Perfect Square

We can apply the formula: $(a-b)^2 = a^2 - 2 a b + b^2$ . In our case $a=\sqrt{k}x, b=3$ .

$k x^2 -3 k x + 9 = (\sqrt{k}x-3)^2 = k x^2 - 6 \sqrt{k}x + 9 \Rightarrow 3 k x = 6 \sqrt{k}x \Rightarrow k = 2 \sqrt{k} \Rightarrow \sqrt{k}=2 \Rightarrow k=\boxed{4}$

Note: $k=0$ is also a solution to the equation, but it would not be quadratic expression.

To be a perfect square, its discriminant must equal $0$ : $9k^2-4\cdot 9k=0 \Rightarrow k(k-4)=0$ So the solutions should be, $0$ or $4$ , but if we use $k=0$ the expression is not quadratic, so the answer is $4$

Kx²-3kx+9 is the expression.Therefore,

Zero of the equation is Kx²-3x+9=0

Let the roots be β&β.so,

β+β=3k/k (sum of roots are -b/a)

Therefore, β=3/2

β²=9/k (product of roots are c/a)

9/4=9/k

Therefore,K=4

(2x-9)². =4x²-12x+9 Is the same as kx²-3kx+9. So k=4