Mathematics

Key unit competence: By the end of this unit, learners should be able to solve
quadratic equations.

Unit outline

• Definition and examples of quadratic equations.

• Solving quadratic equations.

• Problems involving quadratic equations.

Introduction

Unit Focus Activity

1. Use your knowledge of solving equations to solve the following
real- life problem. “A number of people planned to
contribute a total of 800 000 FRW to start a small self help groups
from where they could be getting small loans to boost their business.
They agreed to contribute each an equal amount of money. Four of
them withdrew from the venture. As a result, each of the remaining
people had to contribute 10 000 FRW more to reach the target”.
Determine the:

(a) Original number (x) of people in the venture.

(b) Amount of money each contributed after the withdrawal
    of the four.

2. Compare your answer with those of other classmates during the class
presentations. Financial matters!
Pooling finances together for mutual benefit is one of the most successful ways
of empowering one another financially.

There is strength and security in numbers. Advice your working parents,
siblings, friends and neighbours about it. Consider joining one as soon as you
 finish schooling.

5.1 Definition and examples of quadratic equations

Activity 5.1

1. Expand the following expressions.

(a) 3(5x + 6)

(b) (x – 5)(x – 2)

(c) (x – 1)(x + 2)(x – 3)

2. State the highest power of the unknown in each of the expression
 in (1) above.

3. Which of the expressions in question (1) above are quadratic?

A quadratic expression has the general form ax² + bx + c where a, b and c are
real numbers. It is a polynomial of order 2. Examples of quadratic expressions include
x² + 3x + 1, a² + 2a, y² – 2y + 1, c² – 4 etc. The general form of a quadratic equation
is ax² + bx + c = 0, where a, b and c are real numbers. Examples of quadratic
equations are: x² + 3x + 1 = 0, x² – 9 = 0, x² – 5x + 6 = 0 etc.

Exercise 5.1

Expand each of the expression and state
whether it is quadratic or not:
1. (a) (x + 3)(x + 1)

   (b) (x + 2)(x – 4)(x – 1)
 

  (c) (x – 7)(x + 3) (x – 1)

  (d) x(x – 4)(x – 3)

  (e) (a2 + 8)(a – 3)

  (f) (p + 2)(p – 5)

2. Which of the following belongs to quadratic expressions? Circle the
    correct expression.

(a) (x3 – 5)(0 + 4)

(b) (x + 5)(x + 4)

(c) (x + 2)(5 a + 3)/x – 1

(d) (3x + 2)(x – 4)

(e) (2x – 7)(4x – 3)

(f) (x + 3)(x + 3)

3. Write true/false in front of an expression if it is quadratic.

(a) (a – 4)(a2 – 4)

(b) 6(x + 7)(x)

(c) (5t + 3)(3t + 2)

(d) (2 – x)(3 – x)

(e) (3 + p)(5 – p)

(f) (4 – 2y)(1 – 3y)

(g) (3x + 1)(8 – 2x)

(5 – 3x)(2 – 4x)

4. A man bought a certain number of shirts for 2 000 FRW. If each shirt
had cost 200 FRW less, he could have bought five more for the same money.

(a) Form and simplify an expression from the above information.

(b) What name can be given to the above expression?

5. The perimeter of a rectangle is 42 cm. If the diagonal is 15 cm;

(a) Form an expression that relates the perimeter, diagonal, length
   and width of the rectangle.

(b) What special name is given to the expression in 5(a) above?

5.2 Solving quadratic equations

Quadratic equations are solved using the
 following methods:

• Factorisation method

• Graphical method

• Completing squares method

• Quadratic formula method

5.2.1 Quadratic equations by factorisation method

5.2.1.1 Factorising quadratic expressions

Activity 5.2

1. What do you understand by theterm “factorise”?

2. Factorise the following quadratic expressions:

(a) x2 + x            (b) 4x2 + 2x

(c) x2 – 4

3. Factorise by grouping

   x2 + 2 x + 3x + 6. Hence.
   factorise x2 + 5 x + 6.

