Unit 8 · Lesson 3 of 6
Adding, subtracting, and multiplying radicals
You'll be able to: add, subtract, and multiply square roots, including two-term products like \((3+\sqrt{2})(1-\sqrt{2})\).
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Answer from memory first. Pulling it out of your head is what makes it stick.
Simplifying \(\sqrt{48}\) gives \(a\sqrt{3}\). What number is \(a\)?
Find the biggest perfect square that divides 48: it is 16, because \(48 = 16 \cdot 3\).
\(\sqrt{48} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3}\). Answer: \(a = 4\).
Simplify \(3x + 5y - 7x + 2y\).
Combine the x terms: \(3x - 7x = -4x\). Combine the y terms: \(5y + 2y = 7y\).
Answer: \(-4x + 7y\). Keep this in mind: radicals combine the same way.
Step 2 of 5The idea
Like radicals have the same index and the same radicand (the number under the root). \(3\sqrt{2}\) and \(5\sqrt{2}\) are like radicals. \(\sqrt{2}\) and \(\sqrt{3}\) are not.
You combine like radicals exactly the way you combine like terms. Since \(3x + 5x = 8x\), also \(3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}\). Think of \(\sqrt{2}\) as the "x".
Some radicals only look unlike. Simplify each one first (Lesson 39), then look again.
Multiplying is easier: you can multiply any two square roots. Multiply the outside numbers together and the inside numbers together, then simplify.
Add or subtract only like radicals: add the numbers in front, keep the radical. \(a\sqrt{r} + b\sqrt{r} = (a+b)\sqrt{r}\).
Simplify first: \(\sqrt{8} + \sqrt{18} = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}\).
Multiply: \(\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}\), and \(\sqrt{a} \cdot \sqrt{a} = a\) (for \(a \ge 0\)).
Two binomials: multiply every term by every term (FOIL), then combine. For a conjugate pair, the radicals cancel: \((a+\sqrt{b})(a-\sqrt{b}) = a^2 - b\).
Why this shows up on the placement test: answer choices are always fully simplified, so you have to combine and simplify radicals to recognize the right one.
Step 3 of 5Worked examples
Example 1: \(\sqrt{50} - \sqrt{8} + 3\sqrt{2}\)
- Simplify each radical: \(\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}\) and \(\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}\).
- Now the problem is \(5\sqrt{2} - 2\sqrt{2} + 3\sqrt{2}\). All three are like radicals now.
- Combine the numbers in front: \(5 - 2 + 3 = 6\). Answer: \(6\sqrt{2}\).
Example 2: \((3+\sqrt{2})(1-\sqrt{2})\), then \((4-\sqrt{3})(4+\sqrt{3})\)
- First: \(3 \cdot 1 = 3\). Outer: \(3 \cdot (-\sqrt{2}) = -3\sqrt{2}\). Inner: \(\sqrt{2} \cdot 1 = \sqrt{2}\). Last: \(\sqrt{2} \cdot (-\sqrt{2}) = -2\) (\(\sqrt{2} \cdot \sqrt{2} = 2\)).
- Together: \(3 - 3\sqrt{2} + \sqrt{2} - 2\).
- Combine the plain numbers, \(3 - 2 = 1\), and the like radicals, \(-3\sqrt{2} + \sqrt{2} = -2\sqrt{2}\). Answer: \(1 - 2\sqrt{2}\).
- The second one is a conjugate pair: \(4^2 - (\sqrt{3})^2 = 16 - 3 = 13\). The middle terms \(+4\sqrt{3}\) and \(-4\sqrt{3}\) cancel.
Step 4 of 5Practice
Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went. Give every answer fully simplified; in every problem, assume all variables are positive.
\(4\sqrt{5} + 7\sqrt{5}\)
Like radicals: add the numbers in front, \(4 + 7 = 11\), and keep \(\sqrt{5}\). Answer: \(11\sqrt{5}\).
\(\sqrt{12} + \sqrt{27}\)
Simplify first: \(\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}\) and \(\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}\).
\(2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}\). Answer: \(5\sqrt{3}\).
