Учебное пособие "Quadratic equations"

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QQUUAADDRRAATTIICC EEQQUUAATTIIOO

MOSCOW, 2011

WORKBOOK

OONNSS

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ATLANTIC INTERNATIONAL SCHOOL Ildar Ilyazov Quadratic equations. Workbook / Ildar Ilyazov – Moscow: AIS, 2011. – 32 pages. Reviewed by: Vasekin Sergey Vladimirovich

© Atlantic International School © Ildar Ilyazov

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PREFACE

We believe the key to learning mathematics, at any level, is active participation.

When students are active participants in the learning process, they have the opportunity

to construct their own mathematical ideas and relate new concepts to what they have

studied previously. Such participation leads to understanding, success, and confidence.

We developed this book with this idea in mind, and have incorporated many

features to encourage active student participation. The matched Problems are designed

to keep the students active. Exercise Sets are designed to arouse students' interest.

Book includes Review Exercises and Review Test to give students confidence and

guidance in preparing for exams. Our hope is that every student who uses this book

will be a better mathematical thinker as a result.

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CONTEST

1. QUADRATIC EQUATION AND ITS ROOTS

1.1 DEFINITION OF QUADRATIC EQUATION AND INCOMPLETE QUADRATIC EQUATION 5

1.2 SOLVING QUADRATIC EQUATIONS WHICH ARE SQUARE OF BINOMIALS 7

2. SOLUTION METHODS OF QUADRATIC EQUATION

2.1 SOLVING QUADRATIC EQUATIONS BY FACTORING 9

2.2 SOLVING QUADRATIC EQUATIONS BY COMPLETING PERFECT SQUARE 11

2.3 SOLVING QUADRATIC EQUATIONS BY QUADRATIC FORMULA 13

2.4 SOLVING WORD PROBLEMS BY MEANS OF QUADRATIC EQUATIONS 15

2.5 VIETA’S THEOREM 17

3. RATIONAL EQUATIONS

3.1 SOLVING RATIONAL EQUATIONS 19

3.2 SOLVING WORD PROBLEMS BY MEANS OF RATIONAL EQUATIONS 21

3.3 SOLVING RATIONAL EQUATIONS BY GRAPHICAL METHOD 23

REVIEW EXERCISES 25

REVIEW TEST 28

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1. Quadratic equation and its roots

1.1 Definition of quadratic equation and incomplete quadratic equation

Problem: Find two consecutive positive odd numbers whose product is 143. Solution: let first odd number be x then the next consecutive odd number will be x+2

Equation is illustrated below:

( )2 143x x + =

2 2 143 0x x+ − = (Second degree equation)

( ) ( )13 11 0x x+ − = (Factorize trinomials)

13x = − or 11x =

Answer: The numbers are 11 and 13

e.g.: a) 22 4 3 0x x+ − = is a quadratic equation, where 2a = , 4b = , and 3c = −

b) 25 3 0x x− = is another quadratic equation, where 5a = , 3b = − , and 0c =

Note 1: In a quadratic equation, if at least one of the numbers b or c is zero, then this quadratic equation is called ‘incomplete quadratic equation’ (Remember that 0a ≠ )

e.g.: 23 5 0x− + = , 2 4 0x x− = , and 22 0x− = are incomplete quadratic equations.

There are three types of incomplete quadratic equations. These types and their solutions are given below:

1. 2 0ax = Solution: 0x =

2. 2 0ax bx+ = where 0b ≠ Solution: 1 20,b

x xa

−= =

3. 2 0ax c+ = where 0c ≠ Solution: 1,2

cx

a

−= ± , where 0

c

a

−>

Example1: Solve 23 0x− =

Solution: 2 0x = (divide both sides by -3)

Answer: 0

Definition: An equation in the form of 2 0ax bx c+ + = 0a ≠ (Standard Form),

where x is a variable and a, b, and c are constants is called a quadratic equation

or second degree equation.

Here, a is first or leading, b is second coefficient and c is constant term.

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Example2: Solve 25 8 0x x+ =

Solution: ( )5 8 0x x + = (Take common factor out)

0x = , and 5 8 0x + = (Equate each factor to zero)

Answer:8

0,5

−

Example3: Solve 22 18 0x− + = .

