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For what real value of k is \((13-√131)/4 \) a root of \(2x^2-13x+k\)?

Guest Dec 26, 2017
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The  other root will be

 

(13  +  sqrt (131))/ 4

 

In the form, ax^2  + bx  +  c......the product of the roots  =  c / a

 

So

 

(13  +  sqrt (131))/ 4  *   (13  -  sqrt (131))/ 4  =   k/2

 

(169  - 131) / 16  =  k/2

 

38/16  =  k/2

 

38/8  =  k

 

 

cool cool cool

CPhill  Dec 26, 2017
edited by CPhill  Dec 26, 2017

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