Question

If one root of the equation x^(2)-3kx+(2k+1)=0 is less than 2 and other root is greater than 4 then minimum possible integral value of k is​

Answer 1

Answer:

sorry i am not able to solve this question

Answer 2

Given: Quadratic equation x^(2)-3kx+(2k+1)=0 whose one root is less than 2 and other root is greater than 4

To find: Minimum integral value of k

Solution: Let the quadratic equation be

F(x) = x^(2)-3kx+(2k+1)=0

For real roots to exit for this equation F(x), the discriminant of the expression x^(2)-3kx+(2k+1)=0 has to be negative that is <0 and the coefficient of x^2 should also be positive and here it is 1.

Δ= (-3k)^2 - 4(1)(2k+1)

Δ= 9k^2 - 8k- 4

The discriminant of expression 9k^2 - 8k- 4 is positive and coefficient of k^2is also positive, therefore we can say that Δ will be positive.

That means, for all values of k f(x) will have real and distinct roots.

Through the given equation, we can say that, F(x) is an upward parabola and one of its roots is less than 2 and other root is greater than 4.

Therefore, f(4) <0 , f(2) <0

f(2)= 4-6k+2k+1=5-4k<0

k>5/4

f(4)= 16-12k+2k+1= 17-10k<0

k> 17/10

Hence the Common set for values of k satisfying all the equation will be ( 17/10, ∞).