1. More about Equations (2)12
More about Equations (2)
Case Study
12.1 Equations Reducible to Quadratic Equations
12.2 Solving Simultaneous Equations
12.3 Problems Leading to Quadratic Equations
Chapter Summary

2. Case Study
We have a budget of $40. Go and buy some yellow and red ribbons to decorate the classroom.
I know that the yellow ribbon is $2 cheaper than the red ribbon.
If we use all $40 to buy only one kind of ribbon, there will be one more roll of yellow ribbon than red ribbon and there will be no change.
In order to find out the prices of a roll of yellow ribbon and a roll of red ribbon, we have to set up an equation.
Let the prices of yellow ribbon and red ribbon be $x per roll and $(x  2) per roll respectively.
According to the given conditions,
Using the technique of solving quadratic equations, we can solve for x.

3. 12.1 Equations Reducible to Quadratic Equations
In Book 4, we learnt how to solve quadratic equations.
In fact, many equations are not quadratic in appearance, but they can be transformed into quadratic equations that can be solved easily.
In this section, we will discuss four types of equations that can be transformed into quadratic equations:
1. Fractional equations
2. Higher degree equations
3. Equations with surd form
4. Indicial equations

4. 12.1 Equations Reducible to Quadratic Equations
A. Fractional Equations
Fractional equations are equations that consist of algebraic
fraction in the form or

5. Example 12.1T 12.1 Equations Reducible to Quadratic Equations
A. Fractional Equations
Example 12.1T
Solve
Solution:
In this example, x cannot be zero.
Do not forget to check whether the answers obtained can satisfy the given equation.
Expand the expression first.
Multiply both sides by x.

6. Example 12.2T 12.1 Equations Reducible to Quadratic Equations
A. Fractional Equations
Example 12.2T
Solve
Solution:
If x  2 and x  6, we may multiply both sides by
(x  2)(x  6).
x  2 and x  6

7. 12.1 Equations Reducible to Quadratic Equations
B. Higher Degree Equations
Consider the equation 4x4  15x2  4  0.
We can use the substitution y  x2 to transform the given equation into a quadratic equation 4y2  15y  4  0.
Then solve for y:
4y2  15y  4  0
(4y  1)(y  4)  0
y  4 or
Since we are required to solve for the real roots of 4x4  15x2  4  0, substitute each of the values of y into y  x2:
x2  or 4
are not real numbers

8. Example 12.3T 12.1 Equations Reducible to Quadratic Equations
B. Higher Degree Equations
Example 12.3T
Find the real roots of the equation x6  7x3  8  0.
Solution:
Let y  x3, the equation becomes
y  1 or 8
Do not stop here. We are required to find
the values of x.
Since y  x3,
The equation has 4 unreal roots which are the roots of the equations x2  x  1  0 and x2  2x  4  0.

9. 12.1 Equations Reducible to Quadratic Equations
C. Equations with Surd Form
Consider the equation 2x   6.
We can use the substitution y  to transform the given equation into a quadratic equation 2y2  y  6.

10. Example 12.4T 12.1 Equations Reducible to Quadratic Equations
C. Equations with Surd Form
Example 12.4T
Solve the equation
Solution:
If we put y  , then y2  x.
Let y  , the equation becomes
y  4 or 2
Since y  , we have
should be positive.
∴ The solution is x  16.

11. 12.1 Equations Reducible to Quadratic Equations
D. Indicial Equations
Consider the equation 32x  3x  12  0.
It can be rewritten as (3x)2  3x  12  0.
We can use the substitution y  3x to transform the given equation into a quadratic equation y2  y  12  0.
Then solve for y:
y2  y  12  0
(y  3)(y  4)  0
y  3 or 4
Since y  3x,
Since 3x  0, for all real values of x, so 3x  4 is rejected.
3x  3 or 4 (rejected)
 31
x  1

12. Example 12.5T 12.1 Equations Reducible to Quadratic Equations
D. Indicial Equations
Example 12.5T
Solve the equation 9x  10(3x)  24  0.
(Give the answers correct to 3 significant figures.)
Solution:
Rewrite the equation as (3x)2  10(3x)  24  0.
Let y  3x, the equation becomes
y  4 or 6
Since y  3x, we have: 3x  4 or 3x  6
log 3x  log 4 or log 3x  log 6
x log 3  log 4 or x log 3  log 6 or
(cor. to 3 sig. fig.)