Solution: 22 18 0x− + =

2 18 182 18x − = −− + (Add -18 to both sides)

2 82 1x− = −

2 9x = (Divide both sides by -2)

9x = ± (Take square root of both sides)

Answer: 3; 3−

Matched Problems1: Find solution of given equations

1) 2 0x− = 2) 22 5 0x x− = 3) 23 6 0x − = 4) 23 27 0x− − =

Exercise Set 1.1

1. Which one of the following equations are quadratic.

1) ( ) ( ) ( ) ( )22 3 3 9 2 8 9x x x x x x− + + = − +

2) ( ) ( ) ( )( )22 3 3 9 2 8 7 0x x x x x x+ − + − − − =

3) ( )( ) ( )24 3 4 1 2 3 17y y y− + + − =

4) ( ) ( )22 1 2 6 1 0x x x− − − − =

2. Find the roots of the equations.

1) 22 0x = 2) 210

5x− = 3) 2 360 0y − = 4) 2 16 0x − =

5) 21 80

3 27y − = 6) 23

45 07

y− + = 7) 20,2 45 0y− + = 8) 27 3 0x + =

3. Solve the equations for x.

1) 29 25 0x − = 2) 22 5 0x x+ = 3) 27 3 0x x− + = 4) 226 0

3x x− =

5) ( )22 9 0x − − = 6) ( )2

3 7 0x + + = 7) ( ) ( )22 2 2x x+ = +

8) ( )( )2 1 3 3x x− + = −

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1.2 Solving quadratic equations which are square of binomials

In this part, solution of complete quadratic equation in perfect square forms is indicated.

i)2

x d= , if 0d > then there are two distinct real roots which are:

1x d= , and 2x d= −

If 0d = , there is a real root which is 0. But if 0d < then there is no real root.

ii) ( )2x k d+ = , if 0d > then there are two distinct real roots which are:

1x k d= − + , and 2x k d= − −

If 0d = , there is a real root which is -k. But if 0d < then there is no real root

Example 4: Solve 2 49x =

Solution: 49x = ±

1,2 7x = ±

Answer: 7, 7−

Example 5: Solve 2 6 9 0x x− + =

Solution: ( )23 0x− =

3 0x − =

Answer: 3.

Example6: Solve ( )22 36x − =

Solution: 2 6x − = ±

1 8x = , 1 4x = −

Answer: 4; 8.−

Example7: Solve ( )23 2 5 0x − + =

Solution: ( )23 2 5x − = − ( L.h.s is non negative whereas r.h.s is negative)

There is no real root of given quadratic equation. Answer: no roots

Matched Problems 2: Find solution of given equations

1) 22 32x = 2) 2 4 4 9x x− + = 3) 2 36 0x − = 4) ( )23 2 27x− − = −

5) ( )24 49 0x− − = 6) ( )2

2 5 0x − + =

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Exercise Set 1.2

1. Determine whether the following are quadratic equations or not:

1) 2 1 0x − = 2) 23 2 1 0x x− + = 3) ( )223 2 1 21 0x x− + + =

4) 23 2 1x x− − 5) 2

32 5 0x

x− + = 6) 23 1

52 3

x x− − =

2. Complete the table given below:

3. For which value of k given equations are quadratic:

1) 2 3 0kx− + = 2) ( ) 22 5 3 0k x x k− − + = 3) 23 6 0x kx− + =

4) 2327 0

5 2x k

k

−− =

− 5) 25

27 1 02

x kx− − = 6) 23 21 0

5 2

kx

k x

− −− + =

−

4. Solve the given quadratic equations:

1) 230

5x− = 2) 2 25 0x − = 3) 22 8 0x − = 4) 23

6 02

x − =

5) 2 1 0x + = 6) 2 1 0x − = 7) 22 3 0x + = 8) 2 6 0x − =


1) 2 0x x− = 2) 2 36 0x x− = 3) 228 0

3x x− = 4) ( )23

6 3 22

x x− = −

5) 23 0x x− = 6) 22 6 0x x+ =


1) 20, 25 0x = 2) ( ) 23 2 0x− = 3) 22 3 0x − = 4) ( ) 25 3 6 0x x− − =

Quadratic Equation a (Leading

coefficient)

b (second

coefficient)

c (constant term)

22 0x =

22 5 2 0x x− + =

2

2 1 02

xx

−− + =

2 1 0x + =

( )23 2 1 2x x x− − + =

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2. Solution methods of Quadratic equation

2.1 Solving quadratic equations by factoring

The method of factoring can certainly be applied to any polynomial equation and is not restricted to quadratic equations. In addition to factoring, a major tool used in solving equations is the Zero Product Principle. This principle states the obvious property of the real number system: The only way the product of two numbers can be zero is if at least one of them is zero.

Example 8: Solve 2 25 0x − = .

Solution: ( )( )5 5 0x x− + = (factor by identity ‘differences of the squares’)

1 5x = , 2 5x = −

Answer: 5;5− .

Example9: Solve 23 4 0x x+ =

Solution: ( )3 4 0x x + = (Take the common factor out)

1 0x = , 2

4

3x

−= (By zero product principle)

Answer: 4

;03

− .

