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Surds: Adding and Subtracting
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Surds: Adding and Subtracting

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Surds: Adding and Subtracting

We can also add and subtract surds together, which can be a little more complicated.

In order to add or subtract surds, they need to have the same root. This means that the number inside the square root signs must be the same.

Can you add these surds together as they are?

10+6\sqrt{10}+\sqrt{6}

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Adding surds means that you can have multiples of that surd:

8+8=28\sqrt8 +\sqrt8=2\sqrt8

Similarly:

28+68=882\sqrt8+ 6\sqrt8=8\sqrt8

Let's find:

34+1243\sqrt{4}+12\sqrt{4}

1

Are they the same root? Yes or no

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2

They're the same, so we can add them

To add them, all we need to do is add together the numbers before the square root.

3

What is 3+123+12?

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4

Nice! We've added them together

The final answer is 15415\sqrt{4}.

What is 62+326\sqrt{2} + 3\sqrt{2}?

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Unfortunately, surds don't always have the same number in their root.

In order to add surds with different numbers inside the roots, we can first look to simplify them. This leads to a smaller number inside the root:

1052\sqrt{10} \rightarrow5\sqrt2

For example, write 8+32\sqrt8 + \sqrt{32} in terms of 2\sqrt2

1

We can simplify 8\sqrt{8}

To do this, we need to find two factors: one surd and one perfect square.

2

44 and 22 multiply to make 88

44 is also a perfect square. So, we can simplify 8\sqrt{8} to 222\sqrt{2}

3

We can also simplify 32\sqrt{32}

To do this, we follow the same process as we just did with 8\sqrt{8}.

4

Write 3232 as a product of a square and 22

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5

Express these as different surds

32=16×2=16×2\sqrt{32}=\sqrt{16 \times 2}=\sqrt{16} \times \sqrt{2}

6

What is 16\sqrt{16}?

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7

We're left with 4×24 \times \sqrt{2}

2×4=42\sqrt2\times 4=4\sqrt2. So, 32=42\sqrt{32}=4\sqrt2

8

Add our new surds together

32+8\sqrt{32}+\sqrt8 is the same as 42+224\sqrt{2}+2\sqrt{2}.

9

What is 42+224\sqrt{2}+2\sqrt{2}?

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10

The final answer is 626\sqrt{2}

Good work! We've added two surds with different root numbers 😃

What is 27+12\sqrt{27} +\sqrt{12} in terms of 3\sqrt3

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Give 20+80\sqrt{20} +\sqrt{80} in terms of 5\sqrt{5}

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This means that when we expand the brackets, there will be a +5+\sqrt5and a 5-\sqrt5 which cancel out, leaving us with a rational denominator.

Rationalise 12+5\dfrac{1}{2+\sqrt{5}}

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Surds are an exact way to write irrational numbers.

Let's have a go at solving this one! 45\dfrac{4}{\sqrt5}

1

Multiply the top by the surd

45×55=455×5\dfrac{4}{\sqrt5} \times \dfrac{\sqrt5}{\sqrt5}=\dfrac{4\sqrt5}{\sqrt5 \times \sqrt5}

2

Multiply the bottom together

455×5=455\dfrac{4\sqrt5}{\sqrt5 \times \sqrt5}=\dfrac{4\sqrt5}{5}

3

Nice! 👌

45\dfrac{4}{\sqrt5} can be rationalised to 455\dfrac{4\sqrt5}{5}

If the fraction looks like: 315\dfrac{3}{1-\sqrt{5}}

We need to multiply by: 1+51+5 \dfrac{1+\sqrt{5}}{1+\sqrt{5}}

1

Rationalising the denominator

Multiply the numerator and the denominator by the surd with the OPPOSITE sign

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Rationalise 11+2\dfrac{1}{1+\sqrt{2}}

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Now let's try solving 315\dfrac{3}{1-\sqrt{5}}

1

Multiply by the denominator with the opposite sign

315×1+51+5\dfrac{3}{1-\sqrt{5}} \times \dfrac{1+\sqrt{5}}{1+\sqrt{5}}

2

Multiply the numerators

315×1+51+5=3+35(15)×(1+5)\dfrac{3}{1-\sqrt{5}} \times \dfrac{1+\sqrt{5}}{1+\sqrt{5}}=\dfrac{3+3\sqrt5}{(1-\sqrt{5}) \times (1+\sqrt{5})}

3

Multiply the denominators

3+35(15)×(1+5)=3+354\dfrac{3+3\sqrt5}{(1-\sqrt{5}) \times (1+\sqrt{5})}=\dfrac{3+3\sqrt5}{-4}

4

Make the denominator positive

3+354×11=3354\dfrac{3+3\sqrt5}{-4} \times \dfrac{-1}{-1}=\dfrac{-3-3\sqrt5}{4}

5

Great work! 💪

This can't be simplified any more, so the answer is 3354\dfrac{-3-3\sqrt5}{4}

1

Multiply the numerator and the denominator by the surd

This is a key step!

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Rationalise 113\dfrac{1}{\sqrt{13}}

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For a fraction that looks like 45\dfrac{4}{\sqrt5}, we need to multiply by 55\dfrac{\sqrt5}{\sqrt5} to make the denominator an exact number, as a square root multiplied by itself is a whole number.