Adding and Multiplying Radicals

Adding like radicals

A radical behaves like a letter in algebra. $3\sqrt{2}$3√2 means $3\times \sqrt{2}$3 × √2, three lots of $\sqrt{2}$√2, just as 3x means three lots of x. So $3\sqrt{2}+5\sqrt{2}$3√2 + 5√2 is three lots of $\sqrt{2}$√2 and five more lots of $\sqrt{2}$√2, which make eight lots: $3\sqrt{2}+5\sqrt{2}=8\sqrt{2}$3√2 + 5√2 = 8√2, just as 3x + 5x = 8x.

Terms with the same radical are like terms. To add them, add the numbers in front and keep the radical as it is: $2\sqrt{3}+5\sqrt{3}=7\sqrt{3}$2√3 + 5√3 = 7√3. The root does not change, and the numbers in front are added, not multiplied.

Every part is one $\sqrt{2}$√2. The two bars hold 3 + 5 = 8 parts, so together they make $8\sqrt{2}$8√2.

Different radicals stay apart

$\sqrt{2}+\sqrt{3}$√2 + √3 cannot be written as one radical. It is not $\sqrt{5}$√5: $\sqrt{2}$√2 is about 1.41 and $\sqrt{3}$√3 is about 1.73, so their sum is about 3.15, while $\sqrt{5}$√5 is only about 2.24. Different radicals are not like terms, just as 2x + 3y cannot be collected, so $\sqrt{2}+\sqrt{3}$√2 + √3 stays as it is.

Sometimes simplifying first shows that two radicals are alike after all. $\sqrt{8}=\sqrt{4}\times \sqrt{2}=2\sqrt{2}$√8 = √4 × √2 = 2√2, so $\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}$√2 + √8 = √2 + 2√2 = 3√2.

$\sqrt{2}+\sqrt{3}$√2 + √3 is about 1.41 + 1.73 = 3.15, well past $\sqrt{5}$√5, which is about 2.24.

Multiplying joins the roots

Multiplying works differently from adding. The square root of a product is the product of the square roots, so two roots multiplied together make one root of the product: $\sqrt{2}\times \sqrt{8}=\sqrt{2\times 8}=\sqrt{16}$√2 × √8 = √(2 × 8) = √16. And $\sqrt{16}$√16 is exactly 4.

A number in front of a radical multiplies with the other numbers in front: $3\sqrt{2}\times 4\sqrt{5}=\left(3\times 4\right)\times \left(\sqrt{2}\times \sqrt{5}\right)=12\sqrt{10}$3√2 × 4√5 = (3 × 4) × (√2 × √5) = 12√10. And a radical times itself is the number under it: $\sqrt{7}\times \sqrt{7}=7$√7 × √7 = 7.

Expanding two brackets

Brackets that hold radicals expand in the usual way: multiply each term in the first bracket by each term in the second. $\left(1+\sqrt{3}\right)\left(2-\sqrt{3}\right)$(1 + √3)(2 − √3) gives four products.

1 × 2 = 2 and $1\times \left(-\sqrt{3}\right)=-\sqrt{3}$1 × (−√3) = −√3. Then $\sqrt{3}\times 2=2\sqrt{3}$√3 × 2 = 2√3, and $\sqrt{3}\times \left(-\sqrt{3}\right)=-3$√3 × (−√3) = −3, because $\sqrt{3}\times \sqrt{3}=3$√3 × √3 = 3.

Each piece is one term of the first bracket times one term of the second: 2, $-\sqrt{3}$−√3, $2\sqrt{3}$2√3 and −3.

Collect the like terms

Now collect the four pieces: $2-\sqrt{3}+2\sqrt{3}-3$2 − √3 + 2√3 − 3. The whole numbers collect, 2 − 3 = −1. The $\sqrt{3}$√3 terms collect as like terms, $-\sqrt{3}+2\sqrt{3}=\sqrt{3}$−√3 + 2√3 = √3. So $\left(1+\sqrt{3}\right)\left(2-\sqrt{3}\right)=\sqrt{3}-1$(1 + √3)(2 − √3) = √3 − 1.

The −1 and the $\sqrt{3}$√3 are not like terms, so $\sqrt{3}-1$√3 − 1 cannot be simplified any further.

Common mistakes

Adding like radicals adds the numbers in front; it does not multiply them, and it does not change the root. $2\sqrt{5}+3\sqrt{5}=5\sqrt{5}$2√5 + 3√5 = 5√5, not $6\sqrt{5}$6√5 and not $5\sqrt{10}$5√10.

Multiplying puts the product under the root, not the sum. $\sqrt{3}\times \sqrt{12}=\sqrt{36}=6$√3 × √12 = √36 = 6, not $\sqrt{15}$√15.

When you expand, keep all four pieces and the sign of each. Taking only the whole-number pieces 2 and −3 loses the two $\sqrt{3}$√3 terms.