Surds

Find a square number that is a factor of the number inside the root.

1. Write 20=4\times5, using the square factor 4.
2. Then \sqrt{20}=\sqrt4\times\sqrt5=2\sqrt5.

Simplify the square root before multiplying by the coefficient outside it.

1. Use the square factor 4: \sqrt{12}=\sqrt{4\times3}=2\sqrt3.
2. Multiply by the factor outside the root: 3\times2\sqrt3=6\sqrt3.

Simplify both square roots so that any matching surds can be collected.

1. The square factors give \sqrt8=2\sqrt2 and \sqrt{18}=3\sqrt2.
2. These are like terms, so add the coefficients: 2\sqrt2+3\sqrt2=5\sqrt2.

Write the first root as a multiple of the second root.

1. Since 45=9\times5, \sqrt{45}=3\sqrt5.
2. Subtract the coefficients of the like surds: 3\sqrt5-\sqrt5=2\sqrt5.

Multiply the numbers inside the roots, then look for a square factor.

1. For positive numbers, the product of square roots is the square root of the product: \sqrt6\times\sqrt{15}=\sqrt{90}.
2. Use 90=9\times10 to obtain 3\sqrt{10}.

The quotient of positive square roots is the square root of their quotient.

1. Divide inside the root: \frac{\sqrt{98}}{\sqrt2}=\sqrt{49}.
2. The positive square root of 49 is 7.

Multiply both terms in the bracket by the factor outside, remembering what happens when a root is squared.

1. The first product is 2\sqrt5\times3=6\sqrt5.
2. The second product is 2\sqrt5\times\sqrt5=2\times5=10.
3. Therefore the expanded expression is 10+6\sqrt5.

The brackets have the form of a difference of two squares.

1. Using (a+b)(a-b)=a^2-b^2, the product is (\sqrt7)^2-2^2.
2. This is 7-4=3; the two mixed terms cancel.

Simplify each root before collecting the coefficients of the common surd.

1. The roots simplify to \sqrt{27}=3\sqrt3 and \sqrt{12}=2\sqrt3.
2. Including the coefficients gives 12\sqrt3-4\sqrt3+\sqrt3.
3. Collect the like terms: (12-4+1)\sqrt3=9\sqrt3.

Squaring a bracket produces two equal mixed products as well as the two squares.

1. Write the square as (3+\sqrt2)(3+\sqrt2) and multiply every term in the first bracket by every term in the second.
2. The four products are 3\times3=9, 3\times\sqrt2=3\sqrt2, \sqrt2\times3=3\sqrt2 and \sqrt2\times\sqrt2=2.
3. Collect the whole numbers, 9+2=11, and the surds, 3\sqrt2+3\sqrt2=6\sqrt2, to get 11+6\sqrt2.

The side length is the positive square root of the area.

1. If the side length is s, then s^2=72.
2. Take the positive square root because a length is positive: s=\sqrt{72}.
3. Use 72=36\times2, giving a side length of 6\sqrt2\text{ cm}.

Use Pythagoras' theorem, expanding each squared bracket carefully.

1. By Pythagoras' theorem, \text{hypotenuse}^2=(2+\sqrt3)^2+(2-\sqrt3)^2.
2. Expand the first square: (2+\sqrt3)^2=4+2\sqrt3+2\sqrt3+3=7+4\sqrt3.
3. Expand the second: (2-\sqrt3)^2=4-2\sqrt3-2\sqrt3+3=7-4\sqrt3.
4. Add: (7+4\sqrt3)+(7-4\sqrt3)=14, as the surd terms cancel.
5. So \text{hypotenuse}^2=14 and the hypotenuse is \sqrt{14}\text{ cm}.

Multiply the two side lengths, then collect rational terms and like surds.

1. Area is length times width, so expand (\sqrt5+3)(\sqrt5+1).
2. The four terms are 5, \sqrt5, 3\sqrt5 and 3.
3. Combining them gives an area of (8+4\sqrt5)\text{ cm}^2.

Both numbers are positive, so compare their squares.

1. The square of the first number is 5^2\times2=50.
2. The square of the second number is 3^2\times6=54.
3. For positive numbers, the smaller square comes from the smaller number, so 5\sqrt2<3\sqrt6.

Take square factors and the positive square root of the squared variable outside the root.

1. Write the radicand as 25\times2\times x^2.
2. The square roots of 25 and x^2 are 5 and x, because x>0.
3. The simplified expression is 5x\sqrt2.

Simplifying the roots before multiplying makes both brackets much shorter.

1. The square factors give \sqrt{12}=2\sqrt3 and \sqrt{27}=3\sqrt3.
2. The brackets simplify to 5\sqrt3 and \sqrt3.
3. The product is 5(\sqrt3)^2=15.

Include the two equal negative mixed products when squaring the bracket.

1. Use (p-q)^2=p^2-2pq+q^2 with p=2\sqrt7 and q=\sqrt3.
2. p^2=(2\sqrt7)^2=4\times7=28 and q^2=(\sqrt3)^2=3.
3. 2pq=2\times2\sqrt7\times\sqrt3=4\sqrt{21}, using \sqrt7\times\sqrt3=\sqrt{21}.
4. So the expansion is 28-4\sqrt{21}+3=31-4\sqrt{21}.

Simplify the known root and isolate the unknown root before squaring.

1. Simplify the known root: \sqrt{12}=\sqrt{4\times3}=2\sqrt3, so \sqrt n+2\sqrt3=5\sqrt3.
2. Subtract 2\sqrt3 from both sides: \sqrt n=3\sqrt3.
3. Square both sides: n=(3\sqrt3)^2=9\times3=27.

Simplify the difference inside the square and the separate product before adding.

1. The difference in the bracket is 3\sqrt2-2\sqrt2=\sqrt2.
2. Squaring it gives 2.
3. The remaining product is \sqrt{64}=8.
4. The total is 2+8=10.

Simplify the larger root first, then expand and collect the two types of term.

Part (a)

1. First write \sqrt{20}=2\sqrt5.
2. In (3+2\sqrt5)(2-\sqrt5), the rational products are 3\times2=6 and 2\sqrt5\times(-\sqrt5)=-10.
3. The rational part is 6-10=-4, so a=-4.

Part (b)

1. After simplifying \sqrt{20}, the mixed products are -3\sqrt5 and 4\sqrt5.
2. Their sum is \sqrt5, whose coefficient is 1.
3. Therefore b=1.