Question 3585: I need help
(x^2+2x)^2-5(x^2+2x)+6=0
and
x^1/2-11x^1/4+30
and
6x^3/5-24=0
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! (I) (x^2+2x)^2-5(x^2+2x)+6=0
Let y = x^2+2x, then the given equation is converted to
y^2 - 5y + 6 =0,
Factor: (y-2)(y-3) = 0
so, y= 2 or 3.
That is, x^2+2x = 2 or x^2+2x = 3,
Hence x^2+2x - 2 = 0,
By quadratic formula, x= -1 + sqrt(3) or -1 - sqrt(3)
Or x^2+2x - 3 = 0,
Factor: (x-1)(x+3) =0, so x = 1 or -3 [Four solutions of x]
(II) x^1/2-11x^1/4+30
Let u = x^1/4, then we have
u^2 - 11 u + 30 = 0,
Factor: (u-6)(u-5) =0, so u = 6 or 5.
When u=6 , x^1/4 = 6, x = 6^4 = 36^2 = 1296
When u=5 , x^1/4 = 5, x = 5^4 = 25^2 = 625.
(III) 6x^3/5-24=0
Cancel 6: x^3/5-4=0,
Or x^3/5 = 4
So, x = 4^(5/3) = 4 * 4^(2/3) = 4 * (2^4)^(1/3)
= 4*2 * 2^(1/3) = 8* 2^(1/3)
Try to check the above answers yourself.
Kenny
|
|
|
