Algebra, Year 11: quadratic inequalities and sequences

−2 ≤ x ≤ 4

HintFind where the expression equals zero, then decide which side of those values you need.

The whyx² − 2x − 8 = (x + 2)(x − 4), which is zero at x = −2 and x = 4. The graph is U-shaped, so it is on or below the x-axis between the roots. Test x = 0: −8 ≤ 0, true.

x < −3 or x > 3

HintSketch y = x² and the line y = 9.

The whyx² = 9 at x = −3 and x = 3. The curve y = x² is above 9 outside these values, so x < −3 or x > 3. Test x = −4: 16 > 9, true. Test x = 0: 0 > 9, false.

{x : x < −3} ∪ {x : x > 3}

HintEach inequality gets its own pair of curly brackets, and the two are joined by the symbol that stands for 'or'.

The whyCurly brackets mean 'the set of', and the colon is read 'such that'. ∪ is the union: everything in either set, which matches the word 'or'. {x : x < −3 or x > 3} says the same thing and is also accepted; a single interval such as 2 < x < 3 is written {x : 2 < x < 3}.

below

HintLess than zero means y is negative.

The why'< 0' asks where y is negative, which is where the graph dips under the axis: between the roots, 2 < x < 3. For '> 0' you would want the parts above the axis, x < 2 or x > 3. A quick sketch settles which.

7

HintRemember zero and the negative whole numbers.

The whyx² < 10 when x is between −√10 and √10, about −3.16 to 3.16. The integers in that range are −3, −2, −1, 0, 1, 2 and 3, which is 7 of them.

12

HintGoing from the 1st term to the 5th takes four multiplications.

The why5th term = 3 × (√2)⁴. Since (√2)² = 2, (√2)⁴ = 4, so the term is 3 × 4 = 12. The sequence is 3, 3√2, 6, 6√2, 12.

6√3

HintDivide one term by the term before it to find the multiplier.

The why2√3 ÷ 2 = √3, and 2√3 × √3 = 2 × 3 = 6, so the common ratio is √3. The next term is 6 × √3 = 6√3, and the one after that is 18.

8

HintApply the rule twice, one step at a time.

The whyu₂ = 2² − 1 = 3 and u₃ = 3² − 1 = 8. Rules like this, where each term is worked out from the one before, are given in the question; the skill is to apply them carefully, one term at a time.

n² on its own already has a second difference of 2, so the coefficient of n² is half the second difference

HintLook at how the gaps between 1, 4, 9 and 16 grow.

The whyThe square numbers 1, 4, 9, 16 have differences 3, 5, 7, which go up by 2 each time. Multiplying by a multiplies that 2 by a, so an² has second difference 2a. Here 2a = 6, so a = 3.

n² + 1

HintCompare each term with the square numbers.

The whyThe differences are 3, 5, 7, 9, so the second difference is 2 and the nth term starts with n². Subtracting the square numbers 1, 4, 9, 16, 25 leaves 1 every time, so the nth term is n² + 1.

4; 2; n + 5

HintFind the differences of the differences, then halve.

The whyFirst differences: 7, 11, 15, 19; second differences: 4. Half of 4 is 2, so the rule starts 2n², which gives 2, 8, 18, 32, 50. Taking these away leaves 6, 7, 8, 9, 10, the sequence n + 5, so the nth term is 2n² + n + 5.

3

HintPut n = 2 into the rule and set the result equal to 10.

The whyWhen n = 2, n² + bn = 4 + 2b. Setting 4 + 2b = 10 gives b = 3. The sequence is then 4, 10, 18, 28, …

2n² − n

HintHalve the second difference, then see what is left once that part is taken away from each term.

The whyFirst differences: 5, 9, 13, 17; second differences: 4, so the rule starts 2n², giving 2, 8, 18, 32, 50. Subtracting these from the terms leaves −1, −2, −3, −4, −5, which is −n. So the nth term is 2n² − n.