directions say to find all solutions

1. (x + 4)^1\2 + 5x(x + 4)^3/2 = 0

I don't know how to get rid of the fractions when there are two of them (Wondering)

any help appreciated

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- October 14th 2009, 01:56 PMisundaesolving radical equations
directions say to find all solutions

1. (x + 4)^1\2 + 5x(x + 4)^3/2 = 0

I don't know how to get rid of the fractions when there are two of them (Wondering)

any help appreciated - October 14th 2009, 02:13 PMskeeter
- October 15th 2009, 12:43 AMpacman
(x + 4)^(1/2) + 5x(x + 4)^(1+1/2) = 0, by factoring (x + 4)^(1/2).

(x + 4)^(1/2)(1 + (5x)(x + 4)) = 0,

(x + 4)^(1/2)(1 + 20x + 5x^2) = 0

THESE are the 3 roots:

x_1 = -4,

x_2 = -2 - sqrt (19/5) = -3.95

x_3 = sqrt (19/5) - 2 = -0.05

see graph, (Bow)