find the of k for which the equton x2+k[2x+k+1]+2=0 has equal and real roots
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here is ur answer buddy.....
p(x)=x²+k(2x+k+1)+2=0
p(x)=x²+2kx+k²+k+2=0
p(x)=x²+2kx+(k²+k+2)=0
condition for equal and real roots..
D=b²-4ac=0
=>(2k)²-4(1)(k²+k+2)=0
=>4k²-4k²-k-8=0
=>-k=8
=>k=-8
hence k =-8
hope it helps.......
p(x)=x²+k(2x+k+1)+2=0
p(x)=x²+2kx+k²+k+2=0
p(x)=x²+2kx+(k²+k+2)=0
condition for equal and real roots..
D=b²-4ac=0
=>(2k)²-4(1)(k²+k+2)=0
=>4k²-4k²-k-8=0
=>-k=8
=>k=-8
hence k =-8
hope it helps.......
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