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(3) 3\(m^2\) + 2m – 7 = 0

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Sol.     3\(m^2\) + 2m – 7  = 0
          Comparing with a\(x^2\) + bx + c = 0
          we get, a = 3, b = 2 and c = –7
∴       b2 – 4ac   =  (2)2 – 4 × 3 × –7
                          =   4 + 84
∴       b2 – 4ac   =   88

         \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)                 [Formula]

         m =  \( {-(2) \pm \sqrt{88} \over 2 × 3}\)                     [as variable is m]

             =  \( {-2 \pm \sqrt{4 × 22} \over 2 × 3}\)

             =  \( {-2 \pm2 \sqrt { 22} \over 2 × 3}\)

             =  \( {2( -1\pm \sqrt { 22} )\over 2 × 3}\)

             =  \( { -1\pm \sqrt { 22} \over3}\) 

∴      m  =  \( { -1+ \sqrt { 22} \over3}\)        or        m =   \( { -1- \sqrt { 22} \over3}\) 

∴      \( { -1+ \sqrt { 22} \over3}\)    and   \( { -1- \sqrt { 22} \over3}\)   are roots of the given quadratic equation.

uÉaÉïxÉqÉÏMüUhÉ August 13 , 2018 0 Comments 70 views