solve (9 - x^2)^(1/5) = x^2 + 1 for real x
- Thread starter bedi
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Re: solve: (9-x^2)^(1/5)=x^2+1
In the future please show your work so we know what point you're having trouble with.
Key points:
1. x^(1/5) means the 5th root of x raised to the 1 power.
2. Eliminating roots is as easy as multiplying both sides by that root.
3. Eliminating a power is as easy as taking the root of that power.
4. Any value (x) raised to the 1 power = x.
Begin by eliminating the root and simplifying the power of 1
9-x^2 = 5x^2 + 5
Subtract 5x^2 and 9 from both sides
-6x^2 = -4
Multiply by -1
6x^2 = 4
Divide by 6
x^2 = 2/3
Square to eliminate the power
x = 2/3^1/2
Test answer
9-(2/3) = 5(2/3) + 5
25/3 = 25/3
In the future please show your work so we know what point you're having trouble with.
Key points:
1. x^(1/5) means the 5th root of x raised to the 1 power.
2. Eliminating roots is as easy as multiplying both sides by that root.
3. Eliminating a power is as easy as taking the root of that power.
4. Any value (x) raised to the 1 power = x.
Begin by eliminating the root and simplifying the power of 1
9-x^2 = 5x^2 + 5
Subtract 5x^2 and 9 from both sides
-6x^2 = -4
Multiply by -1
6x^2 = 4
Divide by 6
x^2 = 2/3
Square to eliminate the power
x = 2/3^1/2
Test answer
9-(2/3) = 5(2/3) + 5
25/3 = 25/3
D
Deleted member 4993
Guest
bedi said:
I assume your problem is to solve for x in the following expression:
\(\displaystyle (9\, - \, x^2)^{\frac{1}{5}}\, =\, x^2\, - \, 1\)
What method have you been taught to solve for such equations? Have you been taught Numerical methods?
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- Feb 4, 2004
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One method might be to raise both sides to the fifth power:bedi said:
. . . . .[(9 - x[sup:2663nmi5]2[/sup:2663nmi5])[sup:2663nmi5]1/5[/sup:2663nmi5]][sup:2663nmi5]5[/sup:2663nmi5] = (x[sup:2663nmi5]2[/sup:2663nmi5] + 1)[sup:2663nmi5]5[/sup:2663nmi5]
. . . . .9 - x[sup:2663nmi5]2[/sup:2663nmi5] = x[sup:2663nmi5]10[/sup:2663nmi5] + 5x[sup:2663nmi5]8[/sup:2663nmi5] + 10x[sup:2663nmi5]6[/sup:2663nmi5] + 10x[sup:2663nmi5]4[/sup:2663nmi5] + 5x[sup:2663nmi5]2[/sup:2663nmi5] + 1
. . . . .0 = x[sup:2663nmi5]10[/sup:2663nmi5] + 5x[sup:2663nmi5]8[/sup:2663nmi5] + 10x[sup:2663nmi5]6[/sup:2663nmi5] + 10x[sup:2663nmi5]4[/sup:2663nmi5] + 6x[sup:2663nmi5]2[/sup:2663nmi5] - 8
Then apply the Rational Roots Test, and use synthetic division to try to find any rational (fractional or whole-number) zeroes.
Unfortunately, as alluded to earlier, there are no rational roots, so the above won't work. You will indeed be stuck with numerical methods. Ouch! :shock:
Eliz.