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# Algebra Help

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Suppose a and x satisfy x^2 + (a - (1/a)) x - 1 = 0. Solve for x in terms of a.

Dec 28, 2018

#1
+4810
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$$x^2 +\left(a - \dfrac 1 a\right)x - 1 = 0\\ \text{let }c = a-\dfrac 1 a\\ \text{using the quadratic formula}\\ x = \dfrac{ -c \pm \sqrt{c^2+4}}{2} = \\ \dfrac{\left(\dfrac 1 a - 1\right)\pm \sqrt{\left(a-\dfrac 1 a \right)^2+4}}{2}$$

$$\text{this can be simplified a bit to}\\ x = \dfrac{(1-a) \pm \sqrt{(a-1)^2+4a^2}}{2a}$$

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Dec 28, 2018

#1
+4810
+2

$$x^2 +\left(a - \dfrac 1 a\right)x - 1 = 0\\ \text{let }c = a-\dfrac 1 a\\ \text{using the quadratic formula}\\ x = \dfrac{ -c \pm \sqrt{c^2+4}}{2} = \\ \dfrac{\left(\dfrac 1 a - 1\right)\pm \sqrt{\left(a-\dfrac 1 a \right)^2+4}}{2}$$

$$\text{this can be simplified a bit to}\\ x = \dfrac{(1-a) \pm \sqrt{(a-1)^2+4a^2}}{2a}$$

Rom Dec 28, 2018
#2
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Thank you so much!

sprockit  Dec 30, 2018
edited by sprockit  Dec 30, 2018