1956 AMC 12 Problem 44

Question

If

x\lt a\lt0

means that

x

and

a

are numbers such that

x

is less than

a

and

a

is less than zero, then:

x^2\lt ax\lt0

x^2\gt ax\gt a^2

x^2\lt a^2\lt0

x^2\gt ax

but

ax\lt0

x^2\gt a^2

but

a^2\lt0

Both numbers are negative, but

x

has the larger absolute value

Multiply

x\lt a

once by

x\lt0

and once by

a\lt0,

reversing each inequality

Since

x\lt a\lt0,

multiplying

x\lt a

by the negative number

x

reverses the inequality and gives

x^2\gt ax.

Multiplying the same inequality by the negative number

a

gives

ax\gt a^2.

Therefore

x^2\gt ax\gt a^2.

Thus, the correct answer is B.

Problem 44 in other years