Posted by Algebra on Wednesday, May 27, 2009 at 9:35am.
you could try graphing.
change to 3√x = x+6
then split it into
y = 3√x and y = x+6
a few points on the graph would be (0,0), (1,3), (4,6) and (9,9)
the second is of course a straight line with slope of 1 and y-intercept of 6
it is immediately obvious that the two graphs cannot intersect, so your equation has no real solution.
The only other option is to square both sides of 3√x = x+6 ,
to obtain the quadratic equation x^2 + 3x + 36 = 0
which only has complex roots as shown by our graph above.
I am not sure if this helps.
If we rearrange and square each side we end up with
8x^2-12x-36=0
and by inspection we see that one solution is x=3
to find the other divide
8x^2-12x-36 by x-3
which gives
8x-12=0 as the other solution
but check my maths!!
if you want a "proof" and development of the quadratic equation, it is obtained by completing the square on
Ax + By + C = 0 as shown in
http://mathworld.wolfram.com/QuadraticEquation.html
(BTW, I notice that DrRuss has a small error in his solution.
when you square 3√x you get 9x not 9x^2)
OOOPS That what comes of solving things quickly!!
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