SOLUTION: How do I solve for x when the area of the rectangle is 30 and the length is 3x+1 and the width is x?

Question 346436: How do I solve for x when the area of the rectangle is 30 and the length is 3x+1 and the width is x?
Found 2 solutions by mananth, ewatrrr:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Area of rectangle = L*W
30=(3x+1)*x
30=3x^2+x
3x^2+x-30=0
3x^2+10x-9x-90=0
3x(x+10)-9(x+10)=0
(x+10)(3x-9)=0
3(x+10)(x-3)=0
x=3
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
Hi,
*Note: Area = length*width
The question states the following to be true:
.
(3x + 1)*x = 30
.

.

.
Use the quadratic formual to solve:

.

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=364 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 2.84646400472315, -3.51313067138982. Here's your graph:

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