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5. Sum of the roots of a quadratic equation is double their product. Find k if equation is \(x^2\) –4k\(x\) + k + 3 = 0

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Sol.    \(x^2\) –4k\(x\) + k + 3 = 0
          comparing with a\(x^2\) + b\(x\) + c = 0
          We get, a = 1, b = –4k and c = k + 3
          If α and β are roots of quadratic equation

          α  +   β   =  \(-{b\over a}\)

                        =  \(-({-4k\over 1})\)

                      =  4k   and

          α  ×  β   =  \(c\over a\)

                       =  \(k+3\over 1\)

                      =  k + 3

But      α  +  β  =  2αβ                    [Given]
∴                4k =   2(k + 3)
∴                4k =   2k + 6
∴        4k – 2k =   6
∴                2k =   6
∴                  k = 3 

uÉaÉïxÉqÉÏMüUhÉ August 10 , 2018 0 Comments 15 views