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The roots of the equation X2 - 3X - m(m+3) = 0, where m is a constant, are

1. m, m+3

2. -m, m+3

3. m,-(m+3)

4. -m,-(m+3)

(pls do explain)

my question is wrong....the equation is X2-3x-m(m+3)=0....

  • 12
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HI  AKSHAYA 

 x2-3x-m(m+3)=0

D=b2-4ac
=〖(-3〗2)- 4×1×-m(m+3)
=9 – 4 ×-m2-3m
=9+〖4m〗2+12m
=4m2+12m+9
=〖(2m)〗2+2×(2m)×3+32
=(2m+3)2
√(D   )       =2m+3
                                      x=(-b±√(b2-4ac)) / 2a
                                         =(3+(2m+3))/(2×1)
 
                                         =   (6+2m)/2
                =  6/2+2m/2   = 3+m
                             x         =   (-b±√(b2-4ac))/2a
                                        =  (3-(2m+3))/(2×1)
                                        = (3-2m-3)/2   = -m
ANS    : (-M,M+3)
 
 
  • 49

thanks Akhila....!!

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u r welcome

  • 4

best of luck 4 the xams frnds.........

  • -5
Thank you for u r answer,helped a lot. And thank you meritnation.
  • 3
[IF @,b,y be the roots of the equation x3+2yz+3x+4=0,then find the equation whose roots are 1+1\@,1+1\b,1+1\y.] @ mane alfa b main bita y main gama are equation what is the before number of xin square
  • -6
you can do this in an easier method, x^2-(sum of roots)x+(product of roots)
HERE -m(m+3) is product
therefore -m and (m+3) are roots of equation
  • 4
can't we use factorisation method ???
  • 2
,refer to the give n picture

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