Question

Roots of 2x^2 - x + k are positive then what is the range of k?

Answer 1

The quadratic equation is given as,
2x² - x + k.

Let the roots of the equation be m and n.

Since, for the quadratic equation (ax² + bx + c), we have the relation-

m + n = -b/a
m . n = c/a

Using above two for the given equation,
m + n = 1/2
m . n = k/2

As both the roots are positive,
=> m . n > 0
=> k > 0.

Let us find out the determinant ( D = √(b² - 4.a.c)
So, D = √( (-1)² - 4.2.k)

For real roots, D ≥ 0.
=> 1 - 8k ≥ 0
=> k ≤ 1/8.

Range of k = (0 , 1/8].

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Answer 2

Answer:

k belongs to 0,1/8

Step-by-step explanation:

Finding discriminant

D= 1-8k

For positive roots D>=0----------------1

By x=-b+ or - rootD / 2a

x = 1 + root1-8k/4   (forever +ve)

x= 1-root1-8k/4

for this to be +ve

root 1-8k<0

k>0 ( on further solving)------2

From ----1

k>=1/8-------3

from -----2 and 3

k belongs 0,1/8