Every card in Algebra, Year 10: brackets, formulae and identities

x² − 3x − 28

HintEach term in the first bracket multiplies each term in the second, giving four products before you tidy up.

WhyThe four products are x × x = x², x × (−7) = −7x, 4 × x = 4x and 4 × (−7) = −28. Collecting the two middle terms, −7x + 4x = −3x.

A quadratic expression

HintThe name comes from a Latin word to do with squares, not with the number four.

Whyx² − 9, 3x² + x and x² + 5x + 6 all count, because x² is the highest power in each. Multiplying two brackets such as (x + 2)(x + 3) always produces one.

x² − 6x + 9

HintWrite the bracket out twice, side by side, before multiplying.

Why(x − 3)² means (x − 3)(x − 3). The four products are x², −3x, −3x and +9, because (−3) × (−3) = +9. The two middle terms collect to −6x.

add

HintExpand (x + p)(x + q) and look at where the middle term comes from.

Why(x + p)(x + q) = x² + px + qx + pq, so the number in front of x is p + q and the constant is p × q. For x² + 7x + 12 the pair is 3 and 4: 3 × 4 = 12 and 3 + 4 = 7.

(x + 3)(x − 5)

HintYou need a pair of numbers with a negative product, so one is positive and the other is negative.

WhyThe pairs that multiply to −15 include 3 and −5, and 3 + (−5) = −2, which matches the middle term. Check by expanding: x² − 5x + 3x − 15 = x² − 2x − 15.

x = (5y − 2)/3

HintList what happens to x, in order, then undo those steps starting with the last one.

Whyx is multiplied by 3, then 2 is added, then the result is divided by 5. Undo in reverse: multiply both sides by 5 to get 5y = 3x + 2, subtract 2 to get 5y − 2 = 3x, then divide by 3.

a = (v² − u²)/(2s)

HintGet the term containing the wanted letter on its own first, then deal with what multiplies it.

WhySubtract u² from both sides: v² − u² = 2as. Then divide both sides by 2s. The squared terms do not need to be touched, because the letter you want is not inside them.

r = √(A/π)

HintUndo the multiplication before you undo the squaring.

WhyDivide both sides by π to get A/π = r², then take the square root of both sides. The root covers the whole of A/π. Because a radius is a length, only the positive root is used.

t = d/s

HintThe wanted letter is on the bottom of a fraction, so bring it up to the top first.

WhyMultiply both sides by t to get st = d, then divide both sides by s. With numbers: if 4 = 20/t, then t = 20/4 = 5.

reverse

HintThink of taking off shoes and socks: which went on last?

WhyIn y = 2x + 7, x is doubled and then 7 is added. To free x, subtract 7 first and then halve: x = (y − 7)/2. The last operation applied is the first one to undo.

An identity

HintIn everyday life the same word means who you are, something that stays the same wherever you go.

WhyWhatever number x is, 2(x + 1) and 2x + 2 give the same result, because they are two ways of writing one expression. The three-line sign ≡ is used to show this.

An identity, because it is true for every value of x

HintExpand the bracket on the left and compare the two sides.

WhyExpanding 2(x + 3) gives 2x + 6, exactly the right-hand side. The two sides are the same expression written in two ways, so no value of x can make them differ. It can be written 2(x + 3) ≡ 2x + 6.

equation

HintIt is the kind of statement you solve to find the unknown.

WhySubstituting x = 2 gives 6 + 5 = 11, which is true. Any other value fails: x = 3 gives 14. A statement that holds only for particular values is solved, whereas an identity holds for all values.

7

HintExpand the left-hand side and simplify it fully.

Why5(x + 2) − 3 = 5x + 10 − 3 = 5x + 7. For an identity the two sides must match term by term, so the constant k must be 7.

Expanding gives 3x + 12 − 2x − 2, which simplifies to x + 10

HintMultiply out each bracket, taking care with the minus sign in front of the second one.

WhyThe −2 multiplies both terms of the second bracket, giving −2x and −2. Then 3x − 2x = x and 12 − 2 = 10. To show two expressions are equivalent, work on one of them until it becomes the other.

2n + 1

HintAn odd number is always one away from an even number.

WhyDoubling any whole number gives an even number, and the next whole number after an even one is odd. Writing "any even number" as 2n and "any odd number" as 2n + 1 is the first step of most algebraic arguments about odd and even.

3n + 3; 3(n + 1)

HintCollect the n terms and the plain numbers, then look for a common factor.

Whyn + n + 1 + n + 2 has three lots of n and 1 + 2 = 3 in plain numbers. Taking out the common factor 3 leaves a whole number in the bracket, which shows the total is 3 times a whole number whatever n is.

When x = 1, (x + 3)² = 16 but x² + 9 = 10

HintWork out each expression separately with the given value and compare the results.

Why(1 + 3)² = 4² = 16, while 1² + 9 = 10. Equivalent expressions must agree for every value, so one value where they differ is enough to show they are not equivalent. The correct expansion is x² + 6x + 9.

every

HintAsk how many values would have to be tested before you could be sure.

Whyx² and 2x agree when x = 0 and when x = 2, yet they are different expressions (try x = 3: 9 and 6). Only algebra, such as expanding and simplifying one side until it matches the other, covers all values at once.