• Oct 14th 2009, 01:56 PM
isundae
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
• Oct 14th 2009, 02:13 PM
skeeter
Quote:

Originally Posted by isundae
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

$\displaystyle (x+4)^{1/2} + 5x(x+4)^{3/2} = 0$

factor ...

$\displaystyle (x+4)^{1/2}[1 + 5x(x+4)] = 0$

$\displaystyle (x+4)^{1/2}(5x^2 + 20x + 1) = 0$

set each factor equal to 0 and solve ...

$\displaystyle x = -4$

$\displaystyle x = \frac{-10 \pm \sqrt{95}}{5}$
• Oct 15th 2009, 12:43 AM
pacman
(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)