4. Using the fact you have learnt in 3 above, factorise x2 – x –12.

In order to find the factors of an expression such as ax² + bx, we look for the factors
that are common in both terms. The common factors of ax² and bx is x.
Thus, ax² + bx = x(ax + b). Therefore, the factors of ax² + bx are x
and (ax + b). This means, ax² + bx can be factorised as x(ax + b).
The process of finding the factors of an expression is called factorisation. This
 is the reverse of expansion.

Consider
(x + a)(x + b) = x(x + b) + a(x + b)
                     = x2 + bx + ax + ab
                     = x2 + (a + b)x + ab
The expressions (x + a) and (x + 6) are the factors of x2 + (a + b)x + ab. Similarly:
(x + 3)(x + 4) = x(x + 4) + 3(x + 4)
                     = x2 + 4x + 3x + 12
                     = x2 + 7x + 12.
The last expression shows the relationship between the terms of the quadratic
expression and its factors. Thus, a constant term ab of a quadratic expression is the
product of the number terms a and b. Whereas, the coefficient of x is the sum of
a and b. We can use these facts to factorise quadratic expressions.

Consider the expression x² + 10x + 21.
We can factorise it as follows;
Compare x² + 10x + 21 and(x + __)(x + __), considering
x² + (a + b)x + ab = (x + a)(x + b).
This means
x² + 10x + 21 = x² + (a + b)x + ab.
We need two numbers a and b such that a + b = 10 and ab = 21.

Clearly, a and b are factors of 21 whose
sum is 10. These are 3 and 7.
Rewriting:
x² + 10x + 21 as x² + (3 + 7)x + 21 gives x² + 3x + 7x + 21.
 = x(x + 3) + 7(x + 3) (factoring
common terms)
= (x + 3)(x + 7).
Therefore, x² + 10x + 21 = (x + 3)(x + 7).

When factorising a quadratic expression of the form ax² + bx + c we get two factors
whose sum is b and the product is ac. Consider a quadratic expression
6x² + 5x – 4. It can be factorised as follows:
Comparing 6x² + 5x – 4 to ax² + bx + c, we find out that a = 6, b = 5 and c = –4.
We then get two values whose product is –24 and whose sum is 5 ie ac = –24 and
a + b = 5. These values must be the factors of –24 and they are –3 and 8.
We then write 6x2 + 5x – 4 = 6x2 + (8 – 3)x – 4
                                          = 6x2 + 8x – 3x – 4.
By factorizing out the common factors,
We get
6x² + 8x – 3x – 4 = 2x(3x + 4) – 1(3x + 4).
So, 6x² + 8x – 3x – 4 = (2x – 1)(3x + 4)

Example 5.4

Write down 7 + 3x2 – 22x in terms of its factors.

Solution.

Rearranging the terms,
7 + 3x² – 22x = 3x² – 22x + 7
We have the coefficients a = 3, b = –22
and c = 7.
Sum is -22 and product is 21. The two factors that can add to give –22 andmultiply

to give 21 are –21 and –1.
  We write 3x2 – 22x + 7 = 3x2 – 21x – x + 7.
 By factorising,
3x2 – 21x – x + 7 = 3x(x – 7) – 1(x – 7)
                           = (3x – 1)(x – 7)

Exercise 5.2

1. Factorise the following expressions:

(a) x² + 6x + 8

(b) x² – 8x + 15

(c) x² + 2x – 35

(d) x² – 9

(e) x² – 49

(f) x² + x – 12

2. Can the following expressions  factorise ? Write YES or NO for the
    results.

(a) x2 + 9x + 14

(b) x2 – 11x – 12

(c) 5u² + 6u + 9

(d) 2t² – t – 42

3. A man bought a certain number of shirts for 2 000 FWR. If each shirt
had cost 200 FWR less, he could have bought five more for the same
money.

(a) Form and simplify an expression from the above information.

(b) What name can be given to the above expression?

(c) Factorise the expression obtained in 3(b) above.

