SOLUTION: solve the quadratic equation x squared+4x-5=0 by completing squares. {{{x^2 + 4x - 5 = 0}}}

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: solve the quadratic equation x squared+4x-5=0 by completing squares. {{{x^2 + 4x - 5 = 0}}}      Log On


   

Question 4350: solve the quadratic equation x squared+4x-5=0 by completing squares.
x%5E2+%2B+4x+-+5+=+0
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Begin by adding 5 to each side to set up the "completing the square" process.

x%5E2+%2B+4x+-+5+=+0
x%5E2+%2B+4x++=++5
x%5E2+%2B+4x+%2B+_____+=+5+%2B+_____

Take half of the 4 and square, which is 2 squared, or 4. Add 4 to each side of the equation to create a perfect square trinomial on the left side.

x%5E2+%2B+4x+%2B+4+=+5+%2B+4
+%28x%2B2%29%5E2+=+9+

Take the square root of both sides:

x+%2B+2+=+0+%2B-+3

Subtract 2 from each side:
x+%2B2+-+2+=+-2+%2B-+3
x+=+-2+%2B-+3

In the first case, x = -2 + 3 = 1
In the second case, x = -2 - 3 = -5

Check answers: x%5E2+%2B+4x+-+5+=+0
For x = 1, 1%5E2+%2B+4%2A1+-+5+=+0
For x = -5, %28-5%29%5E2+%2B+4%2A%28-5%29+-+5+=+0

Both answers check!

R^2 at SCC