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Which function has exactly one real solution?

A. [tex]\( f(x) = 6x^2 + 11 \)[/tex]
B. [tex]\( f(x) = -4x^2 + 9x \)[/tex]
C. [tex]\( f(x) = -3x^2 + 30x - 75 \)[/tex]
D. [tex]\( f(x) = 2x^2 + 4x - 5 \)[/tex]

Sagot :

To determine which function has exactly one real solution, we should find the discriminant of each quadratic function. The discriminant, [tex]\(\Delta\)[/tex], for a general quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:

[tex]\[ \Delta = b^2 - 4ac \][/tex]

A quadratic equation has:
- Two real solutions if [tex]\(\Delta > 0\)[/tex]
- One real solution if [tex]\(\Delta = 0\)[/tex]
- No real solutions if [tex]\(\Delta < 0\)[/tex]

Let's analyze the given functions by their discriminants:

### Function A: [tex]\(f(x) = 6x^2 + 11\)[/tex]
The discriminant is:
[tex]\[ \Delta_1 = 11^2 - 4 \cdot 6 \cdot 0 = 121 - 0 = 121 \][/tex]
Since [tex]\(\Delta_1 \ne 0\)[/tex], function A does not have exactly one real solution.

### Function B: [tex]\(f(x) = -4x^2 + 9x\)[/tex]
The discriminant is:
[tex]\[ \Delta_2 = 9^2 - 4 \cdot (-4) \cdot 0 = 81 - 0 = 81 \][/tex]
Since [tex]\(\Delta_2 \ne 0\)[/tex], function B does not have exactly one real solution.

### Function C: [tex]\(f(x) = -3x^2 + 30x - 75\)[/tex]
The discriminant is:
[tex]\[ \Delta_3 = 30^2 - 4 \cdot (-3) \cdot (-75) = 900 - 900 = 0 \][/tex]
Since [tex]\(\Delta_3 = 0\)[/tex], function C has exactly one real solution.

### Function D: [tex]\(f(x) = 2x^2 + 4x - 5\)[/tex]
The discriminant is:
[tex]\[ \Delta_4 = 4^2 - 4 \cdot 2 \cdot (-5) = 16 + 40 = 56 \][/tex]
Since [tex]\(\Delta_4 \ne 0\)[/tex], function D does not have exactly one real solution.

Given these discriminant values, function [tex]\(f(x) = -3x^2 + 30x - 75\)[/tex] (Function C) is the one that has exactly one real solution as its discriminant is [tex]\(0\)[/tex].

Therefore, the correct answer is:
C. [tex]\(f(x) = -3 x^2 + 30 x - 75\)[/tex]