Radicals

Study Sheet

Radicals

Square roots, simplifying, operations, and radical equations

Square Roots & Perfect Squares

Concept
The Big Idea

A square root of a number nn is a value that, when multiplied by itself, gives nn. Because 5×5=255\times 5 = 25, we say 55 is a square root of 2525.

The symbol x\sqrt{\phantom{x}} is called the radical sign. The number under it is the radicand. The small number tucked into the notch is the index (it is a hidden 22 for square roots).

radicand index 25=252=5\sqrt[\,\text{index}\,]{\text{radicand}} \qquad\qquad \sqrt{25}=\sqrt[2]{25}=5

A perfect square is a number whose square root is a whole number (like 1,4,9,16,…1,4,9,16,\dots).

A square root is the positive number whose square is the radicand. Simplify by pulling perfect-square factors out, add only like radicals, multiply radicands together, and rationalize a denominator by multiplying by a well-chosen 11. Solving radical equations means isolating the root and squaring — then checking, because squaring can invent solutions.

The square root is the side of a square with a given area, which is why it is called a square root: 8181 unit tiles make a 99-by-99 square. The radical sign names the principal (non-negative) root because a side length cannot be negative; the negative root, −9-9, is written −81-\sqrt{81} when it is wanted.

Concept
Principal Root

Every positive number really has two square roots: 5×5=255\times 5 = 25 and (−5)×(−5)=25(-5)\times(-5)=25. The symbol  \sqrt{\,} always means the principal (positive) root.

25=5,−25=−5,±25=±5\sqrt{25}=5,\qquad -\sqrt{25}=-5,\qquad \pm\sqrt{25}=\pm 5

Also, 0=0\sqrt{0}=0, and you cannot take a negative number\sqrt{\text{a negative number}} in Algebra 1 (there is no real number that squares to a negative).

Example
Reading a radical
area 81 = 9 × 9, so √81 = 99
Eighty-one tiles form a 99-by-99 square.

Evaluate 81\sqrt{81}. Ask: what positive number times itself is 8181? Since 9×9=819\times 9 = 81, we get 81=9\sqrt{81}=9.

Side note
Where it goes wrong

Believing a+b=a+b\sqrt{a + b} = \sqrt a + \sqrt b. Test it: 9+16=5\sqrt{9 + 16} = 5, but 3+4=73 + 4 = 7. Roots split over products, never over sums.

Perfect Squares to Know by Heart

1.3

Tip

Remember: n2=n\sqrt{n^2}=n for a positive number nn. Squaring and taking a square root are opposite (inverse) operations.

Try it
Try it: roots

Evaluate 169\sqrt{169}, −0.49-\sqrt{0.49}, 2564\sqrt{\tfrac{25}{64}}, and (−6)2\sqrt{(-6)^2}.

Show the answer

1313. −0.7-0.7. 58\tfrac58. (−6)2=36=6\sqrt{(-6)^2} = \sqrt{36} = 6 (not −6-6: the principal root is non-negative).

Estimating Irrational Square Roots

Concept
Trapped Between Two Integers

If a number is not a perfect square, its square root is irrational (a never-ending, non-repeating decimal). We estimate it by finding the two nearest perfect squares.

Larger squares have larger sides, so square roots keep order: 49<50<6449 < 50 < 64 forces 7<50<87 < \sqrt{50} < 8. To go further, square a guess: 7.12=50.417.1^2 = 50.41 is a little too big and 7.072=49.987.07^2 = 49.98 a little too small, so 50\sqrt{50} lies between them. Squaring is easy, so it is the tool for checking any root estimate.

Example
Estimating 50\sqrt{50}
7.07.58.0√50√49√64
50≈7.07\sqrt{50} \approx 7.07, just past 77.

Find perfect squares just below and above 5050:

49<50<64⟹49<50<64⟹7<50<8.49 < 50 < 64 \quad\Longrightarrow\quad \sqrt{49} < \sqrt{50} < \sqrt{64} \quad\Longrightarrow\quad 7 < \sqrt{50} < 8.

Since 5050 is very close to 4949, the value 50≈7.1\sqrt{50}\approx 7.1 is a good estimate.

