Question

Which of the following equations does not have solutions that are integers? A \(x^{2}+21 x+100=-10\) (B) \(x^{2}-169=0\) C \(x^{2}-8 x-105=0\) (D) \(x^{2}-15 x-75=0\)

Short answer

The equation which does not have solutions that are integers is (D) \(x^{2}-15 x-75=0\)

Step-by-step solution

Solve Option A

In \(x^{2}+21x+100=-10\), we need to convert this to a standard format. So it becomes \(x^{2}+21x+110=0\). The discriminant \(b^{2} - 4ac\) is \(21^{2} - 4*1*110 = 441 - 440 = 1\). The square root of 1 is 1, an integer, thus the solutions are integers.

Solve Option B

In \(x^{2}-169=0\), the discriminant \(b^{2} - 4ac\) is \(0^{2} - 4*1*(-169) = 676\). The square root of 676 is 26, which is an integer. Thus, the roots of the equation are integers.

Solve Option C

In \(x^{2}-8x-105=0\), the discriminant \(b^{2} - 4ac\) is \((-8)^{2} - 4*1*(-105) = 64 + 420 = 484\). The square root of 484 is 22, an integer. Thus, the equation yields integer solutions.

Solve Option D

In \(x^{2}-15x-75=0\), the discriminant \(b^{2} - 4ac\) is \((-15)^{2} - 4*1*(-75) = 225 + 300 = 525\). The square root of 525 is not an integer, 22.91287847, we find that the equation does not yield integer solutions.