Math

Question12. Graph f(x)=x2+2f(x)=x^2+2. What is the solution set for x2+2=0x^2+2=0? A) {0}\{0\} B) {2}\{2\} C) {0,2}\{0,2\} D) No real solutions
13. Solve x2+4x+2=0x^2+4x+2=0. Round the greater solution to the nearest hundredth.
14. Solve 2x2+3x2=02x^2+3x-2=0. A) 12\frac{1}{2} and -2 B) 12-\frac{1}{2} and 2 C) 1 and -4 D) -1 and 4
15. What is a solution to 0=(x3)(x+4)0=(x-3)(x+4)?

Studdy Solution

STEP 1

Assumptions for Problem 12
1. The function given is f(x)=x2+2f(x) = x^2 + 2.
2. We are looking for the solution set of the equation x2+2=0x^2 + 2 = 0.

STEP 2

Solve the equation x2+2=0x^2 + 2 = 0 by isolating x2x^2.
x2=2x^2 = -2

STEP 3

Since x2x^2 is always non-negative for real numbers, there is no real number whose square is 2-2. Therefore, there are no real solutions to the equation.
The solution set for the equation x2+2=0x^2 + 2 = 0 is: J No real solutions

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