Side note
Where it goes wrong

Placing 50\sqrt{50} proportionally at "7+1157 + \tfrac{1}{15}" and then trusting more decimal places than that estimate supports. It gives about 7.0677.067; the true value is 7.0711…7.0711\ldots Squares are not evenly spaced, so check a proportional estimate by squaring.

Tip

Tip: The bigger the radicand, the bigger the root, but the root grows slowly. Whenever the radicand sits closer to the lower perfect square, the root sits closer to the lower integer.

Try it
Try it: estimate

Between which consecutive integers is 130\sqrt{130}? Estimate it to one decimal place.

Show the answer

112=121<130<144=12211^2 = 121 < 130 < 144 = 12^2, so 11<130<1211 < \sqrt{130} < 12. Test 11.42=129.9611.4^2 = 129.96 and 11.52=132.2511.5^2 = 132.25: 130130 is just above 129.96129.96, so 130≈11.4\sqrt{130} \approx 11.4.

Simplifying Square Roots (Product Property)

Concept
Product Property of Radicals

For non-negative numbers aa and bb:

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

To simplify, factor the radicand so that one factor is the largest perfect square you can find, then pull its root out front.

The product property holds because (ab)2=(a)2(b)2=ab(\sqrt a \sqrt b)^2 = (\sqrt a)^2(\sqrt b)^2 = ab, so ab\sqrt a \sqrt b is a non-negative number whose square is abab, which is exactly what ab\sqrt{ab} means. Pulling out the largest perfect-square factor finishes in one step; any smaller square factor leaves another to pull out later.

Example
Simplifying 12\sqrt{12}
12=4⋅3(4 is the largest perfect-square factor)=4⋅3product property=23.4=2\begin{aligned} \sqrt{12} &= \sqrt{4\cdot 3} && \text{($4$ is the largest perfect-square factor)}\\ &= \sqrt{4}\cdot\sqrt{3} && \text{product property}\\ &= 2\sqrt{3}. && \text{$\sqrt{4}=2$} \end{aligned}
Side note
Where it goes wrong

Pulling out the factor itself instead of its root: 72=36⋅2=362\sqrt{72} = \sqrt{36 \cdot 2} = 36\sqrt2. Only the square root of the perfect square comes out: 36=6\sqrt{36} = 6, so 626\sqrt2. Check: (62)2=72(6\sqrt2)^2 = 72, while (362)2=2592(36\sqrt2)^2 = 2592.

Example
Simplifying 72\sqrt{72}
66√72 = 6√2
Legs 66 and 66: the hypotenuse 36+36=72=62\sqrt{36 + 36} = \sqrt{72} = 6\sqrt2.
72=36⋅2=36⋅2=62.\begin{aligned} \sqrt{72} &= \sqrt{36\cdot 2} = \sqrt{36}\cdot\sqrt{2} = 6\sqrt{2}. \end{aligned}

If you had used a smaller factor, say 72=9⋅8=38\sqrt{72}=\sqrt{9\cdot 8}=3\sqrt{8}, you are not done: 8=22\sqrt{8}=2\sqrt2, so 38=3⋅22=623\sqrt{8}=3\cdot 2\sqrt2 = 6\sqrt2. Same answer, more steps.

Tip

Fully simplified means the radicand has no perfect-square factors left (other than 11).

Try it
Try it: simplify

Simplify 98\sqrt{98}, 180\sqrt{180} and 3403\sqrt{40}.

Show the answer

49⋅2=72\sqrt{49 \cdot 2} = 7\sqrt2. 36⋅5=65\sqrt{36 \cdot 5} = 6\sqrt5. 34⋅10=3⋅210=6103\sqrt{4 \cdot 10} = 3 \cdot 2\sqrt{10} = 6\sqrt{10}.

Radicals with Variables

Concept
Even Powers Come Out Clean

Since x2=x\sqrt{x^2}=x (for x≥0x\ge 0), any even power under a square root leaves half its exponent outside:

x6=x3,x10=x5.\sqrt{x^{6}}=x^{3},\qquad \sqrt{x^{10}}=x^{5}.

