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Exact Values (Edexcel GCSE Maths: Foundation): Flashcards

Exam code: 1MA1

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FrontExact Values
Define an exact value.

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Define an exact value.
An exact value is one that has not been rounded, truncated or left as an unending decimal.
It is written in a form that is exactly right rather than nearly right, often as a fraction or as a multiple of $π$ or of a root.
When leaving an answer in terms of $π$ , how should you treat $π$ in your working?
Treat $π$ like an algebraic term, doing the arithmetic on the numbers only and leaving $π$ attached to them.
So $π \times 6$ is written as $6 π$ , and $3 π + 4 π = 7 π$ .
True or False?
A terminating decimal such as $0.375$ is an exact value.
True.
The decimal stops, so nothing has been rounded away, and $0.375$ is exactly the same number as $\frac{3}{8}$ .
Why is the perimeter of a semicircle written as $(4 π + 8)$ cm rather than as a single number?
The two parts cannot be combined, because $4 π$ is a multiple of $π$ and $8$ is not.
Adding them would need a decimal for $π$ , and the answer would stop being exact.
Define a surd.
A surd is the square root of a non-square integer.
Writing $5 \sqrt{2}$ rather than $7.071067812$ keeps the answer exact.
Complete the two surd calculations:
$3 \sqrt{5} + 8 \sqrt{5} =______$
$7 \sqrt{3} - 4 \sqrt{3} =______$
The completed calculations are:
$3 \sqrt{5} + 8 \sqrt{5} = 11 \sqrt{5}$
$7 \sqrt{3} - 4 \sqrt{3} = 3 \sqrt{3}$
Multiples of the same surd add and subtract just like letters in algebra.
True or False?
$\sqrt{9} + \sqrt{4} = \sqrt{13}$
False.
$\sqrt{9} + \sqrt{4} = 3 + 2 = 5$ , whereas $\sqrt{13} = 3.6055 \dots$ , so numbers sitting under separate root signs cannot be added together.
What do you get if you square a surd such as $\sqrt{3}$ ?
Squaring a surd gives the number under the root sign, so ${(\sqrt{3})}^{2} = \sqrt{3} \times \sqrt{3} = 3$ .
Squaring undoes a square root.
True or False?
$3.14$ is an exact value for $π$ .
False.
$3.14$ is a rounded approximation, and the decimal for $π$ never ends, so the only exact way to write it is as $π$ itself.

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