13. 12.2 Solving Simultaneous Equations
A. Solving Simultaneous Equations by the
Algebraic Method
In junior forms, we learnt how to solve a pair of simultaneous
 y  2x  1
linear equations in two unknowns, such as 
 x  2y  7
We can make one unknown (x or y) the subject of one of the linear equations, and then substitute the expression of that unknown into the other equation to eliminate that unknown.
Finally, we can solve the linear equation in one unknown.
For a pair of simultaneous equations in two unknowns in which one is
 y  x  6
linear and the other is quadratic, such as  , we can solve it
 y  x2
in a similar way.

14. 12.2 Solving Simultaneous Equations
A. Solving Simultaneous Equations by the
Algebraic Method
 y  x  6
To solve the simultaneous equations  :
 y  x2
Substituting (1) into (2), we have
To find the value of the other unknown, we always substitute the known value into the linear equation.
Substituting x  2 into (1), we have y  2  6  4.
Substituting x  3 into (1), we have y  3  6  9.
It is NOT appropriate to express the solutions as:
x  2 or 3; y  4 or 9
Notes:
The solutions may also be written as (2, 4) or (3, 9).

15. Example 12.6T 12.2 Solving Simultaneous Equations Solution:
A. Solving Simultaneous Equations by the
Algebraic Method
Example 12.6T
Solve x2  x  y  y  3x  6.
Solution:
The given equation can be expressed as:
From (2), we have y  3x  (3)
For the linear equation, we have to make y the subject first and then substitute it into the quadratic equation.
Substituting (3) into (1), we have
x  2 or 6
Substituting x  2 into (3), we have y  3(2)  6  0.
Substituting x  6 into (3), we have y  3(6)  6  24.

16. Example 12.7T 12.2 Solving Simultaneous Equations Solution:
A. Solving Simultaneous Equations by the
Algebraic Method
Example 12.7T
Consider the simultaneous equations:
If the simultaneous equations do not have any real solutions, find the range of the values of k.
Solution:
Substituting (1) into (2), we have
Since the simultaneous equations do not have any real solutions, the discriminant of (3) is D  0.

17. Example 12.8T 12.2 Solving Simultaneous Equations Solution:
A. Solving Simultaneous Equations by the
Algebraic Method
Example 12.8T
Solve the simultaneous equations xy  x  6y  3.
Solution:
The given equation can be expressed as:
From (2), we have x  3  6y (3)
Substituting (3) into (1), we have
Substituting y  1 into (3),
x  3  6(1)
 3
Substituting y  into (3),
x  3  3
 6

18. 12.2 Solving Simultaneous Equations
B. Number of Points of Intersection of a
Parabola and a Straight Line
When solving a pair of simultaneous equations in which one is linear and the other is quadratic by the graphical method (which was discussed in Chapter 11), the graphs of the parabola and the straight line will satisfy one of the following situations:
1. Two distinct points
2. One point only
3. No points of intersection
In actual practice, we can determine the number of points of intersection of the two graphs algebraically.

19. 12.2 Solving Simultaneous Equations
B. Number of Points of Intersection of a
Parabola and a Straight Line
To determine the number of points of intersection of the two graphs algebraically:
Step 1: Use the method of substitution to eliminate one of the variables (usually y) of the simultaneous equations.
Then we get a quadratic equation in one variable.
Step 2: Check the discriminant (D) of the quadratic equation obtained in Step 1.
(a) If D  0, then there are two points of intersection.
(b) If D  0, then there is only one point of intersection.
(c) If D  0, then there are no points of intersection.

20. Example 12.9T 12.2 Solving Simultaneous Equations Solution:
B. Number of Points of Intersection of a
Parabola and a Straight Line
Example 12.9T
Without solving the simultaneous equations, find the number of points of intersection of the parabola y  x2  5x  6 and the line y  x  4.
Solution:
Consider
Substituting (2) into (1),
The discriminant of (*)  62  4(1)(10)
 76
 0
 There are two points of intersection.

21. Example 12.10T 12.2 Solving Simultaneous Equations Solution:
B. Number of Points of Intersection of a
Parabola and a Straight Line
Example 12.10T
If there are no points of intersection of the straight line y  2(x  k) and the parabola y  x2  10x  3, find the range of values of k.
Solution:
Consider the simultaneous equations
Substituting (1) into (2),
Since there are no points of intersection, the discriminant of equation (3) is D  0.

22. 12.3 Problems Leading to Quadratic Equations
A. Practical Problems Involving Quadratic
Equations
In this section, we are going to solve problems involving equations which can be transformed into quadratic equations.
Strategies for Solving Practical Problems
Step 1: Study the problem carefully and understand the objective of the problem. Draw a figure or diagram if necessary.
Step 2: First set a variable, say x, to one of the unknown quantities. Then try to represent all the other unknown quantities in terms of x.
Step 3: Set up an equation that can be transformed into a quadratic equation.
Step 4: Solve the equation.
Step 5: Check all solutions in the context of the original problem.

