Question

Let $k$ be the largest integer such that the equation $(x-1)^2+2kx+11=0$ has no real roots. If $y$ is a positive real number, then the least possible value of $\frac{k}{4y}+9y$ is

Answer 1

Given :
(x - 1)2 + 2kx + 11 = 0 has no real roots , where k is the largest integer
Equation can be stated as follows :
⇒ x2 - 2x + 1 + 2kx + 11 = 0
= x2 - 2(k - 1)x + 12 = 0
As we already know that , for a equation with no real roots ,
D < 0 = b2 - 4ac < 0
Hence , {2(k - 1)}2 - 4×1×12 < 0
= 4(k - 1)2 < 48
= (k - 1)2 < 12
As k is an integer value , it implies that (k-1) is also an integer.
∴ From the above inequality, we can conclude that largest possible of value of (k-1) is 3
So , the value of k is 4.
So , we need to calculate the least possible value of $\frac{k}{4y}+9y$
= $\frac{4}{4y}+9y=\frac{1}{y}+9y$
Now , we can calculate the value by using A.M-G.M inequality as follows :
$\frac{9y+\frac{1}{y}}{2}\ge\sqrt{9y\times\frac{1}{y}}$
⇒ $\frac{9y+\frac{1}{y}}{2}\ge\sqrt{9}$
⇒ $\frac{9y+\frac{1}{y}}{2}\ge3$
⇒ $9y+\frac{1}{y}\ge6$
Therefore , the least possible value for it is 6.