+18 For what real value of \(k\) is \(\frac{13-\sqrt{131}}{4}\) a root of \(2x^2-13x+k\)?
I would like to know the process too, please. Thank you!
+129933 It must be that the discriminant (b^2 - 4ak) = 131
So
(-13)^2 - 4(2)(k) = 131 simplify
169 - 8k = 131 rearrange as
169 - 131 = 8k
38 = 8k divide both sides by 8
38/8 = 19/4 = k
+129933 Welcome!!!
There is another way to solve this
The product of the roots = k/2
13 - √131 13 + √131
If _________ is a root then so is ____________
4 4
So ( 13 - √131) / 4 * ( 13 + √131) / 4 = (169 - 131) / 4 = 38 / 16
So
38/16 = k/ 2 mutiply through by 2
2 * 38 / 16 = k
76/16 = k
19/4 = k