\(3\sqrt{20} - \sqrt{45}\)
\(\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}\), so \(3\sqrt{20} = 3 \cdot 2\sqrt{5} = 6\sqrt{5}\). And \(\sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5}\).
\(6\sqrt{5} - 3\sqrt{5} = 3\sqrt{5}\). Answer: \(3\sqrt{5}\).
Multiply \(2\sqrt{6} \cdot 5\sqrt{3}\). Simplified, the answer has the form \(a\sqrt{2}\). What number is \(a\)?
Outside numbers: \(2 \cdot 5 = 10\). Inside numbers: \(\sqrt{6} \cdot \sqrt{3} = \sqrt{18}\). So the product is \(10\sqrt{18}\).
Simplify: \(\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\), so \(10\sqrt{18} = 10 \cdot 3\sqrt{2} = 30\sqrt{2}\). Answer: \(a = 30\).
Simplify \(2\sqrt{18x} - \sqrt{8x}\).
\(\sqrt{18x} = \sqrt{9 \cdot 2x} = 3\sqrt{2x}\), so \(2\sqrt{18x} = 6\sqrt{2x}\). And \(\sqrt{8x} = \sqrt{4 \cdot 2x} = 2\sqrt{2x}\).
\(6\sqrt{2x} - 2\sqrt{2x} = 4\sqrt{2x}\). Answer: \(4\sqrt{2x}\).
\((2+\sqrt{5})(3-\sqrt{5})\)
FOIL: \(2 \cdot 3 = 6\), \(2 \cdot (-\sqrt{5}) = -2\sqrt{5}\), \(\sqrt{5} \cdot 3 = 3\sqrt{5}\), \(\sqrt{5} \cdot (-\sqrt{5}) = -5\).
\(6 - 2\sqrt{5} + 3\sqrt{5} - 5\). Numbers: \(6 - 5 = 1\). Radicals: \(-2\sqrt{5} + 3\sqrt{5} = \sqrt{5}\). Answer: \(1 + \sqrt{5}\).
Which expression is equal to \((\sqrt{3}+\sqrt{2})^2\)?
- \(5 + \sqrt{6}\)
- \(13 + 2\sqrt{6}\)
- \(5 + 2\sqrt{6}\)
- \(5 + 2\sqrt{5}\)
Write it as two factors: \((\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2})\).
FOIL: \(\sqrt{3}\cdot\sqrt{3} = 3\), \(\sqrt{3}\cdot\sqrt{2} = \sqrt{6}\), \(\sqrt{2}\cdot\sqrt{3} = \sqrt{6}\), \(\sqrt{2}\cdot\sqrt{2} = 2\).
\(3 + \sqrt{6} + \sqrt{6} + 2 = 5 + 2\sqrt{6}\). Answer: (c).
(a) counts the middle term only once. (b) squares 3 and 2 instead of \(\sqrt{3}\) and \(\sqrt{2}\). (d) adds the radicands; \(\sqrt{3}\cdot\sqrt{2}\) is \(\sqrt{6}\), not \(\sqrt{5}\).
A rectangle is \(3 + \sqrt{2}\) inches long and \(3 - \sqrt{2}\) inches wide. What is its area, in square inches?
Area = length × width = \((3+\sqrt{2})(3-\sqrt{2})\).
This is a conjugate pair: \(3^2 - (\sqrt{2})^2 = 9 - 2 = 7\). Answer: 7 square inches.
Step 5 of 5Wrap-up
You can now combine like radicals (after simplifying them), multiply radicals by multiplying outside and inside numbers, and use FOIL and conjugates with square roots.
Read or watch next: OpenStax Intermediate Algebra 2e, 8.4 Add, Subtract, and Multiply Radical Expressions (free textbook, with more practice at the end of the section).
Also good: Khan Academy: Algebra 1, Simplifying square roots (short videos). Quick rules: Radicals and complex numbers cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "Why can I multiply \(\sqrt{2}\) and \(\sqrt{3}\) but not add them?" It knows this course and can explain things another way.