Practice Test on Topic 19 Quadratics
Test on Topic 19 Quadratics
1.
2.
Solve the following equations.
(a) x2 + 7x + 10
=
0
(b) 3x2 – 4x + 1
(c) x3 – 4x2 – 5x
=
0
(d)
=
=
=
=
2x – 16
(c) 2y2 – 8y + 6
0
(b)
x2 – 6x
=
=
0
x–6
0
(b)
x2 – 5x =
32
(b) x2 – 5x + 20 =
0
(c) 2x2 – 5x + 3
0
Solve the following Quadratic equation by using the quadratic formula. x 
(a) x2 – 3x – 10 =
7.
(b) x2 – 6x
For each of the following quadratic equations indicate whether there are 0, 1 or 2 real solutions.
(You do not need to calculate the solutions – you need only say how many there are)
(a) x2 – 4x + 4
6.
0
Solve each of the following quadratic equations by using “Completing the square”
Give your answers t 1 decimal place.
(a) x2 + 4x – 21 =
5.
25
Solve each of the following quadratic equations by using “Completing the square”.
(a) x2 + 6x – 7
4.
=
0
Solve each of the following quadratic equations by using “Completing the square”.
(a) x2 + 6x – 16
3.
x2
=
(b) x2 – 5x – 4
0
=
=
0
 b  b 2  4ac
2a
0
Solve the following equations by any method you choose.
(a)
x2 + 6x + 5 =
(c) 4x2 + 4x – 24
=
0
(b) x2 – x
0
(d)
=
3x2 =
20
6x – 3
Page | 1
Practice Test on Topic 19 Quadratics
8.
Solve the following Quadratic equations by using the quadratic formula.
(a) x2 – 3x – 10
9.
=
0
(b) 0.5x2
=
4x – 8
The quadratic equation x2 – 2x + 2 = 0 has two solutions which are both complex numbers.
Use the quadratic formula to find these solutions.
10. For the quadratic function f(x) = 2x2 – 6x + 10 find the following.
(a) Is the shape of the parabola Concave Down or Convex Up?
(b) Where is its y-intercept?
(c) Where is its vertex?
11. For the quadratic function y = x2 – 6x + 8
(a) Where does the curve cut the y – axis
(b)
Where does the curve cut the x – axis
(c) Draw the parabola with the equation y = x2 – 6x + 8
12. The quadratic y = x2 + 2x – 8
(a) What is the shape of the quadratic?
(b) What is the y-intercept?
(c) What are the x-intercepts?
(d) What are the coordinates of the vertex?
(e) Sketch the quadratic
13. For the quadratic function f(x) = x2 + 6x + 5 find the following.
(a) Is the shape of the parabola Concave Down or convex Up?
(b) Where is its y-intercept?
(c) Where is its vertex?
(d) Where is its x-intercepts?
(e) Draw a graph of the quadratic function.
Page | 2
Practice Test on Topic 19 Quadratics
14. There exists two positive numbers where one is 4 more than the other and their product is 45.
(a) Write down the above information in an algebraic form.
(b) Find an equation and by solving it find the value of these two numbers?
15. A rectangle has its length two more than three times its width.
The area of the rectangle is 85 square feet.
(a) Write down the above information in the form of an algebraic equation.
(b) What is the length and width of the rectangle?
16. John traveled from his own home to his mother’s home 200 miles away and then he returned back home that
same day.During these journey’s he noticed that his speed was 10 mph faster coming home than it was when
he traveled to his mothers home. The total time spent on both parts of the journey was 9 hours. What were his
speeds during this journey?
17. Professor Robertson was jogging one day from his home to the College a distance of 4 miles.
On the return journey he was very tired and jogged 2 miles per hour slower than on the first part of the journey.
The total time he took jogging was 9.6 hours.
What were the going and returning speeds?
18. The formula for the height of a cannon ball after t seconds is H = 20t2 – t +15
(a) What is the height of the ball after 2 seconds?
(b) How long will it take before the cannon ball is at a height of 510 ft?
Page | 3
Practice Test on Topic 19 Quadratics
Test on Topic 18 Quadratics Solutions
1.
Solve the following equations.
(a) x2 + 7x + 10
=
(x + 2)(x + 5) =
0
0
(b) 3x2 – 4x + 1
= 0
(3x – 1)(x – 1) = 0
1
x = and x = 1
3
x = – 2 and x = – 5
(c) x3 – 4x2 – 5x
= 0
2
x(x – 4x – 5) = 0
x(x – 5)(x + 1) = 0
x = 0 and x = 5 and x = – 1
2.
(d)
x2 = 25
x2 – 25 = 0
(x + 5)(x – 5) = 0
x = –5 and x = 5
Solve each of the following quadratic equations by using “Completing the square”.
(a) x2 + 6x – 16
x2 + 6x
(x +3)2
(x +3)2
x+3
=
=
=
=
=
0
16
16 + 9
25
 25
5
2
or
or
(b) x2 – 6x
x2 – 8x
(x – 4)2
(x – 4)2
x–4
x–4
x
=
=
=
=
=
=
=
2x – 16
– 16
– 16 + 16
0
 0
0
4
(c) 2y2 – 8y + 6
2y2 – 8y
y2 – 4y
(y – 2)2
(y – 2)2
y–2
=
=
=
=
=
=
0
–6
–3
–3+4
1
y – 2=
y
or
3
x+3
x
=
=
1
=
x +3 =
x
–5
= –8
 1
or
y – 2=
y
–1
=
1
Page | 4
Practice Test on Topic 19 Quadratics
3.
Solve each of the following quadratic equations by using “Completing the square”.
(a) x2 + 6x – 7
x2 + 6x
(x + 3)2
(x + 3)2
x +3 =
x
= 0
= 7
= 7+9
= 16
 16
= –3 4
Answer: The two solutions are x = – 3 – 4 = 0 – 7 and x = 0 – 3 + 4 = 1
(b)
x2
x2 – 6x
– 6x – x
x2 – 7x
2
7

