Need solution for an expression
- Thread starter roopesh
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mmm4444bot
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One way to evaluate the blue expression is to first solve the equation (red) for x, followed by substituting the solutions (one at a time) for x in the blue expression. :cool:
This problem looks odd for a beginning algebra course.
Is the equation \(\displaystyle \dfrac{x + 1}{x} = - 1\) or \(\displaystyle x + \dfrac{1}{x} = - 1?\)
What topic are you currently studying in your course? Quadratic formula? Complex numbers? Imaginary numbers?
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mmm4444bot
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I intended to click the Intermediate Algebra board, when I moved this thread from the Arithmetic board. My bad.
If the original equation is correct as posted, this student will need experience working with non-Real numbers. :cool:
If the equation is missing grouping symbols, then perhaps the blue expression is missing them also?
If the original equation is correct as posted, this student will need experience working with non-Real numbers. :cool:
If the equation is missing grouping symbols, then perhaps the blue expression is missing them also?
mmm4444bot
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Regardless of givens, I'm thinking that the exercise may be designed to check whether students will conclude (incorrectly) that they simply need to cube -1.
Who knows? I cannot wait for you to post our summary guidelines.
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Helolo, roopesh!
Is there a typo?
I suspect that both binomials should have the same sign . . .
Cube the equation:
. . \(\displaystyle \begin{array}{c}\left(x+\dfrac{1}{x}\right)^3 \:=\: (\text{-}1)^3 \\ \\ x^3 + 3x + \dfrac{3}{x} + \dfrac{1}{x^3} \:=\:\text{-}1 \\ \\ x^3 + \dfrac{1}{x^3} + 3\underbrace{\left(x + \dfrac{1}{x}\right)}_{\text{This is -1}} \:=\:\text{-}1 \\ \\ x^3 + \dfrac{1}{x^3} + 3(-1) \:=\:\text{-}1 \\ \\ x^3 + \dfrac{1}{x^3} - 3 \:=\:\text{-}1 \\ \\ x^3 + \dfrac{1}{x^3} \:=\:2 \end{array}\)
Is there a typo?
I suspect that both binomials should have the same sign . . .
Cube the equation:
. . \(\displaystyle \begin{array}{c}\left(x+\dfrac{1}{x}\right)^3 \:=\: (\text{-}1)^3 \\ \\ x^3 + 3x + \dfrac{3}{x} + \dfrac{1}{x^3} \:=\:\text{-}1 \\ \\ x^3 + \dfrac{1}{x^3} + 3\underbrace{\left(x + \dfrac{1}{x}\right)}_{\text{This is -1}} \:=\:\text{-}1 \\ \\ x^3 + \dfrac{1}{x^3} + 3(-1) \:=\:\text{-}1 \\ \\ x^3 + \dfrac{1}{x^3} - 3 \:=\:\text{-}1 \\ \\ x^3 + \dfrac{1}{x^3} \:=\:2 \end{array}\)