# 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

Algebra ->  Algebra  -> Radicals -> 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       Log On

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Question 190741: This question is from a homework that was handed out in class.
Thanks
Solve for x
(x+4)^1/2 + 5x(x+4)^3/2 =0

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(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}
.