Question

please solve this question​

Answer 1

Answer:

Let's solve for k.

(k−2)(x2)+2(2k−3)x+5k−6=0

Step 1: Add 2x^2 to both sides.

kx2+4kx−2x2+5k−6x−6+2x2=0+2x2

kx2+4kx+5k−6x−6=2x2

Step 2: Add 6x to both sides.

kx2+4kx+5k−6x−6+6x=2x2+6x

kx2+4kx+5k−6=2x2+6x

Step 3: Add 6 to both sides.

kx2+4kx+5k−6+6=2x2+6x+6

kx2+4kx+5k=2x2+6x+6

Step 4: Factor out variable k.

k(x2+4x+5)=2x2+6x+6

Step 5: Divide both sides by x^2+4x+5.

k(x2+4x+5)

x2+4x+5

=

2x2+6x+6

x2+4x+5

k=

2x²+6x+6/x²+4x+5

Answer 2

Step-by-step explanation:

Ok you must de known of the fact that if a quadratic eqn has real roots the discriminant has to be greater than 0. this is an easy inequality solving .here X can be any no between between 0 and 4 . ex- X can be equal to 1,2 and 3. this the the solution set of X .