For an odd power, split off one factor: x5=x4⋅xx^{5}=x^{4}\cdot x, so x5=x2x\sqrt{x^{5}}=x^{2}\sqrt{x}.

A square root halves an exponent because squaring doubles it: (x3)2=x6(x^3)^2 = x^6, so x6=x3\sqrt{x^6} = x^3 (for x≥0x \ge 0). An odd power has one factor left over after pairing, and that single factor stays under the radical.

Example
Simplifying 50x3\sqrt{50x^{3}}
50x3=25⋅2⋅x2⋅x=25⋅x2⋅2x=5x2x.\begin{aligned} \sqrt{50x^{3}} &= \sqrt{25\cdot 2\cdot x^{2}\cdot x}\\ &= \sqrt{25}\cdot\sqrt{x^{2}}\cdot\sqrt{2x}\\ &= 5x\sqrt{2x}. \end{aligned}

Pull out the perfect-square number (25→525\to 5) and the even power (x2→xx^2\to x); leave the rest inside.

Side note
Where it goes wrong

Taking the square root of an exponent instead of halving it: x16=x4\sqrt{x^{16}} = x^4. Halve: x8x^8, since (x8)2=x16(x^8)^2 = x^{16}, while (x4)2=x8(x^4)^2 = x^8.

Try it
Try it: variables

Simplify 18a5b2\sqrt{18a^5b^2} for a,b≥0a, b \ge 0.

Show the answer

18=9⋅218 = 9 \cdot 2, a5=a4⋅aa^5 = a^4 \cdot a, b2b^2 is a perfect square. Pull out 9=3\sqrt9 = 3, a4=a2\sqrt{a^4} = a^2, b2=b\sqrt{b^2} = b: 3a2b2a3a^2b\sqrt{2a}.

Adding & Subtracting Radicals

Concept
Combine LIKE Radicals

Radicals are like when they have the same index and the same radicand. Add or subtract only the numbers in front (the coefficients); the radical stays the same, just like combining like terms.

23+53=73.2\sqrt{3}+5\sqrt{3}=7\sqrt{3}.

Often you must simplify first before you can tell that radicals are alike.

Like radicals combine the same way like terms do, by the distributive property: 23+33=(2+3)32\sqrt3 + 3\sqrt3 = (2 + 3)\sqrt3. Here 3\sqrt3 plays the role of xx. Unlike radicals, such as 2\sqrt2 and 5\sqrt5, are different "units" and cannot be merged, just as 2x+3y2x + 3y cannot.

Example
Simplify first, then combine
0123456789√3√3√3√3√35√3
Two steps of 3\sqrt3 plus three more: five steps, 53≈8.665\sqrt3 \approx 8.66.
12+27=23+33(12=23, 27=33)=53.\begin{aligned} \sqrt{12}+\sqrt{27} &= 2\sqrt{3}+3\sqrt{3} && \text{($\sqrt{12}=2\sqrt3,\ \sqrt{27}=3\sqrt3$)}\\ &= 5\sqrt{3}. \end{aligned}
Side note
Where it goes wrong

Adding radicands: 12+27=39\sqrt{12} + \sqrt{27} = \sqrt{39}. Numerically 39≈6.24\sqrt{39} \approx 6.24, but 12+27≈3.46+5.20=8.66\sqrt{12} + \sqrt{27} \approx 3.46 + 5.20 = 8.66. Simplify each first: 23+33=532\sqrt3 + 3\sqrt3 = 5\sqrt3.

Tip

Warning: a+b≠a+b\sqrt{a}+\sqrt{b}\neq\sqrt{a+b}. For example 9+16=3+4=7\sqrt{9}+\sqrt{16}=3+4=7, but 9+16=25=5\sqrt{9+16}=\sqrt{25}=5. You can only combine radicals that are already alike; unlike radicals such as 22+352\sqrt2+3\sqrt5 stay separate.

Try it
Try it: combine

Simplify 50−18+8\sqrt{50} - \sqrt{18} + \sqrt{8}.

