SOLUTION: Please help! It asks to solve by completing the square, the problem is: x^2-x-1=0, please and Thank you=)

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Please help! It asks to solve by completing the square, the problem is: x^2-x-1=0, please and Thank you=)      Log On


   

Question 131148: Please help! It asks to solve by completing the square, the problem is:
x^2-x-1=0, please and Thank you=)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=1+x%5E2-1+x-1 Start with the given equation



y%2B1=1+x%5E2-1+x Add 1 to both sides



y%2B1=1%28x%5E2-1x%29 Factor out the leading coefficient 1



Take half of the x coefficient -1 to get -1%2F2 (ie %281%2F2%29%28-1%29=-1%2F2).


Now square -1%2F2 to get 1%2F4 (ie %28-1%2F2%29%5E2=%28-1%2F2%29%28-1%2F2%29=1%2F4)





y%2B1=1%28x%5E2-1x%2B1%2F4-1%2F4%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 1%2F4 does not change the equation




y%2B1=1%28%28x-1%2F2%29%5E2-1%2F4%29 Now factor x%5E2-1x%2B1%2F4 to get %28x-1%2F2%29%5E2



y%2B1=1%28x-1%2F2%29%5E2-1%281%2F4%29 Distribute



y%2B1=1%28x-1%2F2%29%5E2-1%2F4 Multiply



y=1%28x-1%2F2%29%5E2-1%2F4-1 Now add %2B1 to both sides to isolate y



y=1%28x-1%2F2%29%5E2-5%2F4 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=1%2F2, and k=-5%2F4. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2-1x-1 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2-1x-1%29 Graph of y=1x%5E2-1x-1. Notice how the vertex is (1%2F2,-5%2F4).



Notice if we graph the final equation y=1%28x-1%2F2%29%5E2-5%2F4 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x-1%2F2%29%5E2-5%2F4%29 Graph of y=1%28x-1%2F2%29%5E2-5%2F4. Notice how the vertex is also (1%2F2,-5%2F4).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.