Find the value of k for which the given equation has equal roots
(k-12) x2+2(k-12)x+2=0
Anonymous:
complete ur que...
check the question
sky my mistake
sry^
Answers
Answered by
0
Roots r real and equal that means they r 0
Also, b²-4ac=0
Here,2k-24+2k-24+2=0
Wait.. Is it x or Multiplication sign
Also, b²-4ac=0
Here,2k-24+2k-24+2=0
Wait.. Is it x or Multiplication sign
please tell if it's x or a Multiplication sign I'll solve it again
Answered by
2
Hey mate thank for the question
Hope it helps
For equal roots
D=0
d=b2-4ac
d=(2(k-12)2-4*(k-12)*2
d=4(k2+144-24k)-8k+96
d=4k2+576-96k-8k+96
d=4k2+672-104k
d=4(k2-26k+168)
d=k2-26k+168
d=k2-(14+12)k+168
d=k2-14k-12k+168
d=k(k-14)-12(k-14)
d=(k-14)(k-12)
k=12,k=14
Hope it helps
For equal roots
D=0
d=b2-4ac
d=(2(k-12)2-4*(k-12)*2
d=4(k2+144-24k)-8k+96
d=4k2+576-96k-8k+96
d=4k2+672-104k
d=4(k2-26k+168)
d=k2-26k+168
d=k2-(14+12)k+168
d=k2-14k-12k+168
d=k(k-14)-12(k-14)
d=(k-14)(k-12)
k=12,k=14
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