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GCSE MATHEMATICS Higher Tier, topic sheet. QUADRATICS

Factorising, completing the square and using the quadratic formula: x =2 4

2b b ac

a− ± −

1. Solve the equation x2 + 4x – 10 = 0 Give your answers to 2 decimal places. You must show your working.

2. Factorise (a) x2 − 4

(b) 3x2 − 12

(c) 5x2 − 17x + 6

3. Find the values of a and b such that x2 – 10x + 18 = (x – a)2 + b.

4. It is given that x2 – 6x + 13 = (x – a)2 + b

(a) Find the values of a and b.

(b) Hence find the minimum value of x2 – 6x + 13.

5. It is given that x2 − 8x + 10 = (x − a)2 + b

(a) Find the values of a and b.

(b) Hence find the minimum value of x2 − 8x + 10

6. (a) Factorise 2x2 – 7x – 15

(b) The graph of y = 2x2 – 7x – 15 is sketched below.

x

y

Not to scale

P Q

Find the equation of the line of symmetry of this graph.

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7. The graph of y = 7 + 5x − 2x2 is sketched below. not drawn accuratelyy C

BA x

(a) Find the coordinates of the points A and B, where the curve crosses the x-axis.

(b) C is the point on the curve where the value of y is a maximum. Use your answer to (a) to find the value of x at C.

8. Solve this equation 53+x − 4

1+x = 2

1

9. Solve the equation 11–2–1 =

+ xxx

10. A rectangle has length (x + 5) cm and width (x – 2) cm. A triangle has base (x + 8) cm and height x cm.

( + 5) cmx

( – 2) cmx

( + 8) cmx

Not to scalex cm

The area of the rectangle is equal to the area of the triangle.

Show that x2 – 2x – 20 = 0 (You are not required to solve this equation.)

11. The base of a triangle is 7 cm longer than its height. The area of the triangle is 32 cm2.

(a) Taking the height to be h cm, show that h2+ 7h − 64 = 0

(b) Solve this equation to find the height of the triangle.

Give your answer to 2 decimal places.

12. Choose one of the following statements to describe the equation (x − 5)2 = x2 − 10x + 25

A It is true for one value of x. B It is true for two values of x. C It is true for all values of x.

Explain your answer.

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SOLUTIONS / ANSWERS.1. x2 + 4x – 10 = 0.

Use the quadratic formula: x =2 4

2b b ac

a− ± −

to get: x =24 4 4 1 ( 10)2 1

− ± − × × −×

= 4 562

− ±

and thus x = −5.74 or x = 1.74 (to 2 decimal places). 2. (a) x2 − 4 = (x − 2)(x + 2). (b) 3x2 − 12 = 3(x2 − 4) = 3(x − 2)(x + 2). (c) 5x2 − 17x + 6 = (5x − 2)(x − 3). 3. In asking us to express x2 − 10x + 18 in the form (x − a)2 + b, the question is simply asking us

to complete the square! Answer: x2 − 10x + 18 = (x − 5)2 − 7. Thus a = 5 and b = 7. (This final part is easily overlooked!)

4. (a) Complete the square to get x2 – 6x + 13 = (x − 3)2 + 4. Thus a = 3 and b = 4. (b) Thus the minimum value of x2 – 6x + 13 is 4. 5. (a) Complete the square to get x2 − 8x + 10 = (x − 4)2 − 6. Thus a = 4 and b = −6. (b) Thus the minimum value of x2 – 8x + 10 is −6. 6. (a) 2x2 − 7x − 15 = (2x + 3)(x − 5). (b) Using the answers from (a), the points P and Q have x-coordinates −1.5 and 5

respectively.

The line of symmetry passes through the mid-point of P and Q which has x-coordinate 1.5 5

2− + = 1.75.

Thus the line of symmetry has equation x = 1.75. 7. (a) The graph cuts the x-axis when y = 0, i.e. 7 + 5x − 2x2 = 0

⇒ 0 = 2x2 − 5x − 70 = (2x − 7)(x + 1)

and thus x = −1 or x = 3.5. This means that A = (−1, 0) and B = (3.5, 0). (b) This is identical to question 6 (b). The required value of x lies half way between −1

and 3.5 and is therefore equal to 1 3.52

− + = 1.25.

x

y

−1.5 5

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8. 3 1 15 4 2x x− =

+ +.

Double both sides: 6 2 15 4x x− =+ +

.Now cross-multiply: 6(x + 4) − 2(x + 5) = (x + 5)(x + 4).

Expand brackets (careful with the negatives) 6x + 24 − 2x − 10 = x2 + 9x + 20

4x + 14 = x2 + 9x + 20 0 = x2 + 5x + 6. Factorise: 0 = (x + 2)(x + 3) and thus x = −2 or x = −3. 9. Cross multiply: x(x − 1) − 2(x + 1) = (x − 1)(x + 1) x2 − x − 2x − 2 = x2 − 1

x2 − 3x − 2 = x2 − 1− 3x − 2 = − 1

⇒ −3x = 1and thus x = − 1

3 .

10. Area of rectangle = length × width = (x + 5)(x − 2) = x2 + 3x − 10. Area of triangle = 1

2 base × height = 12 (x + 8)x = 1

2 (x2 + 8x).

Thus, since the 2 areas are equal, x2 + 3x − 10 = 12 (x2 + 8x)

and hence 2x2 + 6x − 20 = x2 + 8x⇒ x2 − 2x − 20 = 0 as required.

11.

(a) Area = 32 ⇒ 12 base × height = 32.

This means that 12 (h + 7)h = 32

and thus (h + 7)h = 64 ⇒ h2 + 7h = 64 and hence h2 + 7h − 64 = 0 as required. (b) Use the quadratic formula to get h = −12.23 or h = 5.23. This means that the height of the triangle = 5.23 cm. 12. (x − 5)2 = x2 − 10x + 25 is true for all values of x since if you expand the brackets, you will

obtain the expression on the right hand side.

h + 7

h