If x and y are real numbers such that x y^2, which of the following statements must be true? Select all such statements.

$x^3 < y^3$; $\sqrt{x^2} + x = 0$; $x + y < 0$

If x and y are real numbers such that x < 0 < y and x^2 > y^2, which of the following stat… — GRE General Test

If $x$ and $y$ are real numbers such that $x < 0 < y$ and $x^2 > y^2$ , which of the following statements must be true? Select all such statements.

The correct statements are

x^3 < y^3

\sqrt{x^2} + x = 0

, and

x + y < 0

The statement

x^3 < y^3

is correct because cubing a negative number keeps it negative while cubing a positive number keeps it positive. The statement

\sqrt{x^2} + x = 0

is correct because

\sqrt{x^2} = |x| = -x

for negative numbers. The statement

x + y < 0

is correct because

x^2 > y^2

implies

|x| > y

, meaning the negative component

x

has a larger absolute magnitude than the positive component

y

Analyze the signs of odd powers for

x < 0 < y

x^3 < 0

and

y^3 > 0

, which guarantees

x^3 < y^3

Odd powers preserve the original sign of the base.

Apply the definition of principal square roots to negative values

\sqrt{x^2} = |x| = -x

, so

\sqrt{x^2} + x = -x + x = 0

The square root symbol

\sqrt{}

denotes the principal (non-negative) root.

Compare absolute values using

x^2 > y^2

|x| > y \implies -x > y \implies x + y < 0

Since

x < 0

, its magnitude

|x|

-x

, which dominates the positive value

y

Evaluate the false options against exponent and radical rules

\sqrt{x^2 + y^2} \neq |x| + y

due to non-distributivity of roots, and

(-x)^2 = x^2 \neq -x^2

Radicals do not distribute over sums, and even powers eliminate negative signs.

Properties of real exponents, radical expressions, and absolute values for negative bases

Estimated Time:1m 30s