum of roots as $-\dfrac{b}{a}$ and the product of the root as $\dfrac{c}{a}$.

Complete step-by-step solution:
In the question, we have been given a quadratic equation, ${{x}^{2}}-30x+p=0$ which has a variable p whose value we are supposed to find out. Apart from this information, we have been given that one root of this equation is equal to the square of the other root.
So, let us consider root as $\alpha $, so the other root will become ${{\alpha }^{2}}$.
Now, we know that for a general quadratic equation, $a{{x}^{2}}+bx+c=0$, the sum of the roots is equal to $-\dfrac{b}{a}$ and the product of the root is equal to $\dfrac{c}{a}$. We will use this concept to solve our question.
So, in the equation ${{x}^{2}}-30x+p=0$, we have a = 1, b = -30 and c = p.
Thus, we can write the sum of the roots $\alpha \text{ and }{{\alpha }^{2}}$ as
$\alpha +{{\alpha }^{2}}=-\dfrac{b}{a}$
On putting a = 1 and b = -30, we get,
$\begin{align}
  & \alpha +{{\alpha }^{2}}=-\dfrac{-30}{1} \\
 & \alpha +{{\alpha }^{2}}=30 \\
\end{align}$
On taking 30 to the LHS, we get,
$\alpha +{{\alpha }^{2}}-30=0$
We will solve this equation by using the middle term split method, to get the values of $\alpha $.
$\begin{align}
  & {{\alpha }^{2}}+6\alpha -5\alpha -30=0 \\
 & \alpha \left( \alpha +6 \right)-5\left( \alpha +6 \right)=0 \\
 & \left( \alpha -5 \right)\left( \alpha +6 \right)=0 \\
 & \alpha =5,-6 \\
\end{align}$
So, we get the values of $\alpha $ as 5 and -6.
Now, we know that for the given quadratic equation, ${{x}^{2}}-30x+p=0$, we have a = 1, b = -30 and c = p.
Thus, the product of the roots $\alpha \text{ and }{{\alpha }^{2}}$ can be written as,
$\alpha \times {{\alpha }^{2}}=\dfrac{c}{a}$
On putting a = 1 and c = p, we get,
$\begin{align}
  & \alpha \times {{\alpha }^{2}}=\dfrac{p}{1} \\
 & {{\alpha }^{3}}=p \\
\end{align}$
Now, we will substitute the values of $\alpha $ as 5 and -6 in the above equation and get the values of p from it.
Taking $\alpha =5$, we get,
$p={{\left( 5 \right)}^{3}}=125$
Taking $\alpha =-6$, we get,
$p={{\left( -6 \right)}^{3}}=-216$
Therefore, we get the values of p as 125 and -216.
Hence, option (b) is the correct answer.

Note: In this question, we used the direct formula of the sum and product of the roots as $-\dfrac{b}{a}$ and $\dfrac{c}{a}$. We can also solve this question by using the formula of the root as $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. So, we will find the roots $\alpha \text{ and }{{\alpha }^{2}}$, and then on equating them, we will get the value of p.
Also, the students must take care not to make any calculation mistakes.