SOLUTION: the base and height of a triangle are (x+3)cm and (2x+5)cm respectively if the area of a triangle is 20 cm square , find x

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Question 709684: the base and height of a triangle are (x+3)cm and (2x+5)cm respectively if the area of a triangle is 20 cm square , find x
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
the base and height of a triangle are (x+3)cm and (2x+5)cm respectively if the area of a triangle is 20 cm square , find x
.
Since area of a triangle is:
(1/2)bh = area
plugging in the given data:
(1/2)(x+3)(2x+5) = 20
multiplying both sides by 2:
(x+3)(2x+5) = 40
FOIL the left:
2x^2+5x+6x+15 = 40
2x^2+11x+15 = 40
2x^2+11x-25 = 0
applying the "quadratic equation" we get:
x = {1.73, -7.23}
.
details of quadratic follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B11x%2B-25+=+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=%2811%29%5E2-4%2A2%2A-25=321.

Discriminant d=321 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-11%2B-sqrt%28+321+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2811%29%2Bsqrt%28+321+%29%29%2F2%5C2+=+1.72911821679223
x%5B2%5D+=+%28-%2811%29-sqrt%28+321+%29%29%2F2%5C2+=+-7.22911821679223

Quadratic expression 2x%5E2%2B11x%2B-25 can be factored:
2x%5E2%2B11x%2B-25+=+2%28x-1.72911821679223%29%2A%28x--7.22911821679223%29
Again, the answer is: 1.72911821679223, -7.22911821679223. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B11%2Ax%2B-25+%29