23. Example 12.11T 12.3 Problems Leading to Quadratic Equations Solution:
A. Practical Problems Involving Quadratic
Equations
Example 12.11T
In the figure, rectangle ABCD is similar to rectangle PQRS. Find the possible value(s) of x.
Solution:
Since ABCD is similar to PQRS, their corresponding sides are in proportion.

24. Example 12.12T 12.3 Problems Leading to Quadratic Equations Solution:
A. Practical Problems Involving Quadratic
Equations
Example 12.12T
Eric cycles 18 km to a country park. On his return journey, he cycles
4 km/h faster. The total time taken for the whole journey is 99 minutes.
(a) Find the speed of the return journey.
(b) Find the time taken (in minutes) for the return journey.
Solution:
(a) Let x km/h be the speed of the return journey. Then the original speed is (x  4) km/h.
Time taken for:
the journey to the park  hours
the return journey  hours
 The speed of the return journey is 24 km/h.

25. Example 12.12T 12.3 Problems Leading to Quadratic Equations Solution:
A. Practical Problems Involving Quadratic
Equations
Example 12.12T
Eric cycles 18 km to a country park. On his return journey, he cycles
4 km/h faster. The total time taken for the whole journey is 99 minutes.
(a) Find the speed of the return journey.
(b) Find the time taken (in minutes) for the return journey.
Solution:
(b) Time taken for the return journey

26. Example 12.13T 12.3 Problems Leading to Quadratic Equations Solution:
A. Practical Problems Involving Quadratic
Equations
Example 12.13T
Susan is going to make 70 teddy bears in N days. If she makes four more teddy bears each day, then she can take two days less to make 70 teddy bears. Find the value of N.
Solution:
Number of teddy bears that Susan originally makes each day 
New number of teddy bears that Susan makes each day 
New number of days taken  N  2

27. 12.3 Problems Leading to Quadratic Equations
B. Practical Problems Involving Simultaneous
Equations
We are now going to solve problems that lead to simultaneous equations in two unknowns.
It is important to note that the solutions obtained must be checked since some solutions may not satisfy the problem.

28. Example 12.14T 12.3 Problems Leading to Quadratic Equations Solution:
B. Practical Problems Involving Simultaneous
Equations
Example 12.14T
Consider a rectangle with dimensions x cm  y cm. If the perimeter of the rectangle is 28 cm and the area is 49 cm2, find x and y.
Solution:
Since the perimeter of the rectangle is 28 cm, we have
Since the area of the rectangle is 49 cm2, we have
From (1), we have y  14  x (3)
Substituting (3) into (2), we have
Substituting x  7 into (3),
y  14  7
 7

29. Example 12.15T 12.3 Problems Leading to Quadratic Equations Solution:
B. Practical Problems Involving Simultaneous
Equations
Example 12.15T
A 18p cm long wire is cut into two parts for which each part is used to form a circle. If the sum of the areas bounded by the two circles is 53p cm2, find the radii of the two circles.
Solution:
Let the radii of the 2 circles be x cm and y cm respectively, where x  y.
Substituting (1) into (2), we have
Since the sum of the circumferences of the 2 circles is 18 cm, we have
2x  2y  18
x  y  9
y  9  x (1)
x  7 or 2
Since the sum of the areas of the 2 circles is 53p cm2, we have
x2  y2  53
x2  y2  (2)
When x  7, y  2;
When x  2, y  7. (rejected)
 The radii of the two circles are 7 cm and 2 cm.

30. Chapter Summary 12.1 Equations Reducible to Quadratic Equations
1. Fractional equations
2. Higher degree equations
3. Equations with surd form
4. Indicial equations

31. Chapter Summary 12.2 Solving Simultaneous Equations by
the Algebraic Method
1. Use the method of substitution to solve a pair of simultaneous equations in two unknowns in which one is linear and the other is quadratic.
2. The number of points of intersection of a parabola and a line can be determined by the discriminant (D) of the quadratic equation in one of the unknowns (x or y).
(a) If D  0, there are two points of intersection.
(b) If D  0, there is one point of intersection.
(c) If D  0, there are no points of intersection.

32. Chapter Summary 12.3 Problems Leading to Quadratic Equations
Strategies for solving practical problems:
Step 1: Study and understand the problem.
Step 2: Set a variable, say x, to one of the unknown quantities. Then try to represent all the other unknown quantities in terms of x.
Step 3: Set up an equation that can be transformed into a quadratic equation.
Step 4: Solve the equation.
Step 5: Check all solutions in the context of the original problem.