x 
2

=
=
=
x–6
–6
–6
=
7
–6+  
2
2
7

x  =
2

2
–6+
49
4
7

x  =
2

25
4
7
2
7
x
2
=

=

=
7 5

2 2
x
2
25
4
5
2
Answer: The two solutions are x =
4.
7 5
–
= 1 and
2 2
x=
7
5
+ =6
2
2
Solve each of the following quadratic equations by using “Completing the square”
Give your answers t 1 decimal place.
(a) x2 + 4x – 21 = 0
(x + 2)2
= 21 + 4
2
(x + 2)
= 25
x+2
=  25
x+2
= 5
x + 2 = 5 or x + 2 = – 5
x = 3 or x = – 7
(b)
x2 – 5x
(x – 2.5)2
(x – 2.5)2
x – 2.5
x – 2.5
x – 2.5 = 6.2
x = 8.7
= 32
= 32 + 6.25
= 38.25
=  38.25
=  6.2
or x – 2.5 = – 6.2
or x = – 3.7
Page | 5
Practice Test on Topic 19 Quadratics
5.
For each of the following quadratic equations indicate whether there are 0, 1 or 2 real solutions.
(You do not need to calculate the solutions – you need only say how many there are)
(a) x2 – 4x + 4
=
0
a=1 b=–4 c=4
b2 – 4ac = (– 4)2 – 4(1)(4) = 16 – 16 = 0 Number of solutions = 1
(b) x2 – 5x + 20 =
0
a = 1 b = – 5 c = 20
b2 – 4ac = (– 5)2 – 4(1)(20) = 25 – 80 = – 55
(c) 2x2 – 5x + 3 = 0
a=2 b=–5 c=3
6.
Number of solutions = 0
discriminant = b2 – 4ac = (-5)2 – 4(2)(3) = 25 – 24 = 1
since discriminant > 0 there will be two distinct real solutions.
 b  b 2  4ac
Solve the following Quadratic equation by using the quadratic formula. x 
2a
(b) x2 – 3x – 10
=
0
a = 1 b = – 3 c = – 10
 b  b 2  4ac
x
2a
discriminant
= b2 – 4ac = (-3)2 – 4(1)( – 10) = 9 + 40 = 49
3 7
3  49
=
2
2
3  7 10
37  4
The two solutions are x =

