The smallest value of $k$, for which both the roots of the equation \[ x^{2}-8 k x+16\left(k^{2}-k+1\right)=0 \] are real, distinct and have values at least 4, is

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Quadratic Equations and Inequalities

The smallest value of

k

, for which both the roots of the equation

x^{2} - 8 k x + 16 (k^{2} - k + 1) = 0

are real, distinct and have values at least 4, is

Answer: 2

For real distinct roots: discriminant

D > 0

D = 64 k^{2} - 64 (k^{2} - k + 1) = 64 k - 64 > 0 \Rightarrow k > 1

For both roots

\geq 4

: the vertex of the parabola

f (x) = x^{2} - 8 k x + 16 (k^{2} - k + 1)

must be to the right of 4, and

f (4) \geq 0

.

Vertex at

x = \frac{8 k}{2} = 4 k

. For

4 k \geq 4

k \geq 1

f (4) = 16 - 32 k + 16 (k^{2} - k + 1) = 16 k^{2} - 48 k + 32 = 16 (k^{2} - 3 k + 2) = 16 (k - 1) (k - 2) \geq 0

.

This gives

k \leq 1

k \geq 2

.

Combining with

k > 1

k \geq 2

.

The smallest such

k

is 2.

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Integer answer