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Difficulty: MediumExponents, Powers, and Square Roots

If xx and yy are real numbers such that x<−1x < -1 and 0<y<10 < y < 1, which of the following statements must be true? Select all such statements.

  1. x2>y2x^2 > y^2Answer
  2. x2y4=−xy2\sqrt{x^2 y^4} = -x y^2Answer
  3. C
    x3>y3x^3 > y^3
  4. D
    x2+y=x+y\sqrt{x^2} + y = x + y
  5. E
    (x+y)2=x2+y2(x + y)^2 = x^2 + y^2

Answer

The correct statements are the inequality asserting that the square of the first variable is greater than the square of the second, and the identity simplifying the square root of the product of the powers to negative the product of the first variable and the square of the second.
The statement comparing squared values is correct because any real number less than -1 has an absolute value greater than 1, so its square is strictly greater than 1, while any positive number less than 1 has a square strictly less than 1. The radical identity statement is correct because taking the square root of x2x^2 yields ∣x∣|x|, which evaluates to −x-x when xx is negative.

Step-by-Step Solution

1
Analyze the given bounds for both variables.
For the first variable, x<−1x < -1, which implies ∣x∣>1|x| > 1, xx is negative, x2>1x^2 > 1, and x3<−1x^3 < -1. For the second variable, 0<y<10 < y < 1, which implies y>0y > 0, y2<1y^2 < 1, and y3>0y^3 > 0.
Establishing explicit bounds on magnitudes and signs is necessary to evaluate powers and absolute value roots.
2
Evaluate the inequality comparing the squared terms.
Since x2>1x^2 > 1 and y2<1y^2 < 1, it follows directly that x2>y2x^2 > y^2.
Transitive comparison across the threshold value of 1 proves the inequality holds.
3
Simplify the radical expression x2y4\sqrt{x^2 y^4}.
x2y4=x2⋅y4=∣x∣⋅y2\sqrt{x^2 y^4} = \sqrt{x^2} \cdot \sqrt{y^4} = |x| \cdot y^2. Since x<0x < 0, ∣x∣=−x|x| = -x, so the expression simplifies to −xy2-x y^2.
The principal square root of x2x^2 must equal the absolute value ∣x∣|x|, which requires a sign flip when xx is negative.
4
Verify remaining candidate expressions for potential fallacies.
Odd powers preserve negative signs so x3<y3x^3 < y^3; x2\sqrt{x^2} equals −x≠x-x \neq x; and expanding (x+y)2(x+y)^2 produces a nonzero cross-term 2xy2xy.
Eliminating false options confirms that only two statements are universally true.

Key Concept

Principal square roots and even/odd power behaviors under negative variable bounds
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Topics

Exponents, Powers, and Square Roots

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