Question

For what real values of $K$ does $x=K^2(x-1)(x-2)$ have real roots?

none

$-2\lt K\lt1$

$-2\sqrt2\lt K\lt2\sqrt2$

$K\gt1$ or $K\lt-2$

all

Answer: E

Concepts:quadratic algebraic manipulation

Difficulty rating: 1550

Small Hint:

Expand and collect terms to obtain a quadratic in $x$

Big Hint:

Its discriminant simplifies to $K^4+6K^2+1$

Solution:

Rearranging gives $K^2x^2-(3K^2+1)x+2K^2=0.$ Its discriminant is $\begin{aligned} &(3K^2+1)^2-8K^4\\ &\qquad=K^4+6K^2+1\gt0 \end{aligned}$ for every real $K.$ This also covers $K=0,$ when the original equation gives $x=0.$

Therefore, the correct answer is E.

1962 AMC 12 Problem 34

Answer: E

Difficulty rating: 1550

Small Hint:

Big Hint:

Solution:

Problem 34 in Other Years