Show the answer

50=52\sqrt{50} = 5\sqrt2, 18=32\sqrt{18} = 3\sqrt2, 8=22\sqrt8 = 2\sqrt2. All are multiples of 2\sqrt2: 5−3+2=45 - 3 + 2 = 4, so 424\sqrt2.

Multiplying Radicals

Concept
Multiply Outsides, Multiply Insides
(am)(bn)=abmn.\big(a\sqrt{m}\big)\big(b\sqrt{n}\big)=ab\sqrt{mn}.

Multiply the coefficients together and the radicands together, then simplify the result.

Multiplying radicals uses the product property in the combining direction: m⋅n=mn\sqrt m \cdot \sqrt n = \sqrt{mn}, and coefficients multiply as usual because multiplication can be rearranged. Conjugates, (7−2)(7+2)(\sqrt7 - \sqrt2)(\sqrt7 + \sqrt2), are the difference-of-squares pattern, and they are useful precisely because squaring a square root removes it.

Example
A basic product
(26)(32)=612=6⋅23=123.\big(2\sqrt{6}\big)\big(3\sqrt{2}\big)=6\sqrt{12}=6\cdot 2\sqrt{3}=12\sqrt{3}.
Side note
Where it goes wrong

Multiplying a coefficient into the radicand: 32⋅5=303\sqrt2 \cdot \sqrt5 = \sqrt{30}. Coefficients stay outside: 3103\sqrt{10}. Check numerically: 4.24×2.24≈9.49=3104.24 \times 2.24 \approx 9.49 = 3\sqrt{10}, while 30≈5.48\sqrt{30} \approx 5.48.

Example
Distributing and the difference of squares

Distribute: 3(3+5)=9+15=3+15\sqrt{3}\big(\sqrt{3}+\sqrt{5}\big)=\sqrt{9}+\sqrt{15}=3+\sqrt{15}.

Difference of squares (a−b)(a+b)=a2−b2(a-b)(a+b)=a^2-b^2:

(7−2)(7+2)=(7)2−(2)2=7−2=5.\big(\sqrt{7}-\sqrt{2}\big)\big(\sqrt{7}+\sqrt{2}\big)=\big(\sqrt7\big)^2-\big(\sqrt2\big)^2=7-2=5.

The radicals vanish! This trick powers rationalizing below.

Tip

Key fact: (a)2=a\big(\sqrt{a}\big)^{2}=a. Squaring a square root undoes it.

Try it
Try it: multiply

Multiply (2+5)(3−5)(2 + \sqrt5)(3 - \sqrt5) and (43)2(4\sqrt3)^2.

Show the answer

FOIL: 6−25+35−5=1+56 - 2\sqrt5 + 3\sqrt5 - 5 = 1 + \sqrt5. (43)2=16⋅3=48(4\sqrt3)^2 = 16 \cdot 3 = 48.

Dividing & Rationalizing

Concept
Quotient Property & Rationalizing
ab=ab(b>0).\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\quad(b>0).

A denominator should never keep a radical. To rationalize, multiply the top and bottom by the radical in the denominator so it becomes a perfect square.

Rationalizing multiplies by a clever form of 11, 22\tfrac{\sqrt2}{\sqrt2}, so the value does not change, only its appearance. The reason to do it is practical: 22\tfrac{\sqrt2}{2} is easy to compare, combine and estimate (≈0.707\approx 0.707), while dividing by an irrational number is awkward, and it makes equal answers look identical.

Example
Rationalizing 12\dfrac{1}{\sqrt{2}}
12=12⋅22=24=22.\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{4}}=\frac{\sqrt{2}}{2}.
Side note
Where it goes wrong

Multiplying only the denominator: 35=35\tfrac{3}{\sqrt5} = \tfrac{3}{5} ("the root cancels"). That changes the value; the top must be multiplied too: 355≈1.34\tfrac{3\sqrt5}{5} \approx 1.34, matching 35≈1.34\tfrac{3}{\sqrt5} \approx 1.34.

Example
A quotient that simplifies
63=63=2.205=4=2.\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{\frac{6}{3}}=\sqrt{2}. \qquad \frac{\sqrt{20}}{\sqrt{5}}=\sqrt{4}=2.
Try it
Try it: rationalize

Rationalize 63\dfrac{6}{\sqrt3} and 42x\dfrac{4}{\sqrt{2x}} (x>0x > 0).

