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# For what real value of is a root of?

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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

### 1+0 Answers

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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