Question

3) Find K, if K(x-2) +6=0 has real and equal roots​

Answer 1

Answer:

kx

2

−2kx+6=0

Then a=k,b=−2k,c=6

Then b

2

−4ac=0

So (−2k)

2

−4k(6)=0

4k

2

−24k=0

4k(k−6)=0

k=0 and k=6

Put value of k

Then 6x

2

−12x+6=0

6x

2

−6x−6x+6=0

6x(x−1)−6(x−1)=0

∴x=1

Answer 2

Step-by-step explanation:

Kx-2k+6=0

ATQ

real and equal roots possible if √b^2-4ac=0

Here, a=0

√k^2-0=0

k=0