If one of the root of the equation $4{x}^{2}-15x+4p=0$ is the square of the other then the value of $p$ is

Let the roots of the quadratic equation 4x^2 - 15x + 4p = 0 be r and r^2. Using Vieta's formulas, the sum of the roots is r + r^2 = 15/4 and the product of the roots is r * r^2 = r^3 = 4p/4 = p. From the sum equation, 4r^2 + 4r - 15 = 0, which factors to (2r - 3)(2r + 5) = 0, giving r = 3/2 or r = -5/2. Substituting r = 3/2 into the product formula gives p = (3/2)^3 = 27/8. Substituting r = -5/2 into the product formula gives p = (-5/2)^3 = -125/64. The values of p are 27/8 and -125/64.

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