SOLUTION: How do you solve 4x2=96, x2-20=0, and 5(x-2)2=125

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: How do you solve 4x2=96, x2-20=0, and 5(x-2)2=125      Log On


   

Question 60104: How do you solve 4x2=96, x2-20=0, and 5(x-2)2=125
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
How do you solve
4x%5E2=96
4x%5E2%2F4=96%2F4
x%5E2=24
sqrt%28x%5E2%29=%2B-sqrt%2824%29
x=+\-sqrt%2824%29
x=+\-sqrt%284%29%2Asqrt%286%29
x=+\-2%2Asqrt%286%29
:
x%5E2-20=0
x%5E2=20
sqrt%28x%5E2%29=%2B-sqrt%2820%29
x=+\-sqrt%284%29%2Asqrt%285%29
x=+\-2%2Asqrt%285%29
:
5%28x-2%29%5E2=125
5%28x-2%29%5E2%2F5=125%2F5
%28x-2%29%5E2=25
sqrt%28%28x-2%29%5E2%29=%2B-sqrt%2825%29
x-2=+\-5
x=2+\-5
x=2-5 and x=2+5
x=-3 and x=7
Happy Calculating!!!