For what real value of \(k\) is \(\frac{13-\sqrt{131}}{4}\) a root of \(2x^2-13x+k\)?
+240 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,
\(b^2 - 4ac = 169 - 8k = 131\)
\(169 = 131 + 8k\)
\( 38 = 8k\)
\(k = 4.75\)
