1974 AMC 12 Problem 10

Attempt Problem 10 of the 1974 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1974 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

10.

What is the smallest integral value of

k

such that

2x(kx-4)-x^2+6=0

has no real roots?

-1

2

3

4

5

First rewrite the equation in standard quadratic form

No real roots means the discriminant is negative

The equation is

(2k-1)x^2-8x+6=0.

Its discriminant is

64-24(2k-1)=88-48k.

This is negative exactly when

k\gt\frac{11}{6},

so the least integer possible is

2.

Therefore, the correct answer is B.

Problem 10 in other years