GCSE MATHEMATICS Higher Tier, topic sheet.
QUADRATICS
Factorising, completing the square and using the quadratic formula: x =
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
4.
It is given that x2 – 6x + 13 = (x – a)2 + b
5.
6.
x2 – 10x + 18 = (x – a)2 + b.
(a)
Find the values of a and b.
(b)
Hence find the minimum value of x2 – 6x + 13.
It is given that x2
8x + 10 = (x
b ±
a)2 + b
(a)
Find the values of a and b.
(b)
Hence find the minimum value of x2 8x + 10
(a)
Factorise 2x2 – 7x – 15
(b)
The graph of y = 2x2 – 7x – 15 is sketched below.
y
Not to scale
P
Q
Find the equation of the line of symmetry of this graph.
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x
b2
2a
4ac
7.
The graph of y = 7 + 5x
y
2x2 is sketched below.
C
not drawn accurately
x
B
A
(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.
3
x+5
1
1
=
x+4
2
8.
Solve this equation
9.
Solve the equation
10.
A rectangle has length (x + 5) cm and width (x – 2) cm.
A triangle has base (x + 8) cm and height x cm.
x
2
–
=1
x +1 x –1
Not to scale
(x – 2) cm
x cm
(x + 5) cm
(x + 8) cm
The area of the rectangle is equal to the area of the triangle.
Show that x2 – 2x – 20 = 0
11.
12.
(You are not required to solve this equation.)
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.
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:
to get: x =
4 ±
x=
b ±
42 4 × 1× ( 10)
=
2 ×1
b2
2a
4ac
4 ± 56
2
and thus x = 5.74 or x = 1.74 (to 2 decimal places).
x2 4 = (x 2)(x + 2).
3x2 12 = 3(x2 4) = 3(x 2)(x + 2).
5x2 17x + 6 = (5x 2)(x 3).
2.
(a)
(b)
(c)
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)
(b)
5.
(a)
(b)
6.
(a)
(b)
Complete the square to get x2 – 6x + 13 = (x 3)2 + 4.
Thus a = 3 and b = 4.
Thus the minimum value of x2 – 6x + 13 is 4.
Complete the square to get x2 8x + 10 = (x 4)2
Thus a = 4 and b = 6.
Thus the minimum value of x2 – 8x + 10 is 6.
6.
2x2 7x 15 = (2x + 3)(x 5).
Using the answers from (a), the points P and Q have x-coordinates 1.5 and 5
respectively.
y
1.5
5
x
The line of symmetry passes through the mid-point of P and Q which has x1.5 + 5
coordinate
= 1.75.
2
Thus the line of symmetry has equation x = 1.75.
7.
i.e. 7 + 5x 2x2 = 0
0 = 2x2 5x 7
0 = (2x 7)(x + 1)
(a)
The graph cuts the x-axis when y = 0,
(b)
and thus x = 1 or x = 3.5.
This means that A = ( 1, 0) and B = (3.5, 0).
This is identical to question 6 (b). The required value of x lies half way between 1
1 + 3.5
and 3.5 and is therefore equal to
= 1.25.
2
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8.
3
x + 5
1
1
= .
x + 4
2
6
2
= 1.
x + 5
x + 4
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.
Double both sides:
9.
Cross multiply:
and thus
10.
x(x 1) 2(x + 1) = (x
x2 x 2x 2 = x2 1
x2 3x 2 = x2 1
3x 2 =
1
3x = 1
1
x=
.
3
1)(x + 1)
Area of rectangle = length × width = (x + 5)(x 2) = x2 + 3x 10.
1
1
1
Area of triangle = base × height = (x + 8)x = (x2 + 8x).
2
2
2
1
Thus, since the 2 areas are equal,
x2 + 3x 10 = (x2 + 8x)
2
2
and hence
2x + 6x 20 = x2 + 8x
x2 2x 20 = 0 as required.
11.
h
h+7
(a)
(b)
12.
1
base × height = 32.
2
1
(h + 7)h = 32
This means that
2
and thus
(h + 7)h = 64
h2 + 7h = 64
and hence
h2 + 7h 64 = 0 as required.
Use the quadratic formula to get h = 12.23 or h = 5.23.
This means that the height of the triangle = 5.23 cm.
Area = 32
(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.
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