 5 and x =

 2
2
2
2
2
(b) x2 – 5x – 4
=
=
0
a=1 b=–5 c=–4
b2 – 4ac = (– 5)2 – 4(1)( – 4) = 25 + 16 = 41
x
=
x
=
so x =
 b  b 2  4ac
2a
5  41
2
5  6.4
5  6 .4
= 5.7 or x =
= 0.7
2
2
Page | 6
Practice Test on Topic 19 Quadratics
7.
Solve the following equations by any method you choose.
(a)
(b) x2 – x
= 20
2
x – x – 20 = 0
(x – 5)(x + 4)= 0
So x = 5 or x = – 4
x2 + 6x + 5 = 0
(x + 5)(x + 1)
= 0
So x = – 5 or x = – 1
(c) 4x2 + 4x – 24
= 0
4(x2 + x – 6) = 0
4(x + 3)(x – 2) = 0
So x = – 3 or x = 2
(d)
3x2 = 6x – 3
3x – 6x – 3 = 0
a=3b=–6 c=–3
b2 – 4ac =(– 6)2 – 4(3)( – 3) = 36 + 36= 72
2
 b  b 2  4ac
x =
2a
6  72
x =
6
6  8 .5
6  8 .5
So x =
= 2.4 or x =
= – 0.6
6
6
8.
Solve the following Quadratic equations by using the quadratic formula.
(a) x2 – 3x – 10
=
0
a = 1 b = – 3 c = – 10
b2 – 4ac =
x
8.
 b  b 2  4ac
2a
=
So x =
(– 3)2 – 4 x 1 x (– 10)
=
=
9 + 40
=
49
3  49
2
=
3 7
2
16 – 16 =
0
4 0
=
0.5
40
0.5
3 7
37
= 5 or x =
=–2
2
2
(b)
0.5x2
– 4x + 8 =
0.5x2
=
0
4x – 8
a = 0.5 b = – 4 c = 8
b2 – 4ac =
x
=
So x =
(– 4)2 – 4 x 0.5 x 8 =
 b  b 2  4ac
2a
=
40
=8
0.5
Page | 7
Practice Test on Topic 19 Quadratics
9.
The quadratic equation x2 – 2x + 2 = 0 has two solutions which are both complex numbers.
Use the quadratic formula to find these solutions.
a=1 b=–2 c=2
b2 – 4ac =
(– 2)2 – 4 x 1 x (2) =
So
b 2  4ac
x
=
=
4
 b  b 2  4ac
2a
2  2i
2
2  2i
x=
2
So x =
=
or
=
2(1  i )
2
2(1  i )
2
4–8
–4
4  1 =
=
=
=
4  1
=
2i
2  2i
2
=
=
(1  i )
1
(1  i )
1
=
1+i
=
1–i
10. For the quadratic function f(x) = 2x2 – 6x + 10 find the following.
(a) Is the shape of the parabola Concave Down or Convex Up?
Convex Up since positive coefficient on the x2 term.
(b) Where is its y-intercept?
Y –intercept is 10 since f(0) = 2(0)2 – 6(0) + 10 = 10
(c) Where is its vertex?
 b  (6)
6
=
=
= 1.5
2a
4
2( 2)
y-coordinate of vertex is f(1.5) = 2(1.5)2 – 6(1.5) + 10 = 4.5 – 9 + 10 = 5.5
x-coordinate of vertex is x =
Answer: Coordinates of vertex are (1.5,5.5)
Page | 8
Practice Test on Topic 19 Quadratics
11. For the quadratic equation:-
8
y = x – 6x + 8
2
(c) Where does the curve cut the y – axis
2
At y = 8
(d)
Where does the curve cut the x – axis
x2 – 6x + 8 =
(x – 2)(x – 4) =
x = 2 and x = 4
3
4
–2
0
0
(c) Draw the parabola with the equation y = x2 – 6x + 8 on the grid above (choose suitable scales)
 (6)
b
=
= 3
2a
2(1)
y-oordinate of the vertex y = x2 – 6x + 8 = (3)2 – 6(3) + 8 = 9 – 18 + 8
x-coordinate of the vertex x =
12. The quadratic y = x2 + 2x – 8
= –1
y
(a) What is the shape of the quadratic?
Answer = concave up U
(b) What is the y-intercept?
Answer Y-intercept is y = – 8
x
(c) What are the x-intercepts?
Answer Solve x2 + 2x – 8
=0
(x + 4)(x – 2) = 0
x = – 4 and x = 2
(d) What are the coordinates of the vertex?
x-coordinate =
y-coordinate =
b
2
=
= –1
2a
2
(– 1)2 + 2(– 1) – 8 = – 9

