Question

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

Answer 1

k = 4.75

Explanation:

If (13 -√131)/4 is a root of the equation, then its conjugate, (13 +√131)/4 is also a root. The product of the roots is k/2, where 2 is the leading coefficient.

The product is ...

k/2 = ((13 -√131)/4) × ((13 +√131)/4) = (13² -131)/4² = 38/16

Then the value of k is ...

k = 2(38/16) = 19/4

k = 4.75

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The attached graph shows that (13 -√131)/4 is a root of the quadratic when k = 4.75.