Question

Find the value of p for which the quadratic equations x(x-4)+p=0 has real roots

Answer 1

x(x - 4) + p = 0 can be rewritten as x^2 - 4x + p = 0, or (x^2 - 4x + 4) - 4 + p = 0.

x^2 - 4x + 4 = (x - 2)^2

Thus, (x - 2)^2 = 4 - p. This equation yields real solutions if and only if 4 - p >= 0 (since the square of a real number is nonnegative), so p <= 4.