SOLUTION: okay this problem says to solve by completing the square, but the odd numbers are throwing me off, and this problem is not to be graphed, so no problem on here really helped me, th

Algebra ->  Systems-of-equations -> SOLUTION: okay this problem says to solve by completing the square, but the odd numbers are throwing me off, and this problem is not to be graphed, so no problem on here really helped me, th      Log On


   

Question 128955: okay this problem says to solve by completing the square, but the odd numbers are throwing me off, and this problem is not to be graphed, so no problem on here really helped me, the problem is:

1x(1x)+11x-6=0
any help on how to solve it would be greatly appreciated, this is for a college algebra course
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
I think you mean x%5E2%2B11x-6=0

Step 1: Add the additive inverse of the constant term to both sides. For this problem, add 6 to both sides:

x%5E2%2B11x=6

Step 2: Divide through by the coefficient on the x%5E2 term. For this problem, that is 1, so you don't have to do anything.

Step 3: Divide the coefficient on the x term by 2. For this problem, that becomes 11%2F2.

Step 4: Square the result of step 3. For this problem: 121%2F4

Step 5: Add the result of step 4 to both sides of the equation. For this problem:

x%5E2%2B11x%2B%28121%2F4%29=6%2B%28121%2F4%29

x%5E2%2B11x%2B%28121%2F4%29=%2824%2B121%29%2F4

x%5E2%2B11x%2B%28121%2F4%29=145%2F4

Step 6: Factor the left:

%28x%2B%2811%2F2%29%29%5E2=145%2F4

Step 7: Take the square root of both sides:

x%2B%2811%2F2%29=sqrt%28145%2F4%29 or x%2B11%2F2=-sqrt%28145%2F4%29

x=%28-11%2Bsqrt%28145%29%29%2F2 or x=%28-11-sqrt%28145%29%29%2F2

Yes, these numbers are a horror, but certainly prettier than what you generally encounter in real life situations. By the way, sqrt%28145%29 is reduced to simplest terms because there are no perfect square factors of 145 -- the prime factorization is 5 * 29.