## Precalculus (6th Edition) Blitzer

The error in the expression is $\sqrt{-9}\cdot \sqrt{-9}\ne \sqrt{81}$ but the expression ${{\left( \sqrt{-9} \right)}^{2}}=\sqrt{-9}\cdot \sqrt{-9}$ is correct.
Consider the expression, ${{\left( \sqrt{-9} \right)}^{2}}=\sqrt{-9}\cdot \sqrt{-9}=\sqrt{81}$ Take the first part of the expression. ${{\left( \sqrt{-9} \right)}^{2}}=\sqrt{-9}\cdot \sqrt{-9}$ The principal square root of a negative number is such that for any positive real number $b$, $\sqrt{-b}=i\sqrt{b}$ Express all the square roots of a negative numbers in terms of $i$. \begin{align} & {{\left( \sqrt{-9} \right)}^{2}}=\left( i\sqrt{9} \right)\cdot \left( i\sqrt{9} \right) \\ & ={{i}^{2}}{{\left( \sqrt{9} \right)}^{2}} \end{align} The imaginary unit is $i=\sqrt{-1}$, where ${{i}^{2}}=-1$. Replace the value ${{i}^{2}}=-1$. \begin{align} & {{\left( \sqrt{-9} \right)}^{2}}=\left( -1 \right){{\left( \sqrt{9} \right)}^{2}} \\ & =\left( -1 \right)\left( 9 \right) \\ & =-9 \end{align} The value of ${{\left( \sqrt{-9} \right)}^{2}}$ is equal to $-9$. The error in the expression is such that, ${{\left( \sqrt{-9} \right)}^{2}}=\sqrt{-9}\cdot \sqrt{-9}$ and $\sqrt{-9}\cdot \sqrt{-9}\ne \sqrt{81}$ \begin{align} & \sqrt{-9}\cdot \sqrt{-9}\ne \sqrt{\left( -9 \right)\left( -9 \right)} \\ & \ne \sqrt{{{\left( -1 \right)}^{2}}{{\left( 9 \right)}^{2}}} \\ & \ne \sqrt{81} \end{align} Therefore, the error in the expression is $\sqrt{-9}\cdot \sqrt{-9}\ne \sqrt{81}$ but the expression ${{\left( \sqrt{-9} \right)}^{2}}=\sqrt{-9}\cdot \sqrt{-9}$ is correct.