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ESO Y3 · Mathematics
Powers and roots · Roots

Operations with radicals

In geometry, roots crop up all the time: the diagonal of a square with side 1 measures $\sqrt{2}$, the height of an equilateral triangle involves $\sqrt{3}$… To calculate exact perimeters and areas we have to operate with radicals, and the rules are very similar to those of algebra.

Before you start, make sure you know: Operations with monomials (2nd ESO)
Before you start, make sure you know: Powers (2nd ESO)

Adding and subtracting radicals

You can only add like radicals (same index and same radicand). It works just as it does with letters: $2x + 5x = 7x$, and $2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$.

Visual comparison illustrating the addition of like radicals versus unlike radicals and the need to take out factors first. 2√3 + 5√3 − √3 = 6√3 (like 2x + 5x − x = 6x) √3 + √2= ?? not like radicals: leave as it is √12 + √27 = 2√3 + 3√3 = 5√3 take out factors first
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We are going to simplify expressions by grouping like radicals. Imagine that $\sqrt{3}$ is like an "x" or an "apple". Carry out the following additions and subtractions:…

We are going to simplify expressions by grouping like radicals. Imagine that $\sqrt{3}$ is like an "x" or an "apple". Carry out the following additions and subtractions:

  1. $5\sqrt{2} + 3\sqrt{2} - 2\sqrt{2}$

  2. $7\sqrt[3]{5} - 2\sqrt[3]{5}$

  3. $4\sqrt{6} + \sqrt{6} - 8\sqrt{6}$

Solve
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To add and subtract radicals, they must be like radicals (same index and same radicand). Simplify the following radicals by expressing the numbers as products of prime…

To add and subtract radicals, they must be like radicals (same index and same radicand). Simplify the following radicals by expressing the numbers as products of prime factors, then calculate:

  1. a) $3\sqrt{8} - 5\sqrt{2} + \sqrt{50}$
  2. b) $\sqrt{12} + \sqrt{75} - \sqrt{27}$
  3. c) $2\sqrt[3]{16} - \sqrt[3]{54}$
Solve

Multiplying and dividing radicals

With the same index, the radicands are multiplied (or divided) inside a single root:

The area of a rectangle with sides $\sqrt{5}$ and $\sqrt{3}$ is $\sqrt{5} \cdot \sqrt{3} = \sqrt{15}$ cm²:

Properties of multiplying and dividing radicals with the same index, showing how to combine or separate the radicands under a single root. √5 ≈ 2,24 √3 ≈ 1,73 √15 area = √5 · √3 = √15 ≈ 3,87 cm² perimeter = 2√5 + 2√3 (cannot be combined any further)
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Use the property of the product of roots with the same index ($\sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{a \cdot b}$) to group these operations into a single root and, if…

Use the property of the product of roots with the same index ($\sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{a \cdot b}$) to group these operations into a single root and, if possible, work out the exact final result:

  • $\sqrt{2} \cdot \sqrt{32}$

  • $\sqrt[3]{2} \cdot \sqrt[3]{4}$

  • $\sqrt{5} \cdot \sqrt{20}$

  • $\sqrt{3} \cdot \sqrt{27}$

Solve
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Carry out the following multiplications and divisions of radicals. Remember that they must have the same index to operate. If possible, simplify the final result by…

Carry out the following multiplications and divisions of radicals. Remember that they must have the same index to operate. If possible, simplify the final result by extracting factors from the root.

  1. a) $\sqrt{18} \cdot \sqrt{2}$
  2. b) $\sqrt[3]{4} \cdot \sqrt[3]{16}$
  3. c) $\frac{\sqrt{75}}{\sqrt{3}}$
  4. d) $\sqrt{2x} \cdot \sqrt{8x^3}$
Solve
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Calculate the perimeter and area of a rectangular plot of land whose dimensions are given by radicals. The base measures $\sqrt{27}$ metres and the height $\sqrt{3}$…

Calculate the perimeter and area of a rectangular plot of land whose dimensions are given by radicals. The base measures $\sqrt{27}$ metres and the height $\sqrt{3}$ metres. Simplify the results as much as possible.

