SOLUTION: the perimeter is 47, one length is x+4 another one is x+5 the other 3x-2 and the last x^2-2x

Question 433415: the perimeter is 47, one length is x+4 another one is x+5 the other 3x-2 and the last x^2-2x
Answer by jorel1380(3719) (Show Source): You can put this solution on YOUR website!
x2-2x+3x-2+x+5+x+4=47
x2+3x+7=47
x2+3x-40=0
(x+8)(x-5)=0
x=-8,5
Throwing out the negative answer, we get x=5.
The sides are 9,10,13,15.

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=169 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 5, -8. Here's your graph:

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