SOLUTION: find the real solutions. (2x+5)^2-(2+5)-6=0 thanks Caroline

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Question 5221: find the real solutions.
(2x+5)^2-(2+5)-6=0
thanks Caroline
Answer by ichudov(507) About Me  (Show Source):
You can put this solution on YOUR website!
I have a hunch that you forgot an 'x' and meant
%282x%2B5%29%5E2-%282x%2B5%29-6=0
use y instead of 2x+5, to get
y%5E2-y-6+=+0
solve it for y:

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ay%5E2%2Bby%2Bc=0 (in our case 1y%5E2%2B-1y%2B-6+=+0) has the following solutons:

y%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-1%29%5E2-4%2A1%2A-6=25.

Discriminant d=25 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+25+%29%29%2F2%5Ca.

y%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+25+%29%29%2F2%5C1+=+3
y%5B2%5D+=+%28-%28-1%29-sqrt%28+25+%29%29%2F2%5C1+=+-2

Quadratic expression 1y%5E2%2B-1y%2B-6 can be factored:
1y%5E2%2B-1y%2B-6+=+1%28y-3%29%2A%28y--2%29
Again, the answer is: 3, -2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B-6+%29


y=3, y=-2, or, knowing that y=2x+5, you have two equations
2x+5 = 3, or 2x=-2, or x = -1
2x+5 = -2, or 2x=-7, or x = -3.5.
Double check my answer please.
Answer: x=-1, x=-3.5