Question 87940: solve equation, by making appropriate substitution.
a)x^-2+x^-1-56=0
U=
U^2=
b)x^2/3-x^1/3-12=0
x=
x^2=
Answer by madhujiithcharu@yahoo.co.in(28)   (Show Source):
You can put this solution on YOUR website! a) x^-2 + x^-1 - 56=0 => 1/x^2 + 1/x -56 => -56x^2 + x + 1 = 0
i.e, -56x^2 -7x+8x+1 = 0 => -7x(8x+1)+(8x+1)=0 => (8x+1)(-7x+1)=0 => x=-1/8, x=1/7 => x^2=1/64, x^2=1/49
b)x^2/3-x^1/3-12=0 => (x^1/3)^2-x^1/3-12=0
Put x^1/3 = y. Therefore the above equation becomes y^2-y-12 = 0
i.e, y^2+3y-4y-12 =0 => y(y+3)-4(y+3)=0 => (y+3)(y-4)=0 => y=-3,y=4
Therefore x^1/3=-3 & x^1/3=4 => x=-27, x=64
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