Show the answer

63⋅33=633=23\tfrac{6}{\sqrt3} \cdot \tfrac{\sqrt3}{\sqrt3} = \tfrac{6\sqrt3}{3} = 2\sqrt3. 42x⋅2x2x=42x2x=22xx\tfrac{4}{\sqrt{2x}} \cdot \tfrac{\sqrt{2x}}{\sqrt{2x}} = \tfrac{4\sqrt{2x}}{2x} = \tfrac{2\sqrt{2x}}{x}.

The Pythagorean Theorem

Concept
a2+b2=c2a^2+b^2=c^2

In a right triangle, the two shorter sides (legs) aa and bb and the longest side (hypotenuse) cc satisfy

a2+b2=c2.a^{2}+b^{2}=c^{2}.

Solve for the missing side, then leave the answer in simplified radical form unless a decimal is asked for.

Most right triangles have irrational sides, which is why radicals and the Pythagorean theorem arrive together: legs 22 and 33 give a hypotenuse whose square is 1313, and no fraction squares to 1313. Simplified radical form keeps the exact value instead of a rounded decimal.

Example
Finding a hypotenuse

Legs a=2a=2, b=3b=3:

c2=22+32=4+9=13c=13.(13 has no perfect-square factor)\begin{aligned} c^{2}&=2^{2}+3^{2}=4+9=13\\ c&=\sqrt{13}. && \text{($13$ has no perfect-square factor)} \end{aligned}
Side note
Where it goes wrong

Adding when a leg is missing: b2=82+42=80b^2 = 8^2 + 4^2 = 80. The hypotenuse is the longest side, so the leg must be shorter than 88; subtract: b2=64−16=48b^2 = 64 - 16 = 48.

Example
Finding a leg
b = 4√348
Hypotenuse 88, leg 44: the other leg is 48=43≈6.93\sqrt{48} = 4\sqrt3 \approx 6.93.

Hypotenuse c=8c=8, leg a=4a=4:

42+b2=8216+b2=64b2=48b=48=43.\begin{aligned} 4^{2}+b^{2}&=8^{2}\\ 16+b^{2}&=64\\ b^{2}&=48\\ b&=\sqrt{48}=4\sqrt{3}. \end{aligned}
Try it
Try it: exact form

A square has diagonal 1010. Find its side in simplified radical form.

Show the answer

The side ss and the diagonal form a right triangle with two legs ss: s2+s2=100s^2 + s^2 = 100, so s2=50s^2 = 50 and s=50=52≈7.07s = \sqrt{50} = 5\sqrt2 \approx 7.07.

Cube Roots (A Brief Look)

Concept
Index of 33

A cube root n3\sqrt[3]{n} asks: what number cubed gives nn? Because 23=82^3=8, we get 83=2\sqrt[3]{8}=2. Unlike square roots, cube roots of negatives are allowed: −273=−3\sqrt[3]{-27}=-3 since (−3)3=−27(-3)^3=-27.

Perfect cubes: 1, 8, 27, 64, 1251,\,8,\,27,\,64,\,125. Simplify by pulling out perfect-cube factors.

Cube roots allow negative radicands because cubing keeps the sign: (−3)3=−27(-3)^3 = -27, so −273=−3\sqrt[3]{-27} = -3. Square roots cannot, because every real number squared is non-negative. Simplifying uses the largest perfect-cube factor, the same idea as perfect squares.

Example
Simplifying 543\sqrt[3]{54}
543=27⋅23=273⋅23=323.\sqrt[3]{54}=\sqrt[3]{27\cdot 2}=\sqrt[3]{27}\cdot\sqrt[3]{2}=3\sqrt[3]{2}.

Solving Square-Root Equations

Concept
Isolate, Square, Check

To solve an equation with a variable under a radical:

  • Isolate the radical on one side.
  • Square both sides to remove the radical.
  • Solve the resulting equation.
  • Check every answer in the original equation.

