SOLUTION: Solve the following equations algebraically. x3/2=27 3/4=1-3x-2/x+1

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Question 90867: Solve the following equations algebraically.
x3/2=27
3/4=1-3x-2/x+1
Answer by malakumar_kos@yahoo.com(315) About Me  (Show Source):
You can put this solution on YOUR website!

Solve the following equations algebraically.
x3/2=27
3/4=1-3x-2/x+1

1) x^3/2 = 27
square both sides
(x^3/2)^2 = (27)^2

(x)^3 = (3.3.3)^2
(x)^3 = (3^3)^2
(x)^3 = (3)^6
(x)^3 = (3^2)^3
(x)^3 = (9)^3 therefore x = 9

2) 3/4 = 1-3x-2/(x+1)
3x+2/(x+1) = 1-3/4
3x(x+1)+2/(x+1) = 4-3/4
3x^2+3x+2/(x+1) = 1/4
4(3x^2+3x+2) = 1(x+1)
12x^2+12x+8 = x+1
12x^2+12x-x+8-1 = 0
12x^2+11x+7 = 0 since it cannot be factorised use the formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ where a = 12, b= 11 , c = 7
x = [-11+-sq rt (11^2-4.12.7)]/2.12
= [-11+- sq rt (121-336)]/24
= [-11+-Sq rt(-215)]/24 is the solution