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Question: If one of the roots of the equation \(4{x^2} - 15x + 4p = 0\) is the square of the other then, the v...

If one of the roots of the equation 4x215x+4p=04{x^2} - 15x + 4p = 0 is the square of the other then, the value of pp is
A.12564 B.278 C.1258 D.278  A.\dfrac{{125}}{{64}} \\\ B.\dfrac{{ - 27}}{8} \\\ C.\dfrac{{ - 125}}{8} \\\ D.\dfrac{{27}}{8} \\\

Explanation

Solution

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 ax2+bx+c=0a{x^2} + bx + c = 0. This is the basic equation which contains a squared variable xx and three constants a, b and c. The value of the xx 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=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}, where b24ac{b^2} - 4ac tells the nature of the solution.
In the quadratic equation, ax2+bx+c=0a{x^2} + bx + c = 0the sum of the roots is given by ba - \dfrac{b}{a} whereas their products are given by ca\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 4x215x+4p=04{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 4x215x+4p=04{x^2} - 15x + 4p = 0 be mm then, according to question the other root will be m2{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 x0{x^0} and the coefficient of x2{x^2}.
Here, the coefficient of x0{x^0}is 4p and the coefficient of x2{x^2} is 4.
Hence,
m×m2=4p4 m3=p(i)  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 xx and the coefficient of x2{x^2}.
Here, the coefficient of xxis -15 and the coefficient of x2{x^2} is 4.
Hence,

m+m2=(154) m2+m154=0 4m2+4m15=0 m=4±424(4)(15)2(4) =4±16+2408 =4±168 =52,32(ii)  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=52m = \dfrac{{ - 5}}{2}; p=(52)3=1258p = {\left( {\dfrac{{ - 5}}{2}} \right)^3} = \dfrac{{ - 125}}{8}
For m=32m = \dfrac{3}{2} ; p=(32)3=278p = {\left( {\dfrac{3}{2}} \right)^3} = \dfrac{{27}}{8}
Hence, the value of p can either be 1258\dfrac{{ - 125}}{8} or 278\dfrac{{27}}{8}.
Option C and D are correct.

Note: In the quadratic equation if b24ac>0{b^2} - 4ac > 0the equation will have two real roots. If it is equal b24ac=0{b^2} - 4ac = 0 then the equation will have only one real root and when b24ac<0{b^2} - 4ac < 0 then the root is in complex form.