4. The perimeter of a rectangle is 70 cm. If the diagonal is 25 cm:

(a) Form an expression that relates the perimeter, diagonal, length
   and width of the rectangle.

(b) Factorise the expression obtained in 4(a) above.

5.2.1.2 Solving quadratic equations by factorisation method.

Activity 5.3

1. Given that (x – 2)(x + 3) = 0, what are the possible values of (x – 2)
    and (x + 3)? Obtain the values of x for each expression.

2. Given that (x + 4)(x + 6) = 0, find two possible values of x.

3. Factorise completely 4x2 – 2x. Hence find the values of x given
   that 4x2 – 2x = 0. (Hint: Use the idea learnt above.)

In order to solve a quadratic equation, the quadratic expression is factorized
so that the equation is in the form (x + a)(x + b) = 0.
Then either (x + a) = 0 or (x + b) = 0.
Thus x = –a and x = –b.

Example 5.5

Solve: (x – 4)(x + 1) = 0

Solution

If(x – 4)(x + 1) = 0, then either x – 4 = 0
   or x + 1 = 0.
Therefore, x = 4 or x = –1.
Hence the roots of the equation
(x – 4)(x + 1) = 0 are 4 or –1.

Example 5.6

Solve the equation x2 + 7x + 6 = 0.

Solution
x2 + 7x + 6 = 0
Factorizing, x² +7x + 6, gives
(x + 6)(x + 1) = 0.
Therefore, x + 6 = 0 or x + 1 = 0;
Which means x = –6 or x = –1

Example 5.7

Solve: x2 – x – 29 = 1

Solution

Always ensure that the quadratic expression is equated to zero. This is the only time the
method used in these examples can apply.
Thus, x² – x – 29 = 1 should be rewritten
as x² – x – 29 – 1 = 0. That is,
    x² – x – 30 = 0.
The factors of –30, whose sum is –1 and product –30, are –6 and 5.
 We write x² – 6x + 5x – 30 = 0
 We then factorise and get,
 x(x – 6) + 5(x – 6) = (x – 6)(x + 5) = 0.
 Either x – 6 = 0 or x + 5 = 0.
 x = 6 or x = –5.
 The roots are –5 or 6.

Example 5.8

Solve:
(a) x2 – 49 = 0 (b) x2 – 6x = 0

Solutions

(a) x² – 49 = 0 can be written as
     x² – 72 = 0
    (x – 7)(x + 7) = 0
    x – 7 = 0 or x + 7 = 0
    x = 7 or x = –7.
The roots are -7 or 7.

(b) Factorizing x2 – 6x = 0 gives
    x(x – 6) = 0
Either x = 0 or x – 6 = 0
   x = 0 or x = 6
The roots are 0 or 6.

Example 5.9

Solve the quadratic equation.
6x² + 5x – 4 = 0

Exercise 5.3

1. Obtain the values of x from the following expressions.
(a) (x + 4)(x + 2) = 0

(b) (x – 5)(x – 3) = 0

(c) (x – 5)(x + 7) = 0

(d) (x2 – 9) = 0

2. Factorise the following expressions hence find the values of unknowns.
(a) x2 + 9x + 14 = 0

(b) x2 – 11x – 12 = 0

(c) u2 + 6u + 9 = 0

(d) t2 – t – 42 = 0

(e) a2 – 2a + 1 = 0

(f) y2 + 8y = 0

3. Can the following equations be solved by factorisation? Write True/
   False and write down the solution set of equation.
(a) x2 + 10x = 24

(b) x2 = 4x – 3

(c) 6x2 – 29x + 35 = 0

(d) 6x2 – x + 1 = 0

5.2.2 Solving quadratic equations by graphical method

Activity 5.4

Complete the following tables of values
for;

(b) Function y = x2 +3x + 6.

