Question

find all the value of k for which one root of the quadratic equation (k-5)x^2 - 2kx + k - 4 = 0 is smaller than 1 and the other root exceed 2. ​

Answer 1

Answer:

Let f(x)=2x

2

−2kx+k−4,

For real roots to exist for f(x)=0, the discriminant must be positive.

Δ=4k

2

−8(k−4)

Δ=4(k

2

−2k+8)

Δ is always positive, since the discriminant of the expression k

2

−2k+8 is negative and the coefficient of k

2

is also positive.

Hence for all values of k, f(x)=0 has real and distinct roots.

Now f(x) is an upward opening parabola and one of the roots is greater than 2 and the other root is less than 1.

So, f(1)<0 and f(2)<0.

f(1)=2−2k+k−4<0

⇒k>−2

f(2)<0,

⇒2(2)

2

−4k+k−4<0

⇒−3k+4<0

⇒k>

3

4

Hence, the common set for k is (

3

4

,∞).