(e) Sketch the quadratic
Page | 9
Practice Test on Topic 19 Quadratics
13. For the quadratic function f(x) = x2 + 6x + 5 find the following.
(a) Is the shape of the parabola Concave Down or convex Up?
Answer:
Concave Up since it is a positive number of x2.
(b) Where is its y-intercept?
y-intercept is at (0,5)
since cuts the y-axis at f(0) = (0)2 + 6(0) + 5 = 5
(c) Where is its vertex?
b
6
6
=
=  =–3
2a
2
2(1)
y-cordinate of vertex at f(– 4) = (– 3)2 + 6(– 3) + 5 = 9 – 18 + 5 = – 4
x-cordinate of vertex at x = 
Coordinate of vertex is (– 3,– 4)
(d) Where is its x-intercepts?
x-intercepts happen when
x2 + 6x + 5
=
(x + 1)(x + 5) = 0
x = – 1 and x = – 5
0
So x-intercepts are at (– 1 , 0) and (– 5 , 0)
(e) Draw a graph of the quadratic function.
y
0
x
Page | 10
Practice Test on Topic 19 Quadratics
14. There exists two positive numbers where one is 4 more than the other and their product is 45.
(c) Write down the above information in an algebraic form.
x = first number
x + 4 = second number
so x(x + 4) = 45
(d) Find an equation and by solving it find the value of these two numbers?
Solve
x(x + 4)
= 45
x2 + 4x – 45 = 0
( x + 9)(x – 4) = 0
So x = – 9 ( cant use its negative) and x = 4 so the second number is number is x + 4 = 5 + 4 = 9
15. A rectangle has its length two more than three times its width.
The area of the rectangle is 85 square feet.
(c) Write down the above information in the form of an algebraic equation.
Width = x
Length = 3 times width + 2
Area =
x(3x + 2) =
=
3x + 2
85
85
(d) What is the length and width of the rectangle?
x(3x + 2) =
3x2 + 2x =
2
3x + 2x – 85
(3x + 17)(x – 5)
Use only x = 5
85
85
= 0
= 0
So Width = x = 5 feet and Length = 3x + 2 = 3(5) + 2 =
17 feet
Page | 11
Practice Test on Topic 19 Quadratics
16. John traveled from his own home to his mother’s home 200 miles away and then he returned back home that
same day.During these journey’s he noticed that his speed was 10 mph faster coming home than it was when
he traveled to his mothers home. The total time spent on both parts of the journey was 9 hours. What were his
speeds during this journey?
Time (going)
Time (return) =
Dis tan ce
Speed
Dis tan ce
=
Speed
=
Total time
= 9
200
200
+
= 9
x  10
x
200( x  10)
200 x
+
x( x  10)
( x  10) x
200
x
200
x  10
=
=
9
1
200( x  10)  200 x
x( x  10)
=
9
1
200 x  2000  200 x
x( x  10)
=
9
1
=
9
1
9(x2 + 10x)
9x2 + 90x
9x2 – 310x
9x2 – 310x – 2000
(9x + 50)(x – 40)
400 x  2000
x 2  10 x
400x + 2000
400x + 2000
2000
0
0
=
=
=
=
=
So x = 40 (speed going) and (speed returning) = x + 10 = 40 + 10 = 50
Page | 12
Practice Test on Topic 19 Quadratics
17. Professor Robertson was jogging one day from his home to the College a distance of 4 miles.
On the return journey he was very tired and jogged 2 miles per hour slower than on the first part of the journey.
The total time he took jogging was 9.6 hours.
What were the going and returning speeds?
Take the information and make this table
4
4
+
=
x
x2
4
4
x(x – 2) + x(x – 2)
x
x2
4(x – 2) + 4x =
4x – 8 + 4x
8x – 8
0
0
9.6
=
9.6x2 – 19.2x
= 9.6x2 – 19.2x
= 9.6x2 – 19.2x
= 9.6x2 – 19.2x – 8x + 8
= 9.6x2 – 27.2x + 8
We now solve the quadratic equation
the disciminant
9.6x(x – 2)
=
Dis tan ce
Speed
Distance
First
Journey
x
4
4
x
Return
Journey
x–2
4
4
x2
9.6x2 – 27.2x + 8 = 0
a = 9.6 b = –27.2 c = 8
b2 – 4ac =
The two solutions are:
Time =
Speed
(–27.2)2 – 4(9.6)(8)
= 739.84 – 307.2
= 432.64
x
=
x
=
x
=
( so 2 possible solutions)
 b  b2  4ac
2a
27.2  432.64
2(9.6)
27.2  20.8
19.2
Answer: The two solutions are
x=
and
x =
27.2  20.8
=
19.2
6.4
19.2
= 0.3 but x – 2 is negative and so cannot be used.
27.2  20.8
=
19.2
48
19.2
=
2.5 and x – 2 = 0.5 so we can use this solution.
Answer: The first journey the speed was x = 2.5 mph and the return journey was x – 2 = 0.5 mph
Page | 13
Practice Test on Topic 19 Quadratics
18. The formula for the height of a cannon ball after t seconds is H = 20t2 – t +15
(a) What is the height of the ball after 2 seconds?
H = 20t2 – t +15
=
20(2)2 – (2) +15 = 20(4) – 2 + 15 = 80 – 2 + 15 = 93 ft
(b) How long will it take before the cannon ball is at a height of 510 ft?
20t2 – t +15 = 510
20t2 – t – 495 =
0
a = 20 b = – 1 and c = – 495
discriminant
= b2 – 4ac = (-1)2 – 4(20)( – 495) = 1 + 39600 = 39601
There are two real number solutions t 
t
1  199 200

 5 sec
40
40
and
1  39601 1  199

40
40
t
1  199  198

 4.95 sec
40
40
Only one solution t = 5 sec is usable in this solution.
Page | 14