2. State the coordinates in 1 (a) and 1(b) in ordered pairs (x, y).

3. Plot the graphs for 1(a) and 1(b) on different Cartesian planes.

4. For each graph, read and record the x-coordinate(s) of the point(s)
    where the graph cuts x-axis.

5. Solve the equations 2x2 + 3x – 2 = 0 and x2 + 3x + 6 = 0 by factorisation
    method.

In a quadratic function graph, the x-coordinate of the point where the
graph cuts x-axis gives the solution to the quadratic equation represented by the
function.

1. When the graph cuts the x-axis at  one point, then the equation has one
    repeated solution.

2. When the graph cuts x-axis at two points, then the equation has two
     different solutions.
3. When the graph does not cut x-axis at any point, then the equation has no
    solution in the field of real numbers.

The graph of y = –x2

All values of y are numerically the same as the corresponding values of y in y = x2
but are negative. The shape of the curve will be the same but inverted.

The equation x2 = 0 has only one repeated solution since the graph cuts x-axis at only
one point i.e. x = 0.

Exercise 5.4

1. Plot the graph of y = x2 – 4x + 4 for values of x from -1 to +5. Solve from
     your graph the equations:
   (a) x2 – 4x + 4 = 0

   (b) x2 – 4x + 1 = 0

     c) x2 – 4x - 1 = 0.
2. Plot the graph of the function
     x2 – 6x + 5 for –1 < x < 7. Use your
  graph to solve the equation
     x2 – 6x + 5 = 0.

3. Draw the graph of y = 2x2 – 7x – 2 for
    values of x from -3 to +3. Use your graph to solve the equations:
      (a) 2x2 – x = 4,

     (b) 2x2 – x + 6 = 0

     (c) 2x2 – x – 4 = 2x.

4. Plot the graph of y = 1
4 x2 for domain
–4 < x < 4. Using the same scale and
axes, draw the curve of y = 1/4 x² – 3.

5. Plot the graph of the function
  y = 2 – x – x² from domain –3 < x < 3.
Use your graph to solve the equation
 x² + x = 2.

6. Draw the graph of the function
y = 2x² – 7x – 2 for values of x from
–1 to 5. By drawing suitable lines on
the same axes, solve, where possible,
the following equations:
(a) 2x² – 7x = 2,

(b) 2x² – 8x + 4 = 0

7. Copy and complete the following
   table of values for y = 6 + 3x – 2x².

Draw the graph of y = 6 + 3x – 2x²
for domain –2 < x < 3.
Use the graph to obtain solutions of
the equations:

(a) 6 + 3x – 2x² = 0

(b) 2 + 3x – 2x² = 0

(c) 3 + x – 2x² = 0.

8. Plot the graph of y = x2 – x – 2 after completing the following table for
values of x and y.

By drawing a suitable straight line on your graph, solve the equation
x² – 2x = 0.

9. Draw the graph of y = ( 1/2 x – 3)(x + 3)
for domain –3 < x <4.
From your graph find the values of x
for which ( 1/2 x - 3)(x + 3) = 2.

10. Plot the graph of the function
  y = x² – 3x for domain –1 < x < 4.
Use your graph to find:

(a) The range of values of x for which
      the function is negative.

(b) The solutions of the equations
x2 – 3x = 1 and x2 – 2x – 1 = 0.

11. Given that y = (3x + 1)(2x – 5), copy and complete the following table for
    values of x and y.

Plot the graph of y = (3x + 1)(2x – 5) from x = –1 to x = 4.
By considering the points of inter-section of the two graphs, a
certain quadratic equation in x can be solved. Write down and simplify
the equation and obtain its roots from the graphs.

5.2.3 Solving quadratic equations by completing squares
method

5.2.3.1 Perfect squares

Activity 5.5

Determine whether the following
quadratic expressions are perfect
squares or not.
(a) x² + 8x + 16

(b) x² + 10x + 12

(c) 9x² + 6x + 1

(d) 36x² – 8x + 4

Recall that:
1. (a + b)² = a² + b² + 2ab, and

2. (a – b)2 = a² + b² – 2ab

The right hand sides of 1 and 2 are the expansions of the squares on the left hand
sides. Such expressions are called perfect squares. They may be written in words as:

1. The square of the sum of two numbers is equal to the sum of their
  squares plus twice their product.

2. The square of the difference of two numbers is equal to the sum of their
    squares minus twice their product.

Note:

Instead of expanding (2x – 5y)2 in order to
compare the middle term with that of the given expression, we could alternatively
first check if –40xy is equal to 2(2x)(–5y). In this case, it is not. This indicates that
4x2 – 40xy + 25y2 is not a perfect square.

