SOLUTION: use completing the square to solve 2x^2-4x-5=0 I am so rusty but i am sure i will remember once it starts to look familiar again

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: use completing the square to solve 2x^2-4x-5=0 I am so rusty but i am sure i will remember once it starts to look familiar again      Log On


   



Question 904332: use completing the square to solve 2x^2-4x-5=0 I am so rusty but i am sure i will remember once it starts to look familiar again
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: COMPLETING THE SQUARE solver for quadratics
Read this lesson on completing the square by prince_abubu, if you do not know how to complete the square.
Let's convert 2x%5E2%2B-4x%2B-5=0 to standard form by dividing both sides by 2:
We have: 1x%5E2%2B-2x%2B-2.5=0. What we want to do now is to change this equation to a complete square %28x%2Bsomenumber%29%5E2+%2B+othernumber. How can we find out values of somenumber and othernumber that would make it work?
Look at %28x%2Bsomenumber%29%5E2: %28x%2Bsomenumber%29%5E2+=+x%5E2%2B2%2Asomenumber%2Ax+%2B+somenumber%5E2. Since the coefficient in our equation 1x%5E2%2Bhighlight_red%28+-2%29+%2A+x%2B-2.5=0 that goes in front of x is -2, we know that -2=2*somenumber, or somenumber+=+-2%2F2. So, we know that our equation can be rewritten as %28x%2B-2%2F2%29%5E2+%2B+othernumber, and we do not yet know the other number.
We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that %28x%2B-2%2F2%29%5E2+%2B+othernumber is equivalent to our original equation 1x%5E2%2B-2x%2Bhighlight_green%28+-2.5+%29=0.


The highlighted red part must be equal to -2.5 (highlighted green part).

-2%5E2%2F4+%2B+othernumber+=+-2.5, or othernumber+=+-2.5--2%5E2%2F4+=+-3.5.
So, the equation converts to %28x%2B-2%2F2%29%5E2+%2B+-3.5+=+0, or %28x%2B-2%2F2%29%5E2+=+3.5.

Our equation converted to a square %28x%2B-2%2F2%29%5E2, equated to a number (3.5).

Since the right part 3.5 is greater than zero, there are two solutions:

system%28+%28x%2B-2%2F2%29+=+%2Bsqrt%28+3.5+%29%2C+%28x%2B-2%2F2%29+=+-sqrt%28+3.5+%29+%29
, or

system%28+%28x%2B-2%2F2%29+=+1.87082869338697%2C+%28x%2B-2%2F2%29+=+-1.87082869338697+%29
system%28+x%2B-2%2F2+=+1.87082869338697%2C+x%2B-2%2F2+=+-1.87082869338697+%29
system%28+x+=+1.87082869338697--2%2F2%2C+x+=+-1.87082869338697--2%2F2+%29

system%28+x+=+2.87082869338697%2C+x+=+-0.870828693386971+%29
Answer: x=2.87082869338697, -0.870828693386971.