Question

If one of the roots of the equation $4{x^2} - 15x + 4p = 0$ is the square of the other than the value of p is
A) $\dfrac{{125}}{{64}}$
B) $\dfrac{{ - 27}}{8}$
C) $\dfrac{{ - 125}}{8}$
D) $\dfrac{{27}}{8}$

Answer 1

Hint:
It is given in the question that one of the roots of the equation $4{x^2} - 15x + 4p = 0$ is the square of the other.
So, what is the value of p?
Let one of the roots of the equation $4{x^2} - 15x + 4p = 0$ be m.
Since, according to the question another root will be ${m^2}$ .
First, using the property of the quadratic equation that the product of roots is equal to $\dfrac{c}{a}$ , we get ${m^3} = p$ .
Then after, using formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ we will find the roots of the equation and finally we get the answer.

Complete step by step solution:
It is given in the question that one of the roots of the equation $4{x^2} - 15x + 4p = 0$ is the square of the other.
 So, what is the value of p?
Let one of the roots of the equation $4{x^2} - 15x + 4p = 0$ be m.
Since, according to question another root will be ${m^2}$ .
Now, using the property of quadratic equation that the product of roots is equal to $\dfrac{c}{a}$ ,
 $\therefore m \times {m^2} = \dfrac{{4p}}{4}$
 $\therefore {m^3} = p$ (I)
Also, sum of roots of the quadratic equation is $\dfrac{{ - b}}{a}$ ,
 $\therefore m + {m^2} = - \left( {\dfrac{{ - 15}}{4}} \right)$
 $\therefore m + {m^2} - \dfrac{{15}}{4} = 0$
Now, take the L.C.M, we get,
 $\therefore 4{m^2} + 4m - 15 = 0$
Using formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ we will find the roots of above equation,
 $\therefore m = \dfrac{{ - 4 \pm \sqrt {{4^2} - 4\left( 4 \right)\left( { - 15} \right)} }}{{2\left( 4 \right)}}$
 $\therefore m = \dfrac{{ - 4 \pm \sqrt {16 + 240} }}{8}$
 $\therefore m = \dfrac{{ - 4 \pm \sqrt {256} }}{8}$
 $\therefore m = \dfrac{{ - 4 \pm 16}}{8}$
 $\therefore m = - \dfrac{5}{2},\dfrac{3}{2}$ (II)
Now, put the value of m in equation (I), we get,
For $m = - \dfrac{5}{2}$ , $p = {\left( {\dfrac{{ - 5}}{2}} \right)^3} = \dfrac{{ - 125}}{8}$
For $m = \dfrac{3}{2}$ , $p = {\left( {\dfrac{3}{2}} \right)^3} = \dfrac{{27}}{8}$

Hence, the value of p can either be $\dfrac{{ - 125}}{8}$ or $\dfrac{{27}}{8}$ .

Note:
In the quadratic equation if ${b^2} - 4ac > 0$ then the equation will have two real roots.
In the quadratic equation if ${b^2} - 4ac = 0$ then the equation will have only one real root.
In the quadratic equation if ${b^2} - 4ac < 0$ then the roots are in complex form.