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# The A to Z of mathematics a basic guide - Sidebotham T.H.

Sidebotham T.H. The A to Z of mathematics a basic guide - Wiley publishing , 2002. - 489 p.
ISBN 0-471-15045-2
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References: Converse of a Theorem, Hypotenuse, Pythagorean Triples, Right Angle, Square, Square Root, Vertex.
PYTHAGOREAN TRIPLES
A right-angled triangle has three sides which obey the theorem of Pythagoras, c2 = a2 + b2. Any three whole numbers that fit this formula are called Pythagorean triples. For example, the numbers 3, 4, and 5 are Pythagorean triples, because 52 = 32 + 42. So are any multiples of 3,4, and 5, like 6, 8, and 10, or 30,40, and 50.
PYTHAGOREAN TRIPLES 357
There is an infinite number of Pythagorean triples, and some well-known ones are as follows:
♦ 5, 12, 13 (check: 132 = ? + 122)
♦ 7, 24, 25
♦ 8, 15, 17
♦ 9,40,41
♦ 11,60,61
Any multiples of these triples are also Pythagorean triples.
References: Pythagoras’ Theorem, Triangle.
Q
Reference: Circular Functions.
A quadratic equation is an equation of the form
ax2 + bx + c = 0
where a is not equal to zero, and a, b, and c are real numbers. Examples of quadratic equations are
jc2 + 9x + 18 = 0 x2 + x — 6 = 0
x2 — 4x = —4 Which is the same as x2 — 4x + 4 = 0 2x2 — 6x + 4 = 0
x2 = 21 Which is the same as x2 + Ox — 21 = 0
x2 — 9 = 0 Which is the same as x2 + Ox — 9 = 0
4x2 = l(k Which is the same as 4x2 — 10x + 0 = 0
Some of the above equations may not appear to be of the form ax2 + bx + c = 0, but they can be rearranged into that form. To solve quadratic equations we first have to factorize the quadratic expression ax2 + bx + c. Before proceeding it may be necessary for you to study the entry Factorize. The theory behind solving quadratic equations is outlined here:
Suppose two terms A and B multiply together to give zero.
AB = 0
Then it follows that this equation is true if A = 0 or if B = 0.
358
If the two terms A and B are brackets, we can apply this theory in the following way:
Example 1. Solve (a) (x + 1)0 — 5) = 0, (b) 2x(x + 2) = 0.
Solution. For (a), write
0 + 1)0 - 5) = 0 0 + 1) = 0 or (x — 5) = 0 If AB = 0, then either A = 0 or 15 = 0
* + 1=0 or x — 5 = 0 We can dispense with the brackets
x = — 1 or x = 5 Solving linear equations
For (b), Write
2x(x + 2) = 0
2x = 0 or x + 2 = 0 If AB = 0, then either A = 0 or B = 0 x = 0 or x = — 2
In the next example the first stage is to express the quadratic equation in factorized form.
Example 2. Solve *2 + * - 6 = 0.
Solution. Write
x2 + x — 6 = 0 (x - 2)0 + 3) = 0
x — 2 = 0 or x + 3 = 0 x = 2 or x = — 3
Example 3. Solve x2 -4x = -4.
Solution. Write
x2 -4x = -4
x2 - 4x + 4 = 0
(x - 2)0 - 2) = 0
Adding 4 to both sides of equation to make it = 0 Factorizing the quadratic
x — 2 (twice)
When the two brackets are the same there will be only one solution, or we could say there are two equal solutions.
Example 4. Solve 2x2 - 6x + 4 = 0.
Solution. Write
2x2 — 6x + 4 = 0
2(x2 — 3x + 2) = 0 2 is a common factor; always look for these first
2(x — l)(x — 2) = 0 Factorizing the quadratic
(x — l)(x — 2) = 0 Dividing both sides of the equation by 2
x — 1=0 or x—2 = 0 x = 1 or x = 2
Example 5. Solve x2 - 9 = 0.
Solution. Write
x — 3 = 0 or x+3=0 x = 3 or x = — 3
Example 6. Solve 4.x2 = lOx.
Solution. Write
4x2 = lOx
4x2 — lOx = 0 Subtracting lOx from both sides, to
x2 - 9 = 0
(x - 3)(x + 3) = 0
Factorizing by difference of two squares
make equation = 0
2x(2x — 5) = 0
2x = 0 or 2x — 5 = 0
2x is a common factor
x = 0 or x = 2|
Solving the linear equations
Summary:
Always make sure the quadratic equation = 0. Look for common factors first.
♦ Answers may be left as fractions or decimals, but if decimals are used, some may need rounding off.
♦ There are three kinds of factors to know: common, quadratic, and difference of two squares.
There are three types of quadratic equations that are not solved by factorizing, and their solutions are explained here.
Example 7. Solve jc2 = 21.
Solution. Write
JC2 = 21
= ±V2l Reduce x2 to x by taking the square root of both
sides of the equation
x = ±4.58 (to 2 dp) When we take the square root we get ±
Example 8. Solve = 9.
Solution. Write
*Jx = 9
(v^)2 = 92 Make sfx into a: by squaring both sides of the equation
jc = 81
Example 9. Solve (x - 3)2 = 16.
Solution. Write (x - 3 )2 = 16
sj(x — 3)2 = ±VT6 Taking the square root of both sides of the
equation
jc - 3 = ±4
x — 3 = +4 or x — 3 = —4 Separating out into two equations
x =1 or x = — 1
To solve other quadratic equations that do not factorize, see the entry Quadratic Formula for the method.