Inequalities

Get the x term on its own first, and remember what happens to the sign when you divide by a negative number.

1. Subtract 11 from both sides: -3x>15
2. Divide both sides by -3 and reverse the sign
3. Answer: x<-5

Think about negative values of x too: what is (-8)^2?

1. x^2=49 when x=7 or x=-7
2. Squares are bigger than 49 when x is further from 0 than 7, in either direction
3. For example 8^2=64 and (-8)^2=64 both work, but 0^2=0 does not
4. Answer: x<-7 or x>7
5. (x>7 misses the negative values; -7<x<7 is where x^2<49; 24.5 comes from halving 49 instead of square rooting.)

Write -1<\dfrac{2x+7}{3}\leqslant 5 and solve it, doing the same thing to all three parts.

1. -1<\dfrac{2x+7}{3}\leqslant 5
2. Multiply all three parts by 3: -3<2x+7\leqslant 15
3. Subtract 7: -10<2x\leqslant 8
4. Divide by 2: -5<x\leqslant 4
5. Answer: -5<x\leqslant 4

Expand both brackets carefully (watch -2\times -1), then collect the x terms on the side where they stay positive.

1. Expand: 7-6x+2<4x+24
2. Simplify the left side: 9-6x<4x+24
3. Add 6x to both sides: 9<10x+24
4. Subtract 24: -15<10x
5. Divide by 10: -1.5<x
6. Answer: x>-\dfrac{3}{2}
7. (If you divide by a negative number instead, remember to reverse the inequality sign.)

Rearrange to get n^2 on its own, then remember that negative integers and zero can also work.

1. Add 3: 5n^2\leqslant 45
2. Divide by 5: n^2\leqslant 9
3. So -3\leqslant n\leqslant 3 (the sign is \leqslant, so -3 and 3 are included)
4. List the integers, including 0 and the negatives
5. Answer: -3,\ -2,\ -1,\ 0,\ 1,\ 2,\ 3

\mathrm{g}(x)>4 means the part of the curve that is above the line y=4, not above the x-axis.

1. \mathrm{g}(x)>4 where the curve is above the line y=4
2. The curve is above the line between the two crossing points, x=-1 and x=2
3. The inequality is strict, so -1 and 2 are not included
4. Answer: -1<x<2

Multiply every term by 6, the lowest common multiple of 3 and 2, and put brackets round each numerator.

1. Multiply every term by 6: 2(x+4)-3(x-1)<6
2. Expand: 2x+8-3x+3<6
3. Simplify: 11-x<6
4. Add x and subtract 6: 5<x
5. Answer: x>5

First work out the least and greatest values of n^2: n=0 is allowed.

1. n can be 0, so the least value of n^2 is 0
2. The greatest value of n^2 is 4^2=16 (bigger than (-2)^2=4)
3. So 0\leqslant n^2\leqslant 16
4. Subtracting from 10 reverses the order: 10-16\leqslant 10-n^2\leqslant 10-0
5. Answer: -6\leqslant 10-n^2\leqslant 10

Rearrange so one side is 0, factorise, then sketch the parabola and decide whether you want where it is above or below the x-axis.

1. Make one side zero: x^2+3x-28\geqslant 0
2. Factorise: (x+7)(x-4)\geqslant 0
3. Critical values: x=-7 and x=4
4. y=x^2+3x-28 is a U-shaped curve crossing the x-axis at -7 and 4.
5. \geqslant 0 means on or above the x-axis: the two outer parts.
6. Answer: x\leqslant -7 or x\geqslant 4
7. (-7\leqslant x\leqslant 4 is where the curve is below the axis; x\geqslant 4 misses the second region; x\leqslant -4 or x\geqslant 7 comes from factorising with the wrong signs.)

Rearrange so that one side is 0, then factorise into two brackets.

1. Subtract 11x: 3x^2-11x-4\leqslant 0
2. Factorise: (3x+1)(x-4)\leqslant 0
3. Critical values: x=-\dfrac{1}{3} and x=4
4. The curve y=3x^2-11x-4 is U-shaped, so it is on or below the x-axis between the critical values
5. Answer: -\dfrac{1}{3}\leqslant x\leqslant 4

Multiply through by -1 to make the x^2 term positive, remembering to reverse the sign, then factorise.

1. Multiply by -1 and reverse the sign: x^2-3x-10>0
2. Factorise: (x-5)(x+2)>0
3. Critical values: x=-2 and x=5
4. y=x^2-3x-10 is U-shaped, so it is above the x-axis outside the critical values
5. Answer: x<-2 or x>5
6. (-2<x<5 forgets that the sign reversed; x<-5 or x>2 has the signs in the brackets wrong; -2>x>5 is impossible, as no number is both less than -2 and greater than 5.)

Expand both sides fully: the x^2 terms cancel, leaving a linear inequality.