Squaring both sides is a one-way step: if a=ba = b then a2=b2a^2 = b^2, but a2=b2a^2 = b^2 also allows a=−ba = -b. So squaring can admit new "solutions" that were never solutions of the original, which is why every answer must be checked. On a graph, x+3\sqrt x + 3 is always at least 33, so an equation like x+3=1\sqrt x + 3 = 1 has no solution even though squaring produces one.

Example
Solving x+3=7\sqrt{x}+3=7
xy2468101214161812345678(16, 7)
y=x+3y = \sqrt x + 3 meets y=7y = 7 once, at x=16x = 16.
x+3=7x=4isolatex=16square both sides\begin{aligned} \sqrt{x}+3&=7\\ \sqrt{x}&=4 && \text{isolate}\\ x&=16 && \text{square both sides} \end{aligned}

Check: 16+3=4+3=7\sqrt{16}+3=4+3=7. ✓

Side note
Where it goes wrong

Squaring before isolating: x+3=7\sqrt x + 3 = 7 squared as x+9=49x + 9 = 49, giving x=40x = 40. (x+3)2(\sqrt x + 3)^2 is x+6x+9x + 6\sqrt x + 9, not x+9x + 9. Isolate the root first.

Tip

Why check? Squaring can create extraneous solutions, answers that appear valid but fail the original equation. For instance, x=−2\sqrt{x}=-2 has no solution (a principal root is never negative), even though squaring gives x=4x=4. Always test your answers.

Try it
Try it: solve and check

Solve 2x+3=x\sqrt{2x + 3} = x.

Show the answer

Square: 2x+3=x22x + 3 = x^2, so x2−2x−3=0x^2 - 2x - 3 = 0 and (x−3)(x+1)=0(x - 3)(x + 1) = 0. Check x=3x = 3: 9=3\sqrt9 = 3, true. Check x=−1x = -1: 1=1≠−1\sqrt1 = 1 \ne -1, extraneous. Solution: x=3x = 3.

Going Deeper: Advanced Radicals

Concept
nnth Roots and Rational Exponents

The index does not have to be 22 or 33. In general, an\sqrt[n]{a} asks for the number whose nnth power is aa. There is a powerful bridge between roots and exponents:

a1/n=an,am/n=(an)m=amn.a^{1/n}=\sqrt[n]{a},\qquad\quad a^{m/n}=\big(\sqrt[n]{a}\big)^{m}=\sqrt[n]{a^{m}}.

The denominator of the exponent is the index (the root); the numerator is the ordinary power. Once a radical is written this way, every exponent rule you already know applies.

82/3=(83)2=22=4,163/4=(164)3=23=8.8^{2/3}=\big(\sqrt[3]{8}\big)^{2}=2^{2}=4,\qquad 16^{3/4}=\big(\sqrt[4]{16}\big)^{3}=2^{3}=8.
Example
Rational exponents at work

Simplify x1/2⋅x1/3x1/6\dfrac{x^{1/2}\cdot x^{1/3}}{x^{1/6}} and write the answer as a radical.

x1/2⋅x1/3x1/6=x 1/2+1/3−1/6add/subtract exponents=x 3/6+2/6−1/6common denominator 6=x4/6=x2/3reduce=x23.denominator is the index\begin{aligned} \frac{x^{1/2}\cdot x^{1/3}}{x^{1/6}} &= x^{\,1/2+1/3-1/6} && \text{add/subtract exponents}\\ &= x^{\,3/6+2/6-1/6} && \text{common denominator }6\\ &= x^{4/6}=x^{2/3} && \text{reduce}\\ &= \sqrt[3]{x^{2}}. && \text{denominator is the index} \end{aligned}
Tip

Even vs. odd index. An odd index (like  3\sqrt[3]{\ } or  5\sqrt[5]{\ }) accepts negative radicands; an even index (like  \sqrt{\ } or  4\sqrt[4]{\ }) does not allow a negative radicand in the real numbers.

Concept
Rationalizing with a Conjugate

When a denominator is a sum or difference of terms involving radicals, one factor is not enough. Multiply top and bottom by the conjugate, the same two terms with the middle sign flipped. The difference of squares (a−b)(a+b)=a2−b2(a-b)(a+b)=a^{2}-b^{2} then erases the radical:

(a+b)(a−b)=a−b.\big(\sqrt{a}+\sqrt{b}\big)\big(\sqrt{a}-\sqrt{b}\big)=a-b.

