SOLUTION: This question is from a homework that was handed out in class. I don't understand the problem, please help me. Thanks Solve for x (x+4)^1/2 + 5x(x+4)^3/2 =0

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Question 190741: This question is from a homework that was handed out in class.
I don't understand the problem, please help me.
Thanks
Solve for x
(x+4)^1/2 + 5x(x+4)^3/2 =0

Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
(x+4)^1/2 + 5x(x+4)^3/2 =0
.
Begin by factoring out (x+4)^1/2:
(x+4)^1/2 [1 + 5x(x+4)^2/2] = 0
(x+4)^1/2 [1 + 5x(x+4)] = 0
(x+4)^1/2 [1 + 5x^2 + 20x] = 0
(x+4)^1/2 [5x^2 + 20x + 1] = 0
.
Now, the problem is reduced to finding values of 'x' when:
(x+4)^1/2 = 0
and
5x^2 + 20x + 1 = 0
.
Top one:
(x+4)^1/2 = 0
(x+4) = 0
x = -4
.
Bottom one:
5x^2 + 20x + 1 = 0
Applying the quadratic equation yields two solutions:
x = {-0.051, -3.949}
.
Combined solutions, then are:
x = {-0.051, -3.949, -4}
.
Details of the quadratic follows:

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=380 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -0.0506411310382074, -3.94935886896179. Here's your graph: