Algebra

Operations with Radicals

Last updated: March 2026 · Intermediate
Before you start

You should be comfortable with:

Simplifying Radicals
Real-world applications
⚡
Electrical

Voltage drop, wire sizing, load balancing

Once you can simplify individual radicals, the next step is learning to combine them — adding, subtracting, multiplying, and dividing radical expressions. These operations follow specific rules, and many of them require you to simplify first before you can combine terms. The skills in this section show up constantly in the Pythagorean theorem, quadratic formula, and electrical formulas involving impedance.

Adding and Subtracting Radicals

You can only add or subtract radicals that have the same radicand (the expression under the radical sign). These are called like radicals, and they work just like like terms in algebra.

ac+bc=(a+b)ca\sqrt{c} + b\sqrt{c} = (a + b)\sqrt{c}

ac−bc=(a−b)ca\sqrt{c} - b\sqrt{c} = (a - b)\sqrt{c}

Think of it this way: 35+25=553\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}, just as 3x+2x=5x3x + 2x = 5x. The 5\sqrt{5} acts like a variable — you combine the coefficients.

Example 1: Simplify 73+437\sqrt{3} + 4\sqrt{3}

Both terms have 3\sqrt{3}. Add the coefficients:

73+43=(7+4)3=1137\sqrt{3} + 4\sqrt{3} = (7 + 4)\sqrt{3} = 11\sqrt{3}

Answer: 11311\sqrt{3}

Example 2: Simplify 92−529\sqrt{2} - 5\sqrt{2}

92−52=(9−5)2=429\sqrt{2} - 5\sqrt{2} = (9 - 5)\sqrt{2} = 4\sqrt{2}

Answer: 424\sqrt{2}

Example 3: Can you simplify 35+273\sqrt{5} + 2\sqrt{7}?

The radicands are different (5 vs. 7). These are not like radicals.

Answer: 35+273\sqrt{5} + 2\sqrt{7} — cannot be simplified further.

Simplify Before Combining

Often, radicals that look unlike become like radicals after simplification.

Example 4: Simplify 50+18\sqrt{50} + \sqrt{18}

Step 1 — Simplify each radical:

50=25×2=52\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}

18=9×2=32\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}

Step 2 — Now both are like radicals (2\sqrt{2}). Combine:

52+32=825\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}

Answer: 828\sqrt{2}

Example 5: Simplify 312−75+273\sqrt{12} - \sqrt{75} + \sqrt{27}

Simplify each term:

  • 312=3⋅23=633\sqrt{12} = 3 \cdot 2\sqrt{3} = 6\sqrt{3}
  • 75=53\sqrt{75} = 5\sqrt{3}
  • 27=33\sqrt{27} = 3\sqrt{3}

Combine:

63−53+33=(6−5+3)3=436\sqrt{3} - 5\sqrt{3} + 3\sqrt{3} = (6 - 5 + 3)\sqrt{3} = 4\sqrt{3}

Answer: 434\sqrt{3}

Example 6: Simplify 20+45−80\sqrt{20} + \sqrt{45} - \sqrt{80}

  • 20=25\sqrt{20} = 2\sqrt{5}
  • 45=35\sqrt{45} = 3\sqrt{5}
  • 80=45\sqrt{80} = 4\sqrt{5}

25+35−45=(2+3−4)5=15=52\sqrt{5} + 3\sqrt{5} - 4\sqrt{5} = (2 + 3 - 4)\sqrt{5} = 1\sqrt{5} = \sqrt{5}

Answer: 5\sqrt{5}

Multiplying Radicals

Use the product property of radicals:

a⋅b=ab\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}

Multiply the radicands together, then simplify the result.

Example 7: Multiply 3⋅12\sqrt{3} \cdot \sqrt{12}

3⋅12=3×12=36=6\sqrt{3} \cdot \sqrt{12} = \sqrt{3 \times 12} = \sqrt{36} = 6

Answer: 66

Example 8: Multiply 5⋅15\sqrt{5} \cdot \sqrt{15}

5⋅15=75=25×3=53\sqrt{5} \cdot \sqrt{15} = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}

Answer: 535\sqrt{3}

Example 9: Multiply 26⋅3102\sqrt{6} \cdot 3\sqrt{10}

Multiply the coefficients together, then multiply the radicands together:

2⋅3⋅6×10=6602 \cdot 3 \cdot \sqrt{6 \times 10} = 6\sqrt{60}

Simplify: 60=4×1560 = 4 \times 15.

660=6⋅215=12156\sqrt{60} = 6 \cdot 2\sqrt{15} = 12\sqrt{15}

Answer: 121512\sqrt{15}

Multiplying Using the Distributive Property

When one factor has multiple terms, distribute just as you would with variables.

Example 10: Multiply 3(4+6)\sqrt{3}(4 + \sqrt{6})

3⋅4+3⋅6=43+18\sqrt{3} \cdot 4 + \sqrt{3} \cdot \sqrt{6} = 4\sqrt{3} + \sqrt{18}

Simplify 18=32\sqrt{18} = 3\sqrt{2}:

43+324\sqrt{3} + 3\sqrt{2}

These are unlike radicals, so this is the final answer.

Answer: 43+324\sqrt{3} + 3\sqrt{2}

Example 11: Multiply 25(35−10)2\sqrt{5}(3\sqrt{5} - \sqrt{10})

25⋅35−25⋅102\sqrt{5} \cdot 3\sqrt{5} - 2\sqrt{5} \cdot \sqrt{10}

=625−250= 6\sqrt{25} - 2\sqrt{50}

=6(5)−2(52)= 6(5) - 2(5\sqrt{2})

=30−102= 30 - 10\sqrt{2}

Answer: 30−10230 - 10\sqrt{2}

FOIL with Radicals

When multiplying two binomials that contain radicals, use FOIL (First, Outer, Inner, Last).

Example 12: Multiply (3+2)(4−2)(3 + \sqrt{2})(4 - \sqrt{2})

First: 3⋅4=123 \cdot 4 = 12

Outer: 3⋅(−2)=−323 \cdot (-\sqrt{2}) = -3\sqrt{2}

Inner: 2⋅4=42\sqrt{2} \cdot 4 = 4\sqrt{2}

Last: 2⋅(−2)=−4=−2\sqrt{2} \cdot (-\sqrt{2}) = -\sqrt{4} = -2

Combine:

12−32+42−2=10+212 - 3\sqrt{2} + 4\sqrt{2} - 2 = 10 + \sqrt{2}

Answer: 10+210 + \sqrt{2}

Example 13: Multiply (5+3)(5−3)(\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3})

This is a conjugate pair — the same two terms, but the sign between them is opposite. The result follows the difference of squares pattern:

(5)2−(3)2=5−3=2(\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2

Answer: 22

This conjugate property is extremely useful for rationalizing denominators.

Example 14: Expand (7+2)2(\sqrt{7} + 2)^2

Use the pattern (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2:

(7)2+2(7)(2)+22=7+47+4=11+47(\sqrt{7})^2 + 2(\sqrt{7})(2) + 2^2 = 7 + 4\sqrt{7} + 4 = 11 + 4\sqrt{7}

Answer: 11+4711 + 4\sqrt{7}

Dividing Radicals

The quotient property of radicals is:

ab=ab(b≠0)\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} \quad (b \neq 0)

Example 15: Simplify 722\dfrac{\sqrt{72}}{\sqrt{2}}

722=722=36=6\frac{\sqrt{72}}{\sqrt{2}} = \sqrt{\frac{72}{2}} = \sqrt{36} = 6

Answer: 66

Example 16: Simplify 5010\dfrac{\sqrt{50}}{\sqrt{10}}

5010=5010=5\frac{\sqrt{50}}{\sqrt{10}} = \sqrt{\frac{50}{10}} = \sqrt{5}

Answer: 5\sqrt{5}

Example 17: Simplify 64535\dfrac{6\sqrt{45}}{3\sqrt{5}}

Divide the coefficients and use the quotient property:

63⋅455=2455=29=2(3)=6\frac{6}{3} \cdot \frac{\sqrt{45}}{\sqrt{5}} = 2\sqrt{\frac{45}{5}} = 2\sqrt{9} = 2(3) = 6

Answer: 66

Real-World Application: Electrician — Combining Impedance Calculations

In AC circuits, an electrician may need to combine impedance values that involve radicals. The impedance of two components might be:

Z1=32 ohmsandZ2=52 ohmsZ_1 = 3\sqrt{2} \text{ ohms} \quad \text{and} \quad Z_2 = 5\sqrt{2} \text{ ohms}

If these are in series (directly added):

Ztotal=32+52=82≈11.31 ohmsZ_{\text{total}} = 3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2} \approx 11.31 \text{ ohms}

In a more complex calculation, an electrician might encounter:

Z=R2+XL2Z = \sqrt{R^2 + X_L^2}

where R=6R = 6 ohms (resistance) and XL=6X_L = 6 ohms (inductive reactance):

Z=36+36=72=62≈8.49 ohmsZ = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \approx 8.49 \text{ ohms}

Being able to simplify 72\sqrt{72} to 626\sqrt{2} makes the formula result cleaner and easier to verify.

Common Mistakes to Avoid

  1. Adding unlike radicals. 3+5≠8\sqrt{3} + \sqrt{5} \neq \sqrt{8}. You cannot add radicands. Only like radicals (same radicand) can be combined.
  2. Forgetting to simplify before combining. 12+27\sqrt{12} + \sqrt{27} looks like unlike radicals, but after simplifying (23+332\sqrt{3} + 3\sqrt{3}), they combine to 535\sqrt{3}.
  3. Multiplying coefficients with radicands. In 23⋅452\sqrt{3} \cdot 4\sqrt{5}, multiply the coefficients together (2×4=82 \times 4 = 8) and the radicands together (3×5=153 \times 5 = 15) to get 8158\sqrt{15}. Do not mix them.
  4. Forgetting to simplify after multiplying. After computing a⋅b=ab\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}, always check if abab has a perfect square factor.
  5. Misapplying FOIL. Make sure you distribute every term to every other term. The inner and outer products are the ones most commonly forgotten.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Problem 1: Simplify 611−211+116\sqrt{11} - 2\sqrt{11} + \sqrt{11}.

All terms have 11\sqrt{11}:

(6−2+1)11=511(6 - 2 + 1)\sqrt{11} = 5\sqrt{11}

Answer: 5115\sqrt{11}

Problem 2: Simplify 32+50\sqrt{32} + \sqrt{50}.

32=42\sqrt{32} = 4\sqrt{2} and 50=52\sqrt{50} = 5\sqrt{2}

42+52=924\sqrt{2} + 5\sqrt{2} = 9\sqrt{2}

Answer: 929\sqrt{2}

Problem 3: Multiply 7⋅14\sqrt{7} \cdot \sqrt{14}.

7⋅14=98=49×2=72\sqrt{7} \cdot \sqrt{14} = \sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}

Answer: 727\sqrt{2}

Problem 4: Multiply (2+3)(2−3)(2 + \sqrt{3})(2 - \sqrt{3}).

This is a conjugate pair:

(2)2−(3)2=4−3=1(2)^2 - (\sqrt{3})^2 = 4 - 3 = 1

Answer: 11

Problem 5: Expand (6+1)2(\sqrt{6} + 1)^2.

(6)2+2(6)(1)+12=6+26+1=7+26(\sqrt{6})^2 + 2(\sqrt{6})(1) + 1^2 = 6 + 2\sqrt{6} + 1 = 7 + 2\sqrt{6}

Answer: 7+267 + 2\sqrt{6}

Problem 6: Simplify 2008\dfrac{\sqrt{200}}{\sqrt{8}}.

2008=2008=25=5\frac{\sqrt{200}}{\sqrt{8}} = \sqrt{\frac{200}{8}} = \sqrt{25} = 5

Answer: 55

Problem 7: An electrician calculates impedance as Z=R2+XC2Z = \sqrt{R^2 + X_C^2} with R=8R = 8 and XC=6X_C = 6. Find ZZ in simplified radical form.

Z=64+36=100=10 ohmsZ = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ ohms}

Answer: 1010 ohms (this is a Pythagorean triple: 6-8-10).

Key Takeaways

  • Like radicals (same radicand) can be added or subtracted by combining their coefficients — just like combining like terms
  • Always simplify first, then check whether terms can be combined
  • Multiplying radicals uses the product property: a⋅b=ab\sqrt{a} \cdot \sqrt{b} = \sqrt{ab} — then simplify
  • FOIL works for binomials containing radicals; conjugate pairs produce rational results with no radicals
  • Dividing radicals uses the quotient property: a/b=a/b\sqrt{a}/\sqrt{b} = \sqrt{a/b}
  • After every operation, check whether the result can be simplified further

Return to Algebra for more topics in this section.

Next Up in Algebra

Simplifying RadicalsAbsolute Value Equations and InequalitiesThe DiscriminantAdding and Subtracting Polynomials
All Algebra topics

Last updated: March 29, 2026