The conjugate of 3+53+\sqrt5 is 3−53-\sqrt5; the conjugate of 7−2\sqrt7-\sqrt2 is 7+2\sqrt7+\sqrt2.

Example
Rationalizing a two-term denominator

Rationalize 43−5\dfrac{4}{3-\sqrt{5}}.

43−5=43−5⋅3+53+5multiply by the conjugate=4(3+5)32−(5)2difference of squares=4(3+5)9−5=4(3+5)4=3+5.\begin{aligned} \frac{4}{3-\sqrt5} &= \frac{4}{3-\sqrt5}\cdot\frac{3+\sqrt5}{3+\sqrt5} && \text{multiply by the conjugate}\\ &= \frac{4\big(3+\sqrt5\big)}{3^{2}-\big(\sqrt5\big)^{2}} && \text{difference of squares}\\ &= \frac{4\big(3+\sqrt5\big)}{9-5}=\frac{4\big(3+\sqrt5\big)}{4}\\ &= 3+\sqrt5. \end{aligned}

No radical remains in the denominator, so the expression is fully rationalized.

Concept
Denesting a Nested Radical

A radical inside a radical such as 7+43\sqrt{7+4\sqrt3} can sometimes be denested into a+b\sqrt{a}+\sqrt{b}. The idea: assume 7+43=a+b\sqrt{7+4\sqrt3}=\sqrt{a}+\sqrt{b} and square both sides:

7+43=a+b+2ab.7+4\sqrt3=a+b+2\sqrt{ab}.

Match the parts without a radical to each other and the parts with a radical to each other:

a+b=7,2ab=43 ⇒ ab=12.a+b=7,\qquad 2\sqrt{ab}=4\sqrt3\ \Rightarrow\ ab=12.

Find two numbers that add to 77 and multiply to 1212: they are 44 and 33. Hence

7+43=4+3=2+3.\sqrt{7+4\sqrt3}=\sqrt4+\sqrt3=2+\sqrt3.
Tip

Check by squaring. Denesting is only valid if it survives a check: (2+3)2=4+43+3=7+43\big(2+\sqrt3\big)^{2}=4+4\sqrt3+3=7+4\sqrt3. ✓ If no nice pair of numbers exists, the radical simply cannot be denested.

Example
Solving a radical equation with an extraneous root

Solve x+7=x+1\sqrt{x+7}=x+1.

x+7=x+1x+7=(x+1)2square both sidesx+7=x2+2x+10=x2+x−6collect on one side0=(x+3)(x−2)factor\begin{aligned} \sqrt{x+7}&=x+1\\ x+7&=(x+1)^{2} && \text{square both sides}\\ x+7&=x^{2}+2x+1\\ 0&=x^{2}+x-6 && \text{collect on one side}\\ 0&=(x+3)(x-2) && \text{factor} \end{aligned}

So x=−3x=-3 or x=2x=2. Check both in the original equation:

x=2: 2+7=9=3,2+1=3. ✓x=2:\ \sqrt{2+7}=\sqrt9=3,\quad 2+1=3.\ \checkmark
x=−3: −3+7=4=2,−3+1=−2. But 2≠−2. (rejected)x=-3:\ \sqrt{-3+7}=\sqrt4=2,\quad -3+1=-2.\ \text{But }2\neq-2.\ \text{\textbf{(rejected)}}

Only x=2x=2 works; x=−3x=-3 is extraneous and is thrown out.

Concept
The Distance Formula Is the Pythagorean Theorem

Place two points (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2) on the plane. The horizontal gap is a leg of length ∣x2−x1∣|x_2-x_1| and the vertical gap is a leg of length ∣y2−y1∣|y_2-y_1|; the straight-line distance dd between the points is the hypotenuse. Feeding the legs into a2+b2=c2a^{2}+b^{2}=c^{2} gives

d2=(x2−x1)2+(y2−y1)2 ⟹ d=(x2−x1)2+(y2−y1)2.d^{2}=(x_2-x_1)^{2}+(y_2-y_1)^{2}\ \Longrightarrow\ d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}.

