SOLUTION: Solve by completing the square: w^2-2w-15=0

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Question 34537: Solve by completing the square:
w^2-2w-15=0

Answer by lyra(94) 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 1w%5E2%2B-2w%2B-15=0 to standard form by dividing both sides by 1:
We have: 1w%5E2%2B-2w%2B-15=0. What we want to do now is to change this equation to a complete square %28w%2Bsomenumber%29%5E2+%2B+othernumber. How can we find out values of somenumber and othernumber that would make it work?
Look at %28w%2Bsomenumber%29%5E2: %28w%2Bsomenumber%29%5E2+=+w%5E2%2B2%2Asomenumber%2Ax+%2B+somenumber%5E2. Since the coefficient in our equation 1w%5E2%2Bhighlight_red%28+-2%29+%2A+w%2B-15=0 that goes in front of w is -2, we know that -2=2*somenumber, or somenumber+=+-2%2F2. So, we know that our equation can be rewritten as %28w%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 %28w%2B-2%2F2%29%5E2+%2B+othernumber is equivalent to our original equation 1w%5E2%2B-2w%2Bhighlight_green%28+-15+%29=0.


The highlighted red part must be equal to -15 (highlighted green part).

-2%5E2%2F4+%2B+othernumber+=+-15, or othernumber+=+-15--2%5E2%2F4+=+-16.
So, the equation converts to %28w%2B-2%2F2%29%5E2+%2B+-16+=+0, or %28w%2B-2%2F2%29%5E2+=+16.

Our equation converted to a square %28w%2B-2%2F2%29%5E2, equated to a number (16).

Since the right part 16 is greater than zero, there are two solutions:

system%28+%28w%2B-2%2F2%29+=+%2Bsqrt%28+16+%29%2C+%28w%2B-2%2F2%29+=+-sqrt%28+16+%29+%29
, or

system%28+%28w%2B-2%2F2%29+=+4%2C+%28w%2B-2%2F2%29+=+-4+%29
system%28+w%2B-2%2F2+=+4%2C+w%2B-2%2F2+=+-4+%29
system%28+w+=+4--2%2F2%2C+w+=+-4--2%2F2+%29

system%28+w+=+5%2C+w+=+-3+%29
Answer: w=5, -3.

Hope this helps,
lyra