Question

determine k so dat the equation x2-4x+k=0 has

1.....two real roots

2.....two distinct roots

3.....coincident roots

4.....no real roots.

Answer 1

x^2-4x+k=0

D=b^2-4ac=16-4k

Fir,1.)two real roots D>0 or D=0

i.e 16-4k>=0 =>16/4>k

=>k<4 ,k=0or k€(-infinity,4]

For2.) For distinct roots ,D>0.

=>k<4.

For3.)For coincident roots D=0

=>k=4.

4.) For imaginary/no real roots D<0

=>16-4k<0

=>k>4.

Hope it helps

Answer 2

Answer:

Step-by-step explanation

D=b2-4ac = (4)2 - 4 *1*k =-4k (1)

(1)Condition for the two real root  : D=0(2) and D>0 (3)

substitute (1) in (2) ; -4k = 0 ; k =4

substitute (1) in (3) ; -4k>0; k >4

(2)condition for the two distinct roots : D>0 (4)

substitute  (1) in (4) ; -4k >0 = k>4

(3) condition for the coincident roots : D=0(5)

substitute (1) in (5) ; -4k=0 ; k=4

(4) condition for the coincident roots : D<0 (6)

substitute (1) in (6) ;-4k <0 =k<4 .

Hope it will help you all .

THANK YOU