Are \(\sqrt{4\cdot 9}\) and \(\sqrt{4}\cdot\sqrt{9}\) the same or different real numbers? Are \(\sqrt{9 + 16}\) and \(\sqrt{9}+\sqrt{16}\) the same or different real numbers? Explain your answers.
We have \(\sqrt{4\cdot 9} = \sqrt{36} = \sqrt{6^2} = 6\text{,}\) while \(\sqrt{4}\cdot\sqrt{9} = 2 \cdot 3 = 6\) and they agree. In general, \(\sqrt{a\cdot b} = \sqrt{a} \cdot \sqrt{b}\text{,}\) for non-negative \(a\) and \(b\text{.}\)
We have \(\sqrt{9 + 16}=\sqrt{25} = 5\text{,}\) while \(\sqrt{9}+\sqrt{16}= 3+ 4 = 7\) and they do not agree. In general, \(\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}\)
Is \(\sqrt{-4}\) a real number? What about \(-\sqrt{4}\text{?}\) What makes them different? What are the solutions to the equation \(x^2 - 4 = 0\text{?}\) What are the solutions to \(x^2 + 4 = 0\text{?}\) Explain your answer.
You cannot square a real number and get a negative result. Thus, \(\sqrt{-4}\) is not a real number because nothing may be squared to get \(-4\text{.}\) On the otherhand, \(-\sqrt{4}=- 2\) is a real number.
Evaluate the expression \(\sqrt{(-2)^2}\) without using a calculator. What should you be careful about when simplifing expressions of the form \(\sqrt{x^2}\text{?}\) Explain your answer.
