Question

Which of the following equations have real solutions? I. \(6 x^{2}+x-12=0\) II. \(6 x^{2}+7 x-12=0\) III. \(6 x^{2}+7 x-13=0\) A. I only B. II only C. III only D. All

Short answer

All of the given equations have real solutions. So, the answer is (D) All.

Step-by-step solution

Calculate discriminant for Equation I

Substitute a=6, b=1, and c=-12 into the discriminant equation \( b^2 - 4ac \) which gives \( 1^2 - 4*6*(-12) = 289 \) which is greater than 0, indicating two different real roots.

Calculate discriminant for Equation II

Substitute a=6, b=7, and c=-12 into the discriminant equation \( b^2 - 4ac \) which gives \( 7^2 - 4*6*(-12) = 337 \) which is greater than 0, indicating two different real roots.

Calculate discriminant for Equation III

Substitute a=6, b=7, and c=-13 into the discriminant equation \( b^2 - 4ac \) which gives \( 7^2 - 4*6*(-13) = 361 \) which is greater than 0, indicating two different real roots.