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For what real value of \(k\) is \(\frac{13-\sqrt{131}}{4}\) a root of \(2x^2-13x+k\)?

 Jul 2, 2021
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The root given is in the form of the quadratic formula, \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

The part we are interested in is the discriminant, \(b^2-4ac\). Using this,

\(b^2 - 4ac = 169 - 8k = 131\)

\(169 = 131 + 8k\)

\( 38 = 8k\)

\(k = 4.75\)

 Jul 2, 2021

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