SOLUTION: The area of the rectangle shown is 49in.^2 find the length and the width of the rectangle . With the length being x+6 and width 2x+5

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Question 693837: The area of the rectangle shown is 49in.^2 find the length and the width of the rectangle . With the length being x+6 and width 2x+5
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
AreaOfRectangle+=+Length+%2A+Width
49+=+%28x%2B6%29+%2A+%282x%2B5%29
49+=+2x%5E2+%2B+17x+%2B+30
0+=+2x%5E2+%2B+17x+-+19
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B17x%2B-19+=+0) has the following solutons:

x%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=%2817%29%5E2-4%2A2%2A-19=441.

Discriminant d=441 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-17%2B-sqrt%28+441+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2817%29%2Bsqrt%28+441+%29%29%2F2%5C2+=+1
x%5B2%5D+=+%28-%2817%29-sqrt%28+441+%29%29%2F2%5C2+=+-9.5

Quadratic expression 2x%5E2%2B17x%2B-19 can be factored:
2x%5E2%2B17x%2B-19+=+2%28x-1%29%2A%28x--9.5%29
Again, the answer is: 1, -9.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B17%2Ax%2B-19+%29

So x = 1.
Which makes the Length 7 and the width 7. It's a square