Linear and Quadratic Equations in Context

Don't expand the bracket: get (x-3)^2 on its own, then square root both sides, remembering there are two square roots.

1. (x-3)^2=10
2. x-3=\pm\sqrt{10}
3. Answer: x=3+\sqrt{10} or x=3-\sqrt{10}

Write 1.2\times10^3 as an ordinary number first, then undo the division and the square.

1. 1.2\times10^3=1200
2. x^2=3\times1200=3600
3. x=\sqrt{3600} (positive root)
4. Answer: x=60

Cross-multiply to get a single equation with no fractions: each numerator times the other side's denominator.

1. Cross-multiply: 5(x+2)=3(2x-1)
2. 5x+10=6x-3
3. 13=x
4. Check: \dfrac{5}{25}=\dfrac15 and \dfrac{3}{15}=\dfrac15
5. Answer: x=13

An increase of r% means multiplying by 1+\frac{r}{100}, and doing it twice must give the same result as multiplying by 1.44.

1. Each year the value is multiplied by 1+\dfrac{r}{100}
2. Two years: \left(1+\dfrac{r}{100}\right)^2=1.44
3. 1+\dfrac{r}{100}=\sqrt{1.44}=1.2 (the multiplier is positive)
4. \dfrac{r}{100}=0.2
5. Answer: r=20

Expand and rearrange so that one side is 0, then use the quadratic formula.

1. 2x^2-3x-7=0, so a=2, b=-3, c=-7
2. x=\dfrac{3\pm\sqrt{(-3)^2-4\times2\times(-7)}}{2\times2}
3. x=\dfrac{3\pm\sqrt{65}}{4}
4. x=2.7655\ldots or x=-1.2655\ldots
5. Answer: x=2.77 or x=-1.27 (3 s.f.)

The stone is at sea level when h=0: solve that equation, then think about what t measures.

1. At sea level h=0: 30+25t-5t^2=0
2. Divide by -5: t^2-5t-6=0
3. (t-6)(t+1)=0, so t=6 or t=-1
4. t counts the seconds after the throw, so t=-1 is impossible (at t=-1 it is h that is 0, not negative)
5. Answer: after 6 seconds; t=-1 is rejected because time since the throw cannot be negative

Reducing by 30% leaves 70% of the original size, so multiply (4x+10) by the decimal multiplier and set it equal to the compressed size.

1. Reduce by 30%: multiply by 0.7
2. 0.7(4x+10)=2x+19
3. 2.8x+7=2x+19
4. 0.8x=12, so x=15
5. Original size =4\times15+10=70
6. Check: 0.7\times70=49=2\times15+19
7. Answer: 70 MB

Add 8 to both ages, then use the fact that a:b=3:7 means 7a=3b.

1. In 8 years: Sam is x+8, his aunt is 3x+12
2. (x+8):(3x+12)=3:7 so 7(x+8)=3(3x+12)
3. 7x+56=9x+36
4. 20=2x, so x=10
5. Check: in 8 years the ages are 18 and 42, and 18:42=3:7
6. Answer: x=10

Form a quadratic from the area, then complete the square (or use the formula) and keep only the root that can be a length.

1. Rectangle is x by (x+2): x(x+2)=30
2. x^2+2x-30=0
3. Complete the square: (x+1)^2-1-30=0
4. (x+1)^2=31, so x=-1\pm\sqrt{31}
5. -1-\sqrt{31} is negative, so it cannot be a length
6. Answer: x=-1+\sqrt{31}

The angles add up to 180^\circ, which gives a quadratic; check which solution gives three positive angles.

1. x^2+6x+5x=180
2. x^2+11x-180=0
3. (x+20)(x-9)=0, so x=9 or x=-20
4. x=-20 makes 6x negative, so x=9
5. Angles: 81^\circ, 54^\circ and 45^\circ
6. Answer: 81^\circ

Rearrange so that one side is 0, then take out the common factor x before factorising the quadratic: don't divide by x.

1. 3x^3+13x^2-10x=0
2. x(3x^2+13x-10)=0
3. x(3x-2)(x+5)=0
4. Answer: x=0, x=\dfrac23 or x=-5

Use Pythagoras with (x+9) as the hypotenuse to get a quadratic, then reject the solution that gives a negative length.

1. Pythagoras: x^2+(x+1)^2=(x+9)^2
2. x^2+x^2+2x+1=x^2+18x+81
3. x^2-16x-80=0
4. (x-20)(x+4)=0, so x=20 or x=-4
5. A length cannot be negative, so x=20
6. Sides are 20, 21 and 29 cm
7. Perimeter =20+21+29=70
8. Answer: 70 cm

Write each area in terms of a and b, set them equal, then divide through by b.

1. Triangle: \dfrac12\times(3a+2b)\times2b=3ab+2b^2
2. Parallelogram: (5a-b)\times b=5ab-b^2
3. 3ab+2b^2=5ab-b^2
4. 3b^2=2ab; divide by b: 3b=2a
5. Answer: \dfrac{a}{b}=\dfrac32

In similar rectangles the ratio shorter side : longer side is the same, so x:6=6:(x+5).

