x + 1/x =3 what is x^4 + 1/x^4?
- Thread starter brucejin
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brucejin said:
Your instructions are rather meager. I suppose you need to solve the equation for x. Once you find the value of x, plug that into the expression in place of x and evaluate.
To solve x + 1/x = 3, multiply both sides of the equation by the least common denominator, which is x. This will produce a second degree equation. I believe you will have to solve by using the quadratic formula which will produce two complex roots of the form \(\displaystyle x=\frac{m\pm\sqrt{n}}{r}\) where m, n and r are integers. That's what is plugged into the expression and evaluated.
Hello, Bruce!
There is a back-door approach to this problem in which we do not have to solve for x.
\(\displaystyle \text{Raise the equation to the 4th power: }\;\left(x + \frac{1}{x}\right)^4 \:=\:3^4\)
. . \(\displaystyle x^4 + 4x^2 + 6 + \frac{4}{x^2} + \frac{1}{x^4} \;=\;81\)
. . \(\displaystyle \left(x^4 + \frac{1}{x^4}\right) + 4\left(x^2 + \frac{1}{x^2}\right) \;=\;75\)
\(\displaystyle \text{In the second parentheses, add 2 and subtract 2: }\;\left(x^4 + \frac{1}{x^4}\right) + 4\left(x^2 + 2 + \frac{1}{x^2} - 2\right) \;=\;75\)
. . \(\displaystyle \text{and we have: }\;\left(x^4+\frac{1}{x^4}\right) + 4\left(\left[x+\frac{1}{x}\right]^2 -2\right) \;=\;75\)
. . . . . . . . . . . . . . . . . . . . . . . .\(\displaystyle \begin{array}{c}\uparrow \\ ^{\text{This is 3}} \end{array}\)
\(\displaystyle \text{Hence: }\;\left(x^4+\frac{1}{x^4}\right) + 4(3^2-2) \;=\;75 \quad\Rightarrow\quad \left(x^4+\frac{1}{x^4}\right) + 28 \;=\;75\)
. . \(\displaystyle \text{Therefore: }\;\boxed{x^4 + \frac{1}{x^4} \;=\;47} \quad\hdots \text{ ta-}D\!AA!\)
There is a back-door approach to this problem in which we do not have to solve for x.
\(\displaystyle \text{Raise the equation to the 4th power: }\;\left(x + \frac{1}{x}\right)^4 \:=\:3^4\)
. . \(\displaystyle x^4 + 4x^2 + 6 + \frac{4}{x^2} + \frac{1}{x^4} \;=\;81\)
. . \(\displaystyle \left(x^4 + \frac{1}{x^4}\right) + 4\left(x^2 + \frac{1}{x^2}\right) \;=\;75\)
\(\displaystyle \text{In the second parentheses, add 2 and subtract 2: }\;\left(x^4 + \frac{1}{x^4}\right) + 4\left(x^2 + 2 + \frac{1}{x^2} - 2\right) \;=\;75\)
. . \(\displaystyle \text{and we have: }\;\left(x^4+\frac{1}{x^4}\right) + 4\left(\left[x+\frac{1}{x}\right]^2 -2\right) \;=\;75\)
. . . . . . . . . . . . . . . . . . . . . . . .\(\displaystyle \begin{array}{c}\uparrow \\ ^{\text{This is 3}} \end{array}\)
\(\displaystyle \text{Hence: }\;\left(x^4+\frac{1}{x^4}\right) + 4(3^2-2) \;=\;75 \quad\Rightarrow\quad \left(x^4+\frac{1}{x^4}\right) + 28 \;=\;75\)
. . \(\displaystyle \text{Therefore: }\;\boxed{x^4 + \frac{1}{x^4} \;=\;47} \quad\hdots \text{ ta-}D\!AA!\)
D
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These are favorite SAT & GRE problems. There is a long way to do htis and there is a short way to do this.
Another example
Find x[sup:3ozw1r58]2[/sup:3ozw1r58] - 1/x[sup:3ozw1r58]2[/sup:3ozw1r58] if x + 1/x = a
Another example
Find x[sup:3ozw1r58]2[/sup:3ozw1r58] - 1/x[sup:3ozw1r58]2[/sup:3ozw1r58] if x + 1/x = a
mmm4444bot
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brucejin said:
Loren already posted it.
Solve the equation.
x + 1/x = 3
From using the Quadratic Formula, you'll get two solutions for x -- in the form posted by Loren.
(I'm guesing that it's okay to use a scientific calculator to get decimal approximations for these two solutions, at this point in the exercise.)
Substitute these two values for x (one at a time, of course), to evaluate the given expression.
x^4 + 1/x^4
(I'm guessing that it's okay to use the calculator for this evaluation, too.)
You should get the same result, regardless of which solution you use in the evaluation.
Cheers