Flashcards · 8 cards

Surds

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Rule
Simplify $\sqrt{72}$ .
Answer
Take out the largest square factor: $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt2$ .
√72 = 6√236 small squares of area 2: the side is 6 lots of √2
From the lesson: Simplifying surds
Know it
$\sqrt a \times \sqrt b = \,?$
Answer
$\sqrt{ab}$ . And $\sqrt a \times \sqrt a = a$ .
From the lesson: Simplifying surds
Rule
When can you add surds?
Answer
Only like surds, with the same number under the root: $5\sqrt3 + 2\sqrt3 = 7\sqrt3$ . $\sqrt2 + \sqrt3$ stays as it is.
Like surdsCount the √3s, as you would count x's
From the lesson: Simplifying surds
Know it
What is $(3\sqrt2)^2$ ?
Answer
$3^2 \times (\sqrt2)^2 = 9 \times 2 = 18$ . Square both parts.
From the lesson: Simplifying surds
Rule
$(a + \sqrt b)(a - \sqrt b) = \,?$
Answer
$a^2 - b$ , with no surd left.
The difference of two squares(a + √b)(a − √b) = a² − b
From the lesson: Simplifying surds
Rule
Rationalise $\dfrac{6}{\sqrt3}$ .
Answer
Multiply top and bottom by $\sqrt3$ : $\dfrac{6\sqrt3}{3} = 2\sqrt3$ .
Multiply by 1√3 ÷ √3 is 1, so the value doesn't change
From the lesson: Rationalising the denominator
Know it
Rationalise $\dfrac{1}{2 + \sqrt3}$ .
Answer
Multiply top and bottom by the conjugate, $2 - \sqrt3$ . The bottom becomes $4 - 3 = 1$ , so the answer is $2 - \sqrt3$ .
From the lesson: Rationalising the denominator
Which method?
WAEC 2019 · Paper 2 · Q3 (a)
Without using mathematical tables or calculators, simplify $3\sqrt{\frac32} - 4\sqrt{\frac23} - \sqrt{24}$ .
What do you turn every term into?
Answer
A multiple of $\sqrt6$ : $3\sqrt{\tfrac32} = \tfrac{3\sqrt6}{2}$ , $4\sqrt{\tfrac23} = \tfrac{4\sqrt6}{3}$ and $\sqrt{24} = 2\sqrt6$ . Then put them over a common denominator.
From the lesson: Rationalising the denominator Try the question