Example 5.17

Complete 9p² – 12pq + 4q² = ².

Exercise 5.5

1. Write down the squares of the following.

2. State whether each of the following expressions is a perfect square. If it
 is, write it in the form (a + b)² or (a – b)².

5.2.3.2 Completing squares and solving quadratic equations

Activity 5.6

1. What must be done to make the following quadratic expressions

Completing the square when the coefficient of x2 is one

The quantity to be added is the square of half of the coefficient of x (or whatever
letter is involved), i.e. to make x2 + bx a perfect square, we add ( b/2 )²
to get
x2 + bx + ( b/2 )²= (x + b/2 )²
, where b is any positive or negative number.

In some cases, it is not possible to solve a quadratic equation by the method of
factorisation because the LHS does not factorise. In such a case, we first rearrange
the equation to make the LHS a perfect square, i.e. we complete the square on
the LHS.

If LHS of an equation factorises, it is better to use the method of factorisation
instead of completing the square.

Example 5.18

What must be added to x2 + 8x to make
the result a perfect square?

Solution

Suppose x² + 8x +k is a perfect square and
that it is equal to (x + a)2, i.e.
x² + 8x + k = (x + a)2.
We know that

(x + a)² = x² + 2ax + a²

∴ x² + 8x + k = x² + 2ax + a².

The expression on the LHS and that on the
RHS can only be equal if the corresponding
terms are equal.
Comparing coefficients of x:
  2a = 8
  ∴ a = 4.
Comparing the constant terms:
k = a² = 4² = 16.

∴ 16 must be added to the expression x2 + 8x.
Then x² + 8x + 16 = (x + 4)².

Check by opening the bracket on the RHS!

Example 5.19
Find the term that must be added to p2 – 6p
to make the expression a perfect square.
Solution
Suppose p2 – 6p + k is a perfect square.
Then, p2 – 6p + k = (p – a)2
i.e. p2 – 6p + k = p2 – 2ap + a2
Comparing coefficients of p:
–2a = –6
∴ a = 3.
Comparing the constant terms:
k = a² = 32 = 9
∴ 9 must be added to the expression.
Then, p² – 6p + 9 = (p – 3)².

Note:
In Example 5.18, the coefficient of x is 8;
half of 8 is 4; and the square of 4 is 16.
Hence, 16 must be added.

In Example 5.19, the coefficient of p is –6;
half of –6 is –3; and the square of –3 is 9.
Hence, 9 must be added.

Example 5.20

Solve the equation x2 + 8x + 9 = 0.

Example 5.21

Solve the equation q2 – 5q – 2 = 0.

Exercise 5.6

1. What term must be added to each  of the expressions below to make the
    expression a perfect square? Write
   each expression as (a + b)2 or (a – b)2.

(a) y2 + 10y                (b) x2 – 8x

(c) p2 + 6pq                (d) d2 + 5d

(e) a2 – 12ab              (f) q2 – 7q

 (g) n2 + n                    m2 – mn

2. Solve the equations below by factorising where possible, otherwise
   by completing the square. Do not put the answer in decimal form.

(a) x² – 4x – 21 = 0

(b) y² + 3y – 10 = 0

(c) x² + 3x – 11 = 0

(d) d² + 4d – 4 = 0

(e) p² + 3p – 2 = 0

(f) 25 – 10x + x² = 0

(g) n² – 14n + 2 = 0

t² – 15t – 4 = 0

Completing the square when the coefficient of x2 is not equal to one

If the quadratic equation is of the form
ax² + bx + c = 0, where a ≠ 1 and the LHS does not factorise, first divide both sides
by a to make the coefficient of x2 one and then complete the square.

