II. Simplifying negative radicands
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$$ \sqrt{-25} = \red ? $$
A trip down memory lane
The rules for simplifying square roots is critical prerequisite knowledge for this topic. So, let's see if you remember these rules.
(take some good mental notes if you don't)
Review Question 1:
TRUE!, you can do this and you'll need to do it today!
Review Question 2:
FALSE this rule does not apply to negative radicands !
$$ \red{ \sqrt{a} \sqrt{b} = \sqrt{a \cdot b} }$$ only works if a > 0 and b > 0.
In other words, the product of two radicals does not equal the radical of their products when you are dealing with imaginary numbers.
Example 1
$$ \sqrt{-7} \\ \sqrt{\red{-1} \cdot7} \\ \sqrt{\red{-1}} \cdot\sqrt{7} \\ i\sqrt{7} $$
In example 1, the only type of work that you can do is to remove the $$-1$$ from the radicand.
Example 2
$$ \sqrt{-9} \\ \sqrt{-1 \cdot 9} \\ \sqrt{-1} \cdot\sqrt{9} \\ i \cdot 3 = 3i $$
In this one, you can actually reduce the $$\sqrt{ 9 }$$
Example 3
$$ \sqrt{-12} \\ \sqrt{-1 \cdot 4 \cdot 3} \\ \sqrt{-1} \cdot \sqrt{4} \cdot \sqrt{3} \\ i\sqrt{4} \cdot \sqrt{3} \\ 2 i \sqrt{3} $$
Similar to example 2, you can actually reduce the $$\sqrt{ 12 }$$
General Formula
$$ \sqrt{-a} \\ \sqrt{-1 \cdot a} \\ \sqrt{-1} \cdot \sqrt{a} \\ i\sqrt{a}$$
Negative Radical Reducer
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Practice Problems
Problem 1
Step 1
"extract" $$i$$
$$ \sqrt{-5} \\= \red{\sqrt{-1}} \cdot \sqrt{5} \\ = \red{i} \sqrt{5} $$
Step 2
Simplify $$ \red{ \text{radicand}} $$
$$ i \sqrt{ \red{ 5 }} $$
In this case, the radicand is already in simplest form so there's really only 1 step to perform.
Problem 2
Step 1
"extract" $$i$$
$$ \sqrt{-36} \\= \red{\sqrt{-1}} \cdot \sqrt{36} \\ = \red{i} \sqrt{36} $$
Step 2
Simplify $$ \red{ \text{radicand}} $$
$$ i \sqrt {\red{ 36 }} \\ i \cdot 6 \\ 6i $$
As you can see from step 2, most of the work for these kinds of problems involves simplifying the square root !
Problem 3
Step 1
"extract" $$i$$
$$ \sqrt{-48} \\= \red{\sqrt{-1}} \cdot \sqrt{48} \\ = \red{i} \sqrt{48} $$
Step 2
Simplify $$ \red{ \text{radicand}} $$
$$ i \sqrt { \red{ 48 }} \\ i \sqrt { \red{ 16 \cdot 3 }} \\ i \sqrt { \red{ 16}} \cdot \sqrt { \red{ 3 }} \\ i \cdot 4 \cdot \sqrt { \red{3 }} \\ 4 i \sqrt { 3} $$
As you can see from step 2, most of the work for these kinds of problems involves simplifying the square root !
Problem 4
Step 1
"extract" $$i$$
$$ \sqrt{-50} \\ = \red{\sqrt{-1}} \cdot \sqrt{50} \\ = \red{i} \sqrt{50} $$
Step 2
Simplify $$ \red{ \text{radicand}} $$
$$ i \sqrt { \red{ 50 }} \\ i \sqrt { \red{ 25 \cdot 2 }} \\ i \sqrt { \red{ 25}} \cdot \sqrt { \red{2 }} \\ i 5 \cdot \sqrt { \red{2 }} \\ 5 i \sqrt{2} $$
As you can see from step 2, most of the work for these kinds of problems involves simplifying the square root !
Further Reading:
