Find the positive value of k, for which the equation x2 + kx + 64 =0 and x2 - 8x + k=0 will both have real roots.

if the equaion x2+kx+64=0 has real roots, then D>=0

k2-256>=0  => k2>=256  => k2>=(16)2  => k>=16.- (1)

if the equation x2-8x+k=0 has real roots.

then D>=0  => 64-4k>=0  => 4k<=64  => k<=16 - (2)

from (1) and (2) we get k=16

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 b2 - 4ac   ,     a=1 ,b=K  ,c=64

      b2 - 4ac =(k)2  -4(1)(64) >0  real roots

       =    k2 - 256>0

       =  k2>256

        = k>16

        for the second

       X2 - 8x +k

   a=1  ,b= -8 c=  k

    (8)2  - 4(-8)(k) >0

 =   64  +  32k  >

 =   32k >64

 =  k>2

 for first k>16

 second

k >16

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  please thumpz up

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