Hi! I have a question about how to solve an equation with fractions in the exponents. The question is 2x^(1/2)-9x^(1/4)+2=0. I know the answer is (9+/- sqrt(65) /4)^4. I believe I'm supposed to get rid of the 2 at the beginning by dividing the rest of the equation by 2, but I'm not sure where to go from there.
Fractions in the exponents: 2x^(1/2)-9x^(1/4)+2=0
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I'll guess that there were instructions that came along with the equation, and that they said something like "solve" or "solve and check".
This is an equation in quadratic form, but not actually a quadratic. (here) The trick is to notice the form, and figure out how to apply quadratic methods. Since the answer looks to be in the form of something you'd get from the Quadratic Formula, this suggests that we're on the right track.
In your case, the power on the middle term's variable is half of the power on the first term's variable: 1/4 = (1/2)(1/2). Or, more usefully, 1/2 = 2(1/4). So you have:
. . . . .\(\displaystyle 2\,\left(\, (x^{1/4})\, \right)^2\, -\, 9\, \left(\, (x^{1/4})\, \right)\, +\, 2\, =\, 0\)
If they'd given you this:
. . . . .\(\displaystyle 2y^2\, -\, 9y\, +\, 2\, =\, 0\)
...you'd plug into the Quadratic Formula and solve for y. In your case, you'll want to plug into the Quadratic Formula and solve for x^{1/4}. Then back-solve (by raising things to the fourth power) for x.
