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Given: b²-4ac=0
=> (4k-6)²-4(k-2)(5k-6)=0
= (16k²+36-48k)-4(5k²-16k+12) =0
= 16k²+36-48k-20k²-64k-48=0
= -4k²+16k-3=0
i.e., k²-4k+3=0
(k-1) (k-3) =0
k = 1,3
2 5 2
um meghna, i think you missed a number there.. pls recheck the answer thankyou
I'll recheck. I'm so sorry. Though, both values 1 and 3 do hold for k when substituted in the equation. Thank you so much :)
yeah the answer is correct though... i think you went wrong in the calculating part honey Thanks a bunch :D
Oops sorry c: You're welcome :D
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