Question

simplify $xsqrt{x^2}$

Explanation:

Step1: Simplify the square root

Recall that $\sqrt{x^2} = |x|$, the absolute value of \( x \), because the square root of a square gives the non - negative value. So we rewrite the expression \( x\sqrt{x^2} \) as \( x|x| \).

Step2: Analyze cases based on the sign of \( x \)

- Case 1: If \( x\geq0 \), then \( |x| = x \). Substituting into the expression \( x|x| \), we get \( x\times x=x^{2} \).
- Case 2: If \( x < 0 \), then \( |x|=-x \). Substituting into the expression \( x|x| \), we get \( x\times(-x)=-x^{2} \).

We can also write the simplified form using the absolute value property in a combined way. But another way to look at it is to note that \( \sqrt{x^{2}}=|x| \), and when we multiply \( x \) by \( |x| \), we can also think in terms of exponents. We know that \( \sqrt{x^{2}}=x^{2\times\frac{1}{2}} = |x| \), and \( x\times|x| \). But if we consider the domain where we can write it without absolute value (assuming we are dealing with real numbers and the expression is defined), we can also simplify it as follows:

We know that \( \sqrt{x^{2}} = |x| \), so \( x\sqrt{x^{2}}=x|x| \). But if we consider the principal square root and the fact that for real numbers, when we have \( x\sqrt{x^{2}} \), we can also use the property that \( \sqrt{x^{2}}=x \) when \( x\geq0 \) and \( \sqrt{x^{2}}=-x \) when \( x < 0 \). So:

If \( x\geq0 \), \( x\sqrt{x^{2}}=x\times x = x^{2} \)

If \( x < 0 \), \( x\sqrt{x^{2}}=x\times(-x)=-x^{2} \)

But we can also write the simplified expression as \( x|x| \) or, by using the property of exponents, we can note that \( x\times x^{2\times\frac{1}{2}}=x\times x^{1}=x^{2} \) when \( x\geq0 \) and \( x\times(-x)=-x^{2} \) when \( x < 0 \). However, a more concise way (using the definition of absolute value) is:

\( x\sqrt{x^{2}}=x|x|=\begin{cases}x^{2},&x\geq0\\-x^{2},&x < 0\end{cases} \)

But if we consider the expression in terms of simplifying the radical and the monomial, we can also say that \( \sqrt{x^{2}} = |x| \), so \( x\sqrt{x^{2}}=x|x| \). But another approach is to use the exponent rules. We know that \( \sqrt{x^{2}}=x^{2^{\frac{1}{2}}}=|x| \), and \( x\times|x| \). But if we assume that \( x\) is a non - negative real number (for example, in the context of a problem where \( x\) represents a length or a non - negative quantity), then \( \sqrt{x^{2}}=x \), and \( x\sqrt{x^{2}}=x\times x = x^{2} \). If we consider all real numbers, the simplified form is \( x|x| \) or the piece - wise function as above. But a common way to write the simplified form (using the property of square roots) is:

We know that \( \sqrt{x^{2}}=|x| \), so \( x\sqrt{x^{2}}=x|x| \). But if we use the fact that \( |x|=\sqrt{x^{2}} \), and we can also write \( x|x| \) as \( \text{sgn}(x)x^{2} \), where \( \text{sgn}(x) \) is the sign function. However, the most straightforward simplification (using the definition of the square root of a square) is:

\( x\sqrt{x^{2}}=x|x| \), and if we want to write it without the absolute value, we can use the piece - wise definition. But in many cases, when we are dealing with the simplification of algebraic expressions, we can also note that \( \sqrt{x^{2}} = |x| \), so \( x\sqrt{x^{2}}=x|x| \). But if we consider the domain where \( x\) is a real number, and we use the property that \( \sqrt{x^{2}}=x \) when \( x\geq0 \) and \( \sqrt{x^{2}}=-x \) when \( x < 0 \), we can simplify the expression as follows:

For \( x\geq0 \), \( x\sqrt{x^{2}}=x\times x = x^{2} \)

For \( x < 0 \), \( x\sqrt{x^{2}}=x\times(-x)=-x^{2} \)

We can also write the simplified expression as \( x|x| \), which is equivalent to the piece - wise function. But if we consider the problem in the context of algebraic simplification (assuming we are to write it in a simplified form using exponents or without absolute value when possible), we can also use the fact that \( \sqrt{x^{2}}=x^{1} \) when \( x\geq0 \), so \( x\sqrt{x^{2}}=x\times x = x^{2} \) (for \( x\geq0 \)) and \( x\sqrt{x^{2}}=x\times(-x)=-x^{2} \) (for \( x < 0 \)).

Answer:

The simplified form of \( x\sqrt{x^{2}} \) is \( x|x| \) (or \(

$$\begin{cases}x^{2},&x\geq0\\-x^{2},&x < 0\end{cases}$$

\))