Question

Find all values of k for which the equation has one real solution.

2p2
+9p–(k–4)=0
Write your answer starting with k, followed by an equals sign or inequality symbol (for example, k<5). Reduce all fractions.

Answer 1

To find the values of k for which the equation has one real solution, we can examine the discriminant of the quadratic equation.

The discriminant is given by the formula: Δ = b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0.

In this case, the quadratic equation is 2p^2 + 9p - (k - 4) = 0, so a = 2, b = 9, and c = -(k - 4).

The discriminant is:
Δ = (9)^2 - 4(2)(-(k - 4))
= 81 + 8(k - 4)
= 81 + 8k - 32
= 8k + 49.

For the equation to have one real solution, the discriminant Δ must be equal to zero, since a positive discriminant indicates two distinct real solutions, and a negative discriminant indicates two complex solutions.

Setting the discriminant equal to zero and solving for k:
8k + 49 = 0
8k = -49
k = -49/8

Therefore, the value of k for which the equation has one real solution is k = -49/8.