1. Similar, so \dfrac{x}{6}=\dfrac{6}{x+5}
2. x(x+5)=36
3. x^2+5x-36=0
4. (x+9)(x-4)=0, so x=4 (a length cannot be negative)
5. Tile B is 6 cm by 9 cm
6. Answer: 54 cm^2

For a product of three whole numbers to be prime, two of them must each be 1 or -1.

1. A prime has only the factors 1 and itself, so two of the brackets must be \pm1
2. x+9 is more than 9 for a positive integer, so it must be the prime
3. c-3=\pm1 and 2c-7=\pm1 together give c=4
4. f(4)=1\times1\times13
5. Answer: f(c)=13

Find the nth term of B, then set the two nth terms equal and solve the quadratic. Answer part (a) before attempting part (b).

Part (a)

1. The terms go up by 7, so the nth term starts 7n
2. First term: 7\times1+45=52
3. Answer: 7n+45

Part (b)

1. n^2+3n=7n+45
2. n^2-4n-45=0
3. (n-9)(n+5)=0
4. n is a position in a sequence, so it is a positive integer
5. Answer: n=9

Part (c)

1. Sequence A: 9^2+3\times9=108
2. Sequence B: 7\times9+45=108
3. Answer: 108

Write an expression for the time of each crossing using time = distance ÷ speed, then use the fact that they differ by half an hour. Answer part (a) before attempting part (b).

Part (a)

1. Time =\dfrac{\text{distance}}{\text{speed}}, and 30 minutes is \dfrac12 hour
2. \dfrac{36}{v-6}-\dfrac{36}{v}=\dfrac12
3. Multiply by 2v(v-6): 72v-72(v-6)=v(v-6)
4. 432=v^2-6v
5. Answer: v^2-6v-432=0

Part (b)

1. (v-24)(v+18)=0
2. v=24 or v=-18
3. A speed (and the windy speed v-6) must be positive
4. Answer: v=24

Part (c)

1. Windy speed =24-6=18 km/h
2. Time =36\div18=2
3. Check: calm time =36\div24=1.5 hours, half an hour less
4. Answer: 2 hours

25% more means multiply the square's area by 1.25; expand (x+3)^2 fully before collecting terms.

Part (a)

1. Rectangle area =x(2x+1)=2x^2+x
2. 25% more than the square means \times1.25: 2x^2+x=1.25(x+3)^2
3. 1.25(x^2+6x+9)=1.25x^2+7.5x+11.25
4. 2x^2+x-1.25x^2-7.5x-11.25=0
5. 0.75x^2-6.5x-11.25=0; multiply by 4
6. Answer: 3x^2-26x-45=0

Part (b)

1. Use the formula with a=3, b=-26, c=-45
2. x=\dfrac{26\pm\sqrt{(-26)^2-4\times3\times(-45)}}{2\times3}=\dfrac{26\pm\sqrt{1216}}{6}
3. x=10.145\ldots or x=-1.478\ldots
4. x is a length, so it must be positive
5. Answer: x=10.1 (3 s.f.)

Part (c)

1. Perimeter =2(2x+1)+2x=6x+2
2. Use the unrounded value x=10.145\ldots
3. 6\times10.145\ldots+2=62.87\ldots
4. Answer: 62.9 cm (3 s.f.)

Follow the money year by year: multiply by m, add 1000, then multiply by m again. Answer part (a) before attempting part (b).

Part (a)

1. After one year: 4000m; then add 1000 to get 4000m+1000
2. After two years: (4000m+1000)m=4000m^2+1000m
3. 4000m^2+1000m=5500
4. Divide by 500: 8m^2+2m=11
5. Answer: 8m^2+2m-11=0

Part (b)

1. Formula with a=8, b=2, c=-11
2. m=\dfrac{-2\pm\sqrt{2^2-4\times8\times(-11)}}{2\times8}=\dfrac{-2\pm\sqrt{356}}{16}
3. m=1.05424\ldots or m=-1.304\ldots
4. A multiplier must be positive, so m=1.05424\ldots
5. r=100(m-1)=5.424\ldots
6. Answer: r=5.42 (3 s.f.)

Use the cosine rule for BC, taking care with the sign of \cos120^\circ. Answer part (a) before attempting part (b).

Part (a)

1. Cosine rule: 19^2=x^2+(3x+1)^2-2x(3x+1)\cos120^\circ
2. \cos120^\circ=-\dfrac12, so -2x(3x+1)\cos120^\circ=+x(3x+1)
3. 361=x^2+9x^2+6x+1+3x^2+x
4. 361=13x^2+7x+1
5. Answer: 13x^2+7x-360=0

Part (b)

1. (x-5)(13x+72)=0
2. x=5 or x=-\dfrac{72}{13}
3. x is a length, so it must be positive
4. Answer: x=5

Part (c)

1. AB=5 cm and AC=16 cm
2. Area =\dfrac12\times5\times16\times\sin120^\circ
3. \sin120^\circ=\dfrac{\sqrt3}{2}, so area =40\times\dfrac{\sqrt3}{2}
4. Answer: 20\sqrt3 cm^2