Completing the Square

(x+5)^2 is never negative, so y is largest when the bracket is zero.

1. (x+5)^2\geqslant0, so y\leqslant6
2. y=6 when x+5=0, that is when x=-5
3. (Not (5,\ 6): the sign inside the bracket changes.)
4. Answer: (-5,\ 6)

Don't expand: add 11 to both sides, then square root, remembering there are two square roots.

1. (x+2)^2=11
2. x+2=\pm\sqrt{11}
3. x=-2\pm\sqrt{11}
4. Answer: x=-2+\sqrt{11} or x=-2-\sqrt{11}

Halve the coefficient of x to get the number in the bracket, then subtract its square.

1. Half of -14 is -7, so start with (x-7)^2
2. (x-7)^2=x^2-14x+49
3. So x^2-14x+30=(x-7)^2-49+30
4. Answer: (x-7)^2-19

Halve the 9 for the bracket, then work out what you must subtract so the constant term is still 4.

1. Half of 9 is \frac92, so start with \left(x+\frac92\right)^2
2. \left(x+\frac92\right)^2=x^2+9x+\frac{81}{4}
3. So x^2+9x+4=\left(x+\frac92\right)^2-\frac{81}{4}+4
4. -\frac{81}{4}+\frac{16}{4}=-\frac{65}{4}
5. Answer: \left(x+\frac{9}{2}\right)^2-\frac{65}{4}

Complete the square, then think about the smallest value the squared bracket can take.

1. Complete the square: x^2-10x=(x-5)^2-25
2. So x^2-10x+27=(x-5)^2+2
3. (x-5)^2\geqslant0 for every x, so the expression is at least 2
4. (One value of x doesn't show it for all x; (x+5)^2+2 has the wrong sign; (x-5)^2=0 when x=5.)
5. Answer: the second statement

Expand the bracket on the left, then compare the constant terms on each side.

1. (x-3)^2+k=x^2-6x+9+k
2. Compare the x terms: m=-6
3. Compare the constant terms: 9+k=2k
4. Answer: k=9

Complete the square first; the number left outside the bracket is the minimum value of y.

1. Complete the square: x^2+10x=(x+5)^2-25
2. So y=(x+5)^2+k-25
3. (x+5)^2\geqslant 0, so the minimum value of y is k-25
4. k-25=3
5. Answer: k=28

Take the 4 out of the first two terms only, then complete the square inside the bracket.

1. Take out the 4 from the x terms: 4(x^2-10x)+93
2. x^2-10x=(x-5)^2-25
3. So 4\left[(x-5)^2-25\right]+93=4(x-5)^2-100+93
4. Answer: 4(x-5)^2-7

Take the 2 out of the first two terms only, then halve the coefficient of x inside the bracket.

1. Take out the 2 from the x terms: 2(x^2+3x)-1
2. Complete the square inside: x^2+3x=\left(x+\frac32\right)^2-\frac94
3. So 2\left[\left(x+\frac32\right)^2-\frac94\right]-1
4. Multiply out the outer bracket: 2\left(x+\frac32\right)^2-\frac92-1
5. -\frac92-1=-\frac{11}{2}
6. Answer: 2\left(x+\frac32\right)^2-\frac{11}{2}

Take out -3 from the two x terms, so the inside bracket starts x^2+6x; be careful with the signs.

1. Take out -3 from the x terms: 1-3(x^2+6x)
2. x^2+6x=(x+3)^2-9
3. So 1-3\left[(x+3)^2-9\right]=1-3(x+3)^2+27
4. Answer: 28-3(x+3)^2

Write x^2+6x as a squared bracket minus a number, then rearrange and square root both sides.

1. x^2+6x=(x+3)^2-9
2. So (x+3)^2-9-11=0, which gives (x+3)^2=20
3. x+3=\pm\sqrt{20}=\pm2\sqrt5
4. Answer: x=-3+2\sqrt5 or x=-3-2\sqrt5

Write the right-hand side in the form a(x+b)^2+c first; the minimum is where the bracket is zero.

1. Take out the 2: 2(x^2+5x)+9
2. x^2+5x=\left(x+\frac52\right)^2-\frac{25}{4}
3. So y=2\left(x+\frac52\right)^2-\frac{25}{2}+9=2\left(x+\frac52\right)^2-\frac72
4. The bracket is zero when x=-\frac52, and then y=-\frac72
5. Answer: \left(-\frac52,\ -\frac72\right)

The maximum point gives you p and r; then substitute the point where the curve crosses the y-axis to find q.

