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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 164 views