SOLUTION: Solve for all real values of X: Absolute value(x^2+4x-3)=2

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Solve for all real values of X: Absolute value(x^2+4x-3)=2      Log On


   

Question 881178: Solve for all real values of X: Absolute value(x^2+4x-3)=2
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
Either x%5E2%2B4x-3=2 Or -x%5E2-4x%2B3=2.

First case:
x%5E2%2B4x-1=0
x=%28-4%2B-+sqr%2820%29%29%2F2
x=%28-4%2B-+2%2Asqrt%285%29%29%2F2
highlight%28x=-2%2B-+sqrt%285%29%29

Second case:
-x%5E2-4x%2B3-2=0
-x%5E2-4x%2B1=0
x%5E2%2B4x-1=0
x=%28-4%2B-+sqrt%2820%29%29%2F2
Same as the first case.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

Solve for all real values of X: Absolute value(x^2+4x-3)=2


highlight_green%28x+=+-+5%29 OR highlight_green%28x+=+1%29 OR highlight_green%28x+=+-+2+%2B+sqrt%285%29%29 OR highlight_green%28x+=+-+2+-+sqrt%285%29%29

You can do the check!! 



If you need a complete and detailed solution, let me know!!



Send comments, “thank-yous,” and inquiries to “D” at MathMadEzy@aol.com.

Further help is available, online or in-person, for a fee, obviously.