1. The maximum of y=p-q(x-r)^2 is at (r,\ p), so r=2 and p=5
2. So y=5-q(x-2)^2
3. At (0,\ -3): -3=5-q(0-2)^2=5-4q
4. 4q=8, so q=2
5. Answer: y=5-2(x-2)^2

Take the 2 out of the x terms and complete the square, then get the squared bracket on its own before square rooting.

1. Take out the 2: 2(x^2-6x)+5=0
2. x^2-6x=(x-3)^2-9, so 2(x-3)^2-18+5=0
3. 2(x-3)^2=13, so (x-3)^2=\frac{13}{2}
4. x-3=\pm\sqrt{\frac{13}{2}}
5. Answer: x=3\pm\sqrt{\frac{13}{2}}

The fraction is largest when its denominator is smallest, so complete the square on the denominator.

1. Complete the square: x^2+6x+14=(x+3)^2+5
2. (x+3)^2\geqslant0, so the smallest value of the denominator is 5
3. The fraction is largest when the denominator is smallest: \dfrac{15}{5}
4. Answer: 3

Use Pythagoras' theorem for AP^2 in terms of t, then complete the square. Answer part (a) before attempting part (b).

Part (a)

1. Horizontal distance t-5, vertical distance 2t
2. AP^2=(t-5)^2+(2t)^2
3. =t^2-10t+25+4t^2=5t^2-10t+25
4. =5(t^2-2t)+25=5\left[(t-1)^2-1\right]+25
5. Answer: 5(t-1)^2+20

Part (b)

1. AP^2 is smallest when (t-1)^2=0, so t=1
2. P=(t,\ 2t)
3. Answer: (1,\ 2)

Part (c)

1. The least value of AP^2 is 20
2. AP=\sqrt{20}=\sqrt4\times\sqrt5
3. Answer: 2\sqrt5

Differentiate first, then complete the square on the gradient function. Answer part (a) before attempting part (b).

Part (a)

1. \dfrac{\mathrm{d}y}{\mathrm{d}x}=9+6x-3x^2
2. Take out -3 from the x terms: 9-3(x^2-2x)
3. x^2-2x=(x-1)^2-1
4. So 9-3\left[(x-1)^2-1\right]=9-3(x-1)^2+3
5. Answer: 12-3(x-1)^2

Part (b)

1. 3(x-1)^2\geqslant0, so the gradient is greatest (12) when x=1
2. y=5+9+3-1=16
3. Answer: (1,\ 16)

Take out -40 from the x terms before completing the square. Answer part (a) before attempting parts (b) and (c).

Part (a)

1. Take out -40 from the x terms: -40(x^2-14x)-1200
2. x^2-14x=(x-7)^2-49
3. So -40\left[(x-7)^2-49\right]-1200=-40(x-7)^2+1960-1200
4. Answer: 760-40(x-7)^2

Part (b)

1. 40(x-7)^2\geqslant0, so P\leqslant760, with P=760 when x=7
2. Answer: \pounds760

Part (c)

1. 760-40(x-7)^2=0, so (x-7)^2=19
2. x-7=\pm\sqrt{19}, so x=7-\sqrt{19} or x=7+\sqrt{19}
3. The lower price is 7-\sqrt{19}=2.6411\ldots
4. Answer: \pounds2.64

Complete the square to find the minimum point; the curve is symmetrical about the vertical line through B. Answer part (a) before attempting parts (b) and (c).

Part (a)

1. Take out the 3: 3(x^2-4x)+7
2. x^2-4x=(x-2)^2-4
3. So y=3(x-2)^2-12+7=3(x-2)^2-5
4. The minimum is where the bracket is zero
5. Answer: B=(2,\ -5)

Part (b)

1. A=(0,\ 7)
2. The curve is symmetrical about x=2, so C is as far to the right of x=2 as A is to the left
3. Answer: C=(4,\ 7)

Part (c)

1. Base AC=4-0=4
2. Height = vertical distance from B to AC =7-(-5)=12
3. Area =\frac12\times4\times12
4. Answer: 24

The minimum point of (x+p)^2+q is (-p,\ q); use the point on the curve to replace b, then substitute the minimum point into the line.

Part (a)

1. Halve the coefficient of x: -2a\div2=-a
2. x^2-2ax=(x-a)^2-a^2
3. Answer: (x-a)^2+b-a^2

Part (b)

1. Substitute (2,\ 7) into the curve: 4-4a+b=7, so b=4a+3
2. From part (a), the minimum point is (a,\ b-a^2)
3. So the minimum point is (a,\ 4a+3-a^2)
4. It lies on y=x-1: 4a+3-a^2=a-1
5. Rearrange: a^2-3a-4=0
6. Factorise: (a-4)(a+1)=0
7. Answer: a=4 or a=-1