1. Expand the left: (x-5)^2=x^2-10x+25
2. Expand the right: x^2-4x+1
3. So x^2-10x+25>x^2-4x+1
4. Subtract x^2 from both sides: -10x+25>-4x+1
5. Add 10x and subtract 1: 24>6x
6. Answer: x<4

Rearrange to x^2-7x+6>0, factorise, and decide whether the solution is between or outside the critical values.

1. Rearrange: x^2-7x+6>0
2. Factorise: (x-1)(x-6)>0
3. Critical values: x=1 and x=6
4. U-shaped curve, >0: outside the critical values, so x<1 or x>6
5. No positive integer is less than 1, and 6 itself does not work (36+6=42, not more than 42)
6. Answer: x=7

Find the nth term, then solve the inequality 'nth term <300' and think about which whole numbers n can be.

1. The terms go up by 7, so the nth term is 7n-3
2. Solve 7n-3<300
3. 7n<303, so n<43.28\ldots
4. n is a whole number, so n=1,2,\ldots,43 (check: the 43rd term is 298 and the 44th is 305)
5. Answer: 43

QR is opposite angle QPR and PR is the hypotenuse, so write down \sin x^\circ and find the range of a-2.

1. QR is opposite angle x and PR is the hypotenuse: \sin x^\circ=\dfrac{a-2}{12}
2. 2<a<8, so 0<a-2<6
3. So 0<\sin x^\circ<\dfrac{6}{12}=\dfrac12
4. \sin x^\circ=\dfrac12 when x=30, and x is an acute angle
5. Answer: 0<x<30

Multiply every term by x (allowed without reversing the sign because x>0), then rearrange and factorise; remember x must be positive.

1. x>0, so multiply every term by x without reversing the sign: 5x^2-4x<12
2. Rearrange: 5x^2-4x-12<0
3. Factorise: (5x+6)(x-2)<0
4. Critical values: x=-\dfrac{6}{5} and x=2
5. U-shaped curve, <0: between the critical values, -\dfrac{6}{5}<x<2
6. But x>0, so the lower limit is 0
7. Answer: 0<x<2

Area of a parallelogram is base times perpendicular height. Once you have solved the quadratic inequality, check which values of x make every length positive.

Part (a)

1. Area of a parallelogram = base \times perpendicular height
2. (x+4)(3x-2)<24
3. Expand: 3x^2-2x+12x-8<24
4. 3x^2+10x-8<24
5. Subtract 24: 3x^2+10x-32<0

Part (b)

1. Factorise: (3x+16)(x-2)<0
2. Critical values: x=-\dfrac{16}{3} and x=2
3. The curve is U-shaped, so it is below the x-axis between the roots: -\dfrac{16}{3}<x<2
4. But every length must be positive.
5. Height: 3x-2>0, so x>\dfrac{2}{3} (base: x+4>0 gives x>-4, which is weaker)
6. Combine: \dfrac{2}{3}<x<2

Form an inequality for the area and solve it as a quadratic inequality, then use the fact that every length must be positive. Answer part (a) before attempting part (b).

Part (a)

1. (2x-5)(x+3)>63
2. Expand: 2x^2+x-15>63
3. Rearrange: 2x^2+x-78>0
4. Factorise: (2x+13)(x-6)>0
5. U-shaped curve, >0: x<-\dfrac{13}{2} or x>6
6. The length 2x-5 must be positive, so x>\dfrac52: reject x<-\dfrac{13}{2}
7. Answer: x>6

Part (b)

1. Perimeter =2(2x-5)+2(x+3)=6x-4
2. 6x-4<50, so 6x<54 and x<9
3. Combine with x>6 from part (a)
4. Answer: 6<x<9

For part (b), write the nth term <700, solve it as a quadratic inequality by factorising, then decide which whole numbers n can be. Answer part (a) before attempting part (b).

Part (a)

1. First differences: 6,\ 8,\ 10,\ 12
2. Second difference is 2, so the sequence starts n^2
3. Sequence minus n^2: 3,\ 6,\ 9,\ 12,\ 15, which is 3n
4. Answer: n^2+3n

Part (b)

1. Solve n^2+3n<700
2. Rearrange: n^2+3n-700<0
3. Factorise: (n+28)(n-25)<0
4. So -28<n<25
5. n is a positive whole number, so n=1,2,\ldots,24 (the 25th term is exactly 700, which is not less than 700)
6. Answer: 24

A function is decreasing where its gradient is negative, so form and solve the inequality \mathrm{f}'(x)<0. Answer part (a) before attempting part (b).

Part (a)

1. Differentiate each term: 2x^3\to 6x^2, -9x^2\to -18x, -24x\to -24, 7\to 0
2. Answer: \mathrm{f}'(x)=6x^2-18x-24

Part (b)

1. f is decreasing when \mathrm{f}'(x)<0
2. 6x^2-18x-24<0
3. Divide by 6: x^2-3x-4<0
4. Factorise: (x-4)(x+1)<0
5. U-shaped curve, <0: between the critical values
6. Answer: -1<x<4