Question

Difficulty: EasyExponents, Radicals, and Algebraic Expressions

For all real numbers xx, the identity x2=x\sqrt{x^2} = x holds true.

Answer: Answer

Answer

The statement is false because the principal square root x2\sqrt{x^2} is defined as the absolute value of xx, x|x|, which equals x-x when x<0x < 0.
The statement is false because the expression x2\sqrt{x^2} represents the principal (non-negative) square root of x2x^2, which simplifies to x|x|. When xx is negative, x=x|x| = -x, which is not equal to xx.

Step-by-Step Solution

1
Recall the definition of the principal square root radical symbol x\sqrt{\vphantom{x}}
By definition, a0\sqrt{a} \ge 0 for any real number a0a \ge 0.
The principal square root function always yields a non-negative output.
2
Test a negative value for the variable xx
Let x=3x = -3. Then (3)2=9=3\sqrt{(-3)^2} = \sqrt{9} = 3.
Evaluating a specific negative number tests if the equality holds universally for all real numbers.
3
Compare the evaluated result with the original value of xx
Since 333 \neq -3, x2x\sqrt{x^2} \neq x when x<0x < 0.
A single counterexample disproves a universal mathematical identity statement.

Key Concept

Principal Square Root and Absolute Value Property (\sqrt{x^2} = |x|)
Estimated Time:45s
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