Question

plz solve Q no. 7, 8, 9 only.​

Answer 1

6)  Kx2 - 5x +K =0

so if it has real root then b2 - 4ac = 0

25 -4k2=0

4k2 = 25

k2 = 25/4

K = 5/2 ,,-5/2

7)  x2 +K(4x+K-1) = 0

x2+ 4Kx +K2 -K +2 = 0

so if it has real root then b2 - 4ac = 0

16K2 - 4K2 + 4K -8 = 0

12K2 + 4K +8  = 0

12K2+12K -8K -8 = 0

12K ( K+1)-8(k+1) = 0

(12K - 8) (K+1) =0

k= 8/12 ,, -1

Answer 2

Here is ur ans hope it is correct

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