2. Simplify without the use of a calculator: \( \begin{array}{ll}\text { a) } \sqrt{-4}+\sqrt{-16}-\sqrt{81} & \text { b) }-\sqrt{-9}-\sqrt{-25}+\sqrt{-64}-\sqrt{-1} \\ \text { c) } \sqrt{-3}+\sqrt{-48} & \text { d) } \sqrt{-12}+\sqrt{-36}+\sqrt{-28} \\ \text { e) } \frac{\sqrt{-18}}{\sqrt{-2}}+\sqrt{-144} & \end{array} \)
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Bonus Knowledge
Let's simplify each expression one by one. a) \( \sqrt{-4}+\sqrt{-16}-\sqrt{81} \) First, recall that \( \sqrt{-1} = i \), so we can express each term involving the square root of a negative number in terms of \( i \): - \( \sqrt{-4} = \sqrt{4} \sqrt{-1} = 2i \) - \( \sqrt{-16} = \sqrt{16} \sqrt{-1} = 4i \) - \( \sqrt{81} = 9 \) Putting this together: \[ 2i + 4i - 9 = 6i - 9 \] So, the answer for (a) is: \[ 6i - 9 \] b) \( -\sqrt{-9}-\sqrt{-25}+\sqrt{-64}-\sqrt{-1} \) Using the same method: - \( -\sqrt{-9} = -\sqrt{9} \sqrt{-1} = -3i \) - \( -\sqrt{-25} = -\sqrt{25} \sqrt{-1} = -5i \) - \( +\sqrt{-64} = \sqrt{64} \sqrt{-1} = 8i \) - \( -\sqrt{-1} = -i \) Now putting these together: \[ -3i - 5i + 8i - i = (-3 - 5 + 8 - 1)i = -1i = -i \] Thus, the answer for (b) is: \[ -i \] c) \( \sqrt{-3}+\sqrt{-48} \) Now simplifying: - \( \sqrt{-3} = \sqrt{3} \sqrt{-1} = i\sqrt{3} \) - \( \sqrt{-48} = \sqrt{48} \sqrt{-1} = \sqrt{16 \cdot 3} i = 4i\sqrt{3} \) Combining these gives: \[ i\sqrt{3} + 4i\sqrt{3} = (1 + 4)i\sqrt{3} = 5i\sqrt{3} \] Therefore, the solution for (c) is: \[ 5i\sqrt{3} \] d) \( \sqrt{-12}+\sqrt{-36}+\sqrt{-28} \) Breaking it down: - \( \sqrt{-12} = \sqrt{12}\sqrt{-1} = 2i\sqrt{3} \) - \( \sqrt{-36} = \sqrt{36}\sqrt{-1} = 6i \) - \( \sqrt{-28} = \sqrt{28}\sqrt{-1} = 2i\sqrt{7} \) Now, summing up gives: \[ 2i\sqrt{3} + 6i + 2i\sqrt{7} = (2\sqrt{3} + 6 + 2\sqrt{7})i \] Therefore, the result for (d) is: \[ (2\sqrt{3} + 6 + 2\sqrt{7})i \] e) \( \frac{\sqrt{-18}}{\sqrt{-2}}+\sqrt{-144} \) First, simplifying each term: - \( \sqrt{-18} = \sqrt{18}\sqrt{-1} = 3i\sqrt{2} \) - Thus, \( \frac{\sqrt{-18}}{\sqrt{-2}} = \frac{3i\sqrt{2}}{i\sqrt{2}} = 3 \) - \( \sqrt{-144} = 12i \) Combining them gives: \[ 3 + 12i \] Lastly, the answer for (e) is: \[ 3 + 12i \] Final answers: a) \( 6i - 9 \) b) \( -i \) c) \( 5i\sqrt{3} \) d) \( (2\sqrt{3} + 6 + 2\sqrt{7})i \) e) \( 3 + 12i \)
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