√27 m √3 m Area = ? Perimeter = ?
Solve

The root of a root and the power of a root

It all makes sense with fractional exponents: $\sqrt{\sqrt[3]{a}} = \left(a^{\frac13}\right)^{\frac12} = a^{\frac16}$.

Step-by-step calculation of the square root of the cube root of 64 by multiplying the indices to obtain the sixth root of 64, which is 2. √ ( ∛64 ) =⁶√64 =2 2 · 3 = 6: the indices are multiplied (2⁶ = 64)
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When we have a root inside another root, we can simplify it by multiplying their indices ($\sqrt[n]{\sqrt[m]{a}} = \sqrt[n \cdot m]{a}$). Simplify the following…

When we have a root inside another root, we can simplify it by multiplying their indices ($\sqrt[n]{\sqrt[m]{a}} = \sqrt[n \cdot m]{a}$). Simplify the following expressions to a single root:

  • $\sqrt{\sqrt{81}}$

  • $\sqrt[3]{\sqrt{64}}$

  • $\sqrt{\sqrt[3]{x}}$

Solve
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Fractional powers are another way of writing radicals. Use this relationship to simplify the following expressions and give the final answer in radical form: a) Write…

Fractional powers are another way of writing radicals. Use this relationship to simplify the following expressions and give the final answer in radical form:

  1. a) Write $5^{3/4}$ in radical form.
  2. b) Write $\sqrt[5]{2^3}$ in index form.
  3. c) Calculate by first expressing as a power: $\sqrt[3]{\sqrt{64}}$
  4. d) Simplify: $\frac{\sqrt[4]{x^3}}{\sqrt{x}}$
Solve
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Apply the rules of radicals (root of a root and power of a root) to simplify the following expressions to a single irreducible radical: a) $\sqrt{\sqrt[3]{64}}$ b)…

Apply the rules of radicals (root of a root and power of a root) to simplify the following expressions to a single irreducible radical:

  1. a) $\sqrt{\sqrt[3]{64}}$
  2. b) $(\sqrt[3]{2})^6$
  3. c) $\sqrt{2 \cdot \sqrt{2}}$ (Hint: Move the first 2 inside the second square root)
Solve

Radicals in problems

In geometry, Pythagoras' theorem almost always produces roots. The usual practice is to give the simplified exact value and, if asked, the approximate one.

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You are going to buy a new television, and the dimensions are given in inches along the diagonal. The screen is rectangular, with a width of 40 cm and a height of 30 cm.…

You are going to buy a new television, and the dimensions are given in inches along the diagonal. The screen is rectangular, with a width of 40 cm and a height of 30 cm. Using Pythagoras' theorem ($h = \sqrt{a^2 + b^2}$), calculate the length of the screen's diagonal in centimetres.

Note: The result is an exact root.

h = √(a² + b²) a = 40 cm b = 30 cm h = ?
Solve
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We want to attach a guy wire to support a vertical post 4 metres high. The wire will be anchored to the ground at a distance of $\sqrt{20}$ metres from the base of the…

We want to attach a guy wire to support a vertical post 4 metres high. The wire will be anchored to the ground at a distance of $\sqrt{20}$ metres from the base of the post. Calculate the exact length of cable needed using Pythagoras' theorem. Express the answer as a simplified radical.

4 m √20 m Cable c = ?
Solve
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We have a cubic water tank. We know that its total volume is $128 \text{ m}^3$. Calculate the length of the edge of the tank. Express the answer as a simplified radical…

We have a cubic water tank. We know that its total volume is $128 \text{ m}^3$. Calculate the length of the edge of the tank. Express the answer as a simplified radical (by taking factors out of the root) and do not use decimals.

(Hint: Remember that the volume of a cube is $V = \text{arista}^3$)

edge = ? a a V = 128 m³
Solve

Summary

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