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Which function has exactly 1 real solution?

A. f(x)=2x²+4x-5
B. 6x²+11
C. -4x²+9x
D. -3x²+30x-75

1 Answer

5 votes

Final answer:

A quadratic function has exactly 1 real solution when the discriminant is equal to zero. Option B is the correct answer.

Step-by-step explanation:

A quadratic function has exactly 1 real solution when the discriminant of the quadratic equation is equal to zero. The discriminant is found by using the formula: Δ = b^2 - 4ac where a, b, and c are the coefficients of the quadratic equation. If the discriminant is equal to zero, then the function has exactly 1 real solution. Let's calculate the discriminant for each option: A. Δ = (4^2) - (4*2*(-5)) = 16 + 40 = 56 (has 2 real solutions). B. Δ = 0 (has 1 real solution). C. Δ = (9^2) - (4*(-4)*(9)) = 81 + 144 = 225 (has 2 real solutions). D. Δ = (30^2) - (4*(-3)*(-75)) = 900 - 900 = 0 (has 1 real solution). Therefore, option B has exactly 1 real solution.

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