Example 5.22

Solve the equation 2x² + 14x + 9 = 0,
giving your answer correct to 3 d.p.

Solution

2x² + 14x + 9 = 0
First make the coefficient of x2 one by dividing both sides by 2.

Exercise 5.7

1. Solve the following quadratic equations by completing squares
       method.
(a) x² – 2x – 8 = 0

(b) x² – 5x + 2 = 0

(c) x² + 2x – 8 = 0

(d) x² + 2x – 1 = 0

(e) 2x² – 10x = 0

(f) 2x² + 6x + 1 = 0

2. Complete the squares in the following expressions.
(a) x² – 8x – 7 = 0

(b) 2x² + x – 6 = 0

(c) 2x² – 5x + 2 = 0

(d) x² + 4x = 0

(e) 3x² – 2x – 6 = 0

(f) x² + 1/2 x = 0

3. Solve the following equations by factorisation if possible, otherwise by
   completing the square. Give your answers to 2 d.p. where necessary.
(a) 2x² – 3x = 9

(b) 2y² – 4y + 1 = 0

(c) 2b² + b + 1 =0

(d) 4x² – 8x = – 1

(e) 4p² – 8p + 3 = 0

(f) 4p² = 8p + 3

(g) 3e² = 9e – 2

3a² – 2 = 12a

(i) 5x² = – 1 – 15x

(j) 2z² + 10z + 5 = 0

5.2.4 Solving quadratic equations by using formula method

Activity 5.7

1. Use the steps that are applied in solving quadratic equations by
   completing squares to obtain the values of x in terms of a, b and c
   for which px² + qx + r = 0

(a) First divide the whole equation by p to make the coefficient of x2 be one.

(b) Solve for the value of x by completing the square.

(c) Make x the subject of the formula.

2. Use the expression you obtained in step 1 to solve the equation
    2x² – 3x – 4 = 0 by substituting
   p = 2, q = –3 and r = –4.

3. Solve the quadratic equation
  2x²– 5x + 3 = 0 by factorization method.

4. Compare your results for step 2 and step 3. What do you notice?

Consider the quadratic equation  ax² + bx + c = 0. Let us solve by
completing square method.

1. Make sure the coefficient of x2 is one. This is done by dividing by a
     throughout the given equation.

2. Half the coefficient of x, square it and subtract it from the given expression.

3. Take the constant part to the right hand side and find the square root
     both sides and solve for the values of x.

Simplifying the expression under the square root using LCM and taking square
root of denominator we get

This formula is called the quadratic formula and it is used to solve any
quadratic equation provided the coefficients a, b and c are all known.

Exercise 5.8

1. Solve the following with the aid of quadratic formula, giving answers to
    two decimal places where necessary:
   (a) 2x² + 11x + 5 = 0

   (b) 3x² + 11x + 6 = 0

   (c) 6x² + 7x + 2=0

   (d) 3x² – 10x + 3 = 0

  (e) 5x² – 7x + 2 = 0

  (f) 6x² – 11x + 3 = 0

  (g) 2x² + 6x + 3 = 0

  x² + 4x + 1 = 0

2. From the questions below, circle the correct solution.

(a) When the quadratic equation
    2x² + 5x – 3 = 0 is solved, the results obtained are;
  (i) –3 and 2         (ii) –3 and 1/2

(iii) 4 and – 2         (iv) 6 and –1

(b) The equation 2x2 – x – 6 = 0 when solved gives the values of x as;

(i) 2 and –3         (ii) –1 and 3

(iii) 3 and –2

(iv) None of the above

(c) Which of the solutions below is suitable solution for
  2x² + 3x – 5 = 0?

