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 any help appreciated
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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 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 ... $\displaystyle x = -4$ $\displaystyle x = \frac{-10 \pm \sqrt{95}}{5}$
(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,
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