Question

7. Given that 2x² + 4x + 3 = k where k is a constant, find the range of values for k for which the equation has no real roots. (b) The function f is defined on the set R of real numbers by f(x) →kx² + kx + 1, where k is a member of R. Calculate the range of values of k for which the equation f(x) = 0 hasreal roots. Given that k = -2, solve the equation f(x) = 0. 1/2​

Answer 1

Answer: (a) The equation 2x^2 + 4x + 3 = k has real roots if and only if its discriminant, which is 4^2 - 4(2)(3), is nonnegative. The discriminant is equal to 0, so the equation has real roots if and only if k = 3.

(b) For the equation f(x) = kx^2 + kx + 1 to have real roots, its discriminant, which is 1 - 4k, must be nonnegative. This gives us k >= 1/4. The range of values of k for which the equation has real roots is k >= 1/4.

(c) When k = -2, the equation f(x) = -2x^2 - 2x + 1 = 0. Solving this equation using the quadratic formula, we get x = (-1 ± √(1 + 8))/4. So, the roots are x = (-1 ± √(9))/4 = (-1 ± 3)/4 = 1/2, -2.