(i) –2.5 and 1

(ii) –1.5 and 3

(iii) –3.4 and –2

(iv) None of the above

3. Write true/false for the statements
below.
(a) 1 and 9 are the solutions to x2 – 10x + 9 = 0

(b) 3x² – 10x + 3 = 0 is a perfect square

(c) x² + 9x + 14 = 0 has no solution in the field of real numbers

(d) x² + 5x – 500 = 0 has two solutions one is positive and
     another is negative

(e) The equation x² – x – 12 = 0 cannot be solved by factorisation
     method

5.2.5 Solving Quadratic equations by synthetic division method

Activity 5.8

Use internet or library to research about factorisation of quadratic equations and
polynomials by using Horner’s rule. List the steps that are followed using
your own words.

Consider the quadratic equation
ax² + bx + c = 0.
To solve this quadratic equation, we proceed as follows:

Steps

3. Identify all the factors of the constant c both positive and negative. Let the
    factors of c be e, f, g, h etc. Choose the factor of c which can
    reduce the quadratic expression to zero. Let the factor of c be f. This
    means that f is one of the solutions of the quadratic equation provided. i.e.
    x = f and the linear factor is (x – f)

4. Arrange the coefficients and the factor f as shown below.

For vertical patterns, add terms while for the diagonal patterns, multiply by factor f.
Hence on dividing the expression ax² + bx + c by (x – f), we should get the
other factor (quotient) as ax + (b + af). Thus the quadratic equation ax2 + bx + c
= (x – f)(ax + (b + af) in factorised form.
We therefore solve the values of x from the factors (x – f)(ax + (b + af)) = 0, meaning
 x – f = 0 and ax + (b + af) = 0

Example 5.25

Solve the quadratic equation x² – x – 6 = 0 using synthetic division method.

Solution

The factors of 6 are ±1, ±2, ±3 and –6. The  factor of -6 that can reduce the quadratic
equation to zero is 3.

Example 5.26

Solve the quadratic equation
3x2 – 5x – 2 = 0 by synthetic division method.

Exercise 5.9

Solve the following quadratic equations by synthetic division method.
1. x² + x – 12 = 0

2. x² + 9x + 14 = 0

3. u² + 6u + 9 = 0

4. a² – 2a + 1 = 0

5. x² + 10x = 24

6. v² – 36 = 0

7. x² – 11x – 12 = 0

8. t² – t – 42 = 0

9. y2 + 8y = 0

10. x² = 4x – 3

11. 4x² – 9 = 0

5.3 Problems involving quadratic equations

Activity 5.9

A 3 hour cruise ship goes 15 km upstream and then back again. The
water in the river has a speed of 2 km/h. Let x represent the speed of the ship.

(a) Write down an expression for the speed of the ship upstream.

(b) Write down an expression for the speed of the ship downstream.

(c) Find expression for the time taken by the ship to move up and
     downstream.

(d) Write down and simplify an expression for the total time taken by the ship to move up and
   downstream.

(e) What name can be given to the expression obtained in (d) above?

(f) What is the ship’s speed and how long was the upstream journey?

Quadratic equation can be modelled out of a well given data. There is no clear
and specific method for this apart from reading and understanding the problem
word by word and then applying what you have understood to model out a correct
mathematical expression.

Example 5.27

The perimeter of a rectangle is 46 cm. If the diagonal is 17 cm, find the width of
the rectangle.

Exercise 5.10

Solve by forming a quadratic equation:

1. Two numbers which differ by 3,have a product of 88. Find the two
     numbers.

2. The product of two consecutive odd numbers is 143. Find those two odd
    numbers.

3. The length of a rectangle exceeds the  width by 7 cm. If the area is 60 cm2,
    find the length of the rectangle.

4. The length of a rectangle exceeds the width by 2 cm. If the diagonal is 10 cm
     long, find the width of the rectangle.

5. The area of the rectangle exceeds the area of the square by 24 m2. Find x.

6. The perimeter of a rectangle is 68 cm. If the diagonal is 26 cm, find the
   dimensions of the rectangle.

7. A man walks a certain distance due North and then the same distance
  plus a further 7 km due East. If the final distance from the starting point
  is 17 km, find the distances he walks north and East.

8. A farmer makes a profit of x FRW on
   each of the (x + 5) eggs her hen lays. If her total profit was 84,000 FRW,
   find the number of eggs the hen lays.

