Find value of k for which (k+4)xx+(k+1)x+1=0 with equal roots
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answer is either -3 or 5, hope it helps.
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harshrajput26:
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Answered by
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for equal root D=0
since 0=Bsquare -4AC
0=(k+1)(k+1) -4×(k+4)(1)
o=ksquare +1+2k-4k-16
0=kquare-2k -15
ksquare-5k+3k-15=0
k(k-5)+3(k-5)=0
(k+3)(k-5)=,0
since k=-3 or k=5
since 0=Bsquare -4AC
0=(k+1)(k+1) -4×(k+4)(1)
o=ksquare +1+2k-4k-16
0=kquare-2k -15
ksquare-5k+3k-15=0
k(k-5)+3(k-5)=0
(k+3)(k-5)=,0
since k=-3 or k=5
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