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In the quadratic equation

 

 

 

\( x^2 + \left(k-\frac{1}{k}\right)x - 1 = 0,\)

 

 

solve for x in terms of k .
 

 Dec 24, 2019
 #1
avatar+22096 
+1

Use Quadratic Formula

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)      a = 1   b = (k- 1/k)   c = -1

 

 

[(1/k - k) +- sqrt ((k-1/k)^2 + 4 ) ] / 2

 Dec 24, 2019
 #2
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By the qudaratic formula

 

\(x = \dfrac{k - 1/k \pm \sqrt{(k - 1/k)^2 - 4}}{2} = \dfrac{k - 1/k \pm \sqrt{k^2 - 8 + 1/k^2}}{2}\)

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 Dec 24, 2019
 #3
avatar+22096 
+1

Check the -b part of the Quad Form in your answer.......AND the -4ac part too   cheeky

ElectricPavlov  Dec 24, 2019
edited by Guest  Dec 24, 2019

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