9. A number exceeds four times its reciprocal by 3. Find the number.

10. Two numbers differ by 3. The sum of their reciprocals is 7/10 ; find the
 numbers.

Unit Summary

• A quadratic equation: This is an equation that is of the form
  ax² + bx + c = 0 where a, b and c are constants and a ≠ 0.
  Examples of quadratic equations are
 x² + 9x + 14 = 0, u2 – 5u + 4 = 0,
 7 – 6r + r2 = 0 etc.

• Quadratic equations can be solved by the following methods.

• By factorisation method

• Graphical method

• Completing squares method

• Quadratic formula method

• Synthetic division method

• In order to solve a quadratic equation,
  the quadratic expression is factorized
 so that the equation is in the form
(x + a)(x + b) = 0.
Then either (x + a) = 0 or (x + b) = 0
Thus x = –a and x = –b

• In a quadratic function graph, the x-coordinate of the point where the
  graph cuts x-axis gives the solution to the quadratic equation represented
  by the function.

(a) When the graph cuts the x-axis at one point, then the equation
   has one repeated solution.

(b) When the graph cuts x-axis at two points, then the equation has
      two different solutions.

(c) When the graph does not cut x-axis at any point, then the
     equation has no solution in the field of real numbers.

• Before solving any quadratic equation by completing squares method, it is
   better to understand what perfect squares are.
   For example x² + 6x + 9 = (x + 3)² is a perfect square.
   Remember that (a + b)² = a2 + 2ab +b2 and that (a – b)² = a² – 2ab + b2

• The quadratic formula

ax² + bx + c = 0 provide the coefficients a, b and c are all known.

• The quadratic equation
  ax² + bx + c = 0 can also be solved by
  synthetic division method as long as the value x = f is the factor of constant
  c. We follow the trend below.

For vertical patterns, add terms while for the diagonal patterns, multiply by
 factor f.
Hence, on dividing the expression
ax² + bx + c by (x – f), we should get the other factor (quotient) as
ax + (b + af)
Thus the quadratic equation
 ax² + bx + c = (x – f)(ax + (b + af)) in factorised form.
We therefore solve the values of x from
the factors (x – f)(ax + (b + af)) = 0.
Meaning x – f = 0 and ax + (b + af) = 0

Unit 5 Test

1. A cyclist travels 40 km at a speed of x km/h. Find the time taken in terms of
    x. Find the time taken when his speed is reduced by 2 km/h. If the difference
    between the times is 1 hour, find the value of x.

2. A train normally travels 240 km at a certain speed. One day, due to bad
    weather, the train’s speed is reduced by 20 km/h so that the journey takes two
    hours longer. Find the normal speed.

3. An aircraft flies a certain distance on a bearing of 135º and then twice
    the distance o a bearing of 225º. Its distance from the starting point is
    then 350 km. Find the length of the first part of the journey.

4. In the following figure, ABCD is a rectangle with AB = 12 cm and
    BC = 7 cm. AK = BL = CM = DN =
    x cm. If the area of KLMN is 54 cm2 find x.

5. The numerator of a fraction is 1 less
    than the denominator. When both numerator and denominator are
    increased by 2, the fraction is increased1/12
.     Find the original fraction.

6. The perimeters of a square and a
    rectangle are equal. One side of the rectangle is 11 cm and the area of the
   square is 4 cm2 more than the area of the rectangle. Find the side of the square.

7. In a right angled triangle PQR,∠Q = 90º, QR = x cm, PQ = (2x – 2)
    cm and PR = 30 cm. Find x.

8. Draw the graph of y = x² – 4x + 4 for  values of x from –1 to +5. Solve from
  your graph the equations:
(a) x² – 4x + 4 = 0

(b) x² – 4x + 1 = 0

(c) x² – 4x – 1 = 0.

9. Solve the following quadratic equations by synthetic division method.
(a) x² + x – 12 = 0

(b) x² + 9x + 14 = 0.