The distance formula is nothing new to memorize, it is just the Pythagorean theorem wearing coordinates.

Reminder — Distance and midpoint:  d=(x2−x1)2+(y2−y1)2,M=(x1+x22, y1+y22)d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, \qquad M=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)
Reminder — The Pythagorean theorem:  a2+b2=c2a^2+b^2=c^2
Example
Distance in simplest radical form

Find the distance between (1,2)(1,2) and (4,8)(4,8).

d=(4−1)2+(8−2)2=32+62=9+36=45=9⋅5=35.\begin{aligned} d&=\sqrt{(4-1)^{2}+(8-2)^{2}}\\ &=\sqrt{3^{2}+6^{2}}=\sqrt{9+36}\\ &=\sqrt{45}=\sqrt{9\cdot5}\\ &=3\sqrt5. \end{aligned}
Concept
Combining a±b\sqrt{a}\pm\sqrt{b} Cleverly

Sums and differences of unlike radicals will not merge, but you can still simplify products and squares of them. Treat a\sqrt{a} and b\sqrt{b} like binomial terms and use (a)2=a\big(\sqrt{a}\big)^{2}=a:

(a+b)2=a+2ab+b,(a+b)(a−b)=a−b.\big(\sqrt{a}+\sqrt{b}\big)^{2}=a+2\sqrt{ab}+b,\qquad \big(\sqrt{a}+\sqrt{b}\big)\big(\sqrt{a}-\sqrt{b}\big)=a-b.

The first stays irrational (a nested-looking form); the second collapses to a plain integer whenever aa and bb are.

Example
Expanding a squared sum of radicals

Simplify (6+2)2\big(\sqrt6+\sqrt2\big)^{2}.

(6+2)2=(6)2+2(6)(2)+(2)2=6+212+2=8+2⋅2312=23=8+43.\begin{aligned} \big(\sqrt6+\sqrt2\big)^{2} &= \big(\sqrt6\big)^{2}+2\big(\sqrt6\big)\big(\sqrt2\big)+\big(\sqrt2\big)^{2}\\ &= 6+2\sqrt{12}+2\\ &= 8+2\cdot 2\sqrt3 && \text{$\sqrt{12}=2\sqrt3$}\\ &= 8+4\sqrt3. \end{aligned}

Why It Works

ab=ab\sqrt{ab} = \sqrt a\sqrt b for non-negative a,ba, b because (ab)2=ab(\sqrt a\sqrt b)^2 = ab and ab≥0\sqrt a\sqrt b \ge 0 — it is the positive number whose square is abab, which is the definition of ab\sqrt{ab}. Squaring an equation is not reversible: x=−2x = -2 becomes x2=4x^2 = 4, which also has the solution x=2x = 2. That is why every radical equation needs its candidates checked; the extraneous ones are the ghosts of the sign the squaring erased.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Radical rules and rational exponents
ab=ab,ab=ab,am/n=amn\sqrt{ab}=\sqrt{a}\sqrt{b},\qquad \sqrt{\frac{a}{b}}=\frac{\sqrt a}{\sqrt b},\qquad a^{m/n}=\sqrt[n]{a^{m}}

What it means. Roots split over multiplication and division, and a fractional exponent is a root.

Example. 50=252=52\sqrt{50}=\sqrt{25}\sqrt2=5\sqrt2, and 82/3=(81/3)2=48^{2/3}=(8^{1/3})^2=4.

Why it works. a1/na^{1/n} is defined as the number whose nnth power is aa, so (a1/n)n=an/n=a\left(a^{1/n}\right)^n=a^{n/n}=a — exactly the definition of an\sqrt[n]{a}. The product rule then follows from (ab)1/2=a1/2b1/2(ab)^{1/2}=a^{1/2}b^{1/2}.

Tip. Roots do not split over addition: a+b≠a+b\sqrt{a+b}\neq\sqrt a+\sqrt b. Test with a=b=9a=b=9 if you are ever tempted.