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

Verified
147.9k+ views
Hint: Quadratic equations are the equation that contains at least one squared variable which is equal to zero. Quadratic equations are useful in our daily life; they are used to calculate areas, speed of the objects, projection, etc.
Quadratic equation is given as $a{x^2} + bx + c = 0$. This is the basic equation which contains a squared variable $x$ and three constants a, b and c. The value of the $x$ in the equation which makes the equation true is known as the roots of the equation. The numbers of roots in the quadratic equations are two as the highest power on the variable of the equation is x. The roots of the equation are given by the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where ${b^2} - 4ac$ tells the nature of the solution.
In the quadratic equation, $a{x^2} + bx + c = 0$the sum of the roots is given by $- \dfrac{b}{a}$ whereas their products are given by $\dfrac{c}{a}$.
In this question, it is already mentioned that one of the roots is the square of the other and so we need to carry out the calculation by taking only variable for the root of the equation $4{x^2} - 15x + 4p = 0$ and determining the relation between the roots and p.

Complete step by step solution: Let one of the roots of the equation $4{x^2} - 15x + 4p = 0$ be $m$ then, according to question the other root will be ${m^2}$.
Now, following the property of the quadratic equation that the product of the roots is equal to the ratio of the coefficient of ${x^0}$ and the coefficient of ${x^2}$.
Here, the coefficient of ${x^0}$is 4p and the coefficient of ${x^2}$ is 4.
Hence,
$m \times {m^2} = \dfrac{{4p}}{4} \\ {m^3} = p - - - - (i) \\$
Also, the sum of the roots of the quadratic equation is the negation of the ratio of the coefficient of $x$ and the coefficient of ${x^2}$.
Here, the coefficient of $x$is -15 and the coefficient of ${x^2}$ is 4.
Hence,
$m + {m^2} = - \left( {\dfrac{{ - 15}}{4}} \right) \\ {m^2} + m - \dfrac{{15}}{4} = 0 \\ 4{m^2} + 4m - 15 = 0 \\ m = \dfrac{{ - 4 \pm \sqrt {{4^2} - 4(4)( - 15)} }}{{2(4)}} \\ = \dfrac{{ - 4 \pm \sqrt {16 + 240} }}{8} \\ = \dfrac{{ - 4 \pm 16}}{8} \\ = - \dfrac{5}{2},\dfrac{3}{2} - - - - (ii) \\$
By equation (i) and (ii) 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 $\dfrac{{27}}{8}$.
Option C and D are correct.

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