Example10: Solve 2 5 14 0x x+ − =

Solution: let’s factorize the trinomial 2 5 14 0x x+ − =

x 7 7x=

↓ ↓ 7 2 5x x x− =

x 2−

2x= −

Middle term

2x x x⋅ =

first term

2 7 14− ⋅ = −

last term

Therefore, ( )( )2 5 14 7 2x x x x+ − = + −

( )( )7 2 0x x+ − =

1 7x = − , 2 2x =

Answer: 7;2− .

Zero-Product Principle: 0 0a b a⋅ = ⇒ = or 0b =

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Example11: Solve 23 4 7 0x x− + + =

Solution: 23 4 7 0x x− + + =

3x−

7 7x=

↓ ↓ 7 3 4x x x− =

x 1 3x= − Middle term 23 3x x x− ⋅ = −

first term

1 7 7⋅ =

last term

Thus, ( )( )23 4 7 3 7 1x x x x− + + = − + +

( )( )3 7 1 0x x− + + =

1

7

3x = , 2 1x = −

Answer:7

1;3

− .

Matched Problems 3: Find the solution of given equations by factoring

1) 249 4 0x− = 2) 2 53 0

3x x− = 3) 2 6 0x x− − = 4) 26 7 2 0x x− + − =

Exercise Set 2.1

1. Solve the given quadratic equations by factoring:

1) 21 4 0x− = 2) 2 250

9x − = 3) 2 5 0x − =

4) 224 6 0x− = 5) ( )21 4 0x− − = 6) ( )2

1 5 0x + − =


1) 2 4 0x x− = 2) 2 250

9x x− = 3) 22 0x x− + = 4) 212 3 0x x− =


1) 2 3 2 0x x− − = 2) 2 12 0x x− − = 3) 2 3 18 0x x+ − =

4) 2 12 13 0x x+ − = 5) 2 2 24 0x x− − = 6) 2 5 6 0x x− + =


1) 23 4 1 0x x− + = 2) 22 1 0x x− − = 3) 26 2 0x x+ − =

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4) 24 13 3 0x x+ + = 5) 2 2 24 0x x− − + = 6) 212 71 6 0x x− − + =

2.2 Solving quadratic equations by completing perfect square

In addition to factorization methods, the technique of completing the square can also be used to solve a quadratic equation. Even though this technique will be used to solve equation, completion of the square has certain uses in other kinds of mathematical problems. The methods of square root and factoring are generally fast when they apply This method is based on the process of transforming the standard quadratic equation

2 0ax bx c+ + = into the form ( )2x A B+ = , where A and B are constants. The last

equation can easily be solved by using the square root property. 2 2 2 2 Examples 12: Complete the square for each of the following:

a) 2 3x x+

b) 2x kx+

Solutions:

a)

2 2

2 3

2

33

2x x x

+ + = +

b)

2 2

2

2 2

k kx kx x

+ + = +

Examples 13: Solve the given equations by completing perfect square

a) 2 6 5 0x x+ + =

b) 2 7 12 0x x+ + =

c) 22 5 2 0x x+ + =

Solutions:

a) 2 4 46 5x x+ + =+ (Add 4 to both sides)

( )23 4x + = (Perfect square)

3 2x + = ± (Take square root of both sides)

1 5x = − , and 2 1x = −

Answer: 5− ; 1−

Completing the Square: To complete the square of a quadratic of the form 2x bx+ ,

add the square of one-half the coefficient of x; that is, add

2

2

b

. Thus,

2x bx+ +

2

2

b

=

2

2

bx

+

12

b)

2 2

2 7 77 12

2 2x x

+ + + =

2 2

2 7 77 12

2 2x x

+ + = −

27 49 1

122 4 4

x

+ = − =

7 1

2 2x + = ± 1 2

7 14, 3

2 2x x x⇒ = − ± ⇒ = − = −

Answer:-4;-3

c) 2 51 0

2x x+ + = (Divide both sides by 2, that is leading coefficient)

2 2

2 5 5 51

2 4 4x x

+ + + =

(Add both sides

25

4

, that is half of square of second

coefficient) 2

5 25 91

4 16 16x

+ = − =

5 3

4 4x + = ± 1 2

5 3 12,

4 4 2x x x⇒ = − ± ⇒ = − = −

Answer: 1

2;2

− −

Matched problems 4: Solve the given equations by completing perfect square

1) 2 5 0x x+ = 2) 2 6 0x x+ − = 3) 23 7 2 0x x+ + =

Exercise Set 2.2

1. For which values of m given trinomials are perfect squares:

1) 2 2 2 0x x m+ + = 2) 2 3 0x x m+ − = 3) 2 14 0x x m− + =

4) 2 9 0x mx− + = 5) 24 10 0x x m− + = 6) 29 4 0x mx− + =

2. Solve the given quadratic equations by completing perfect square::

1) 2 2 10 0x x− − = 2) 2 4 5 0x x− − = 3) 2 7 18 0x x+ − =

4) 2 5 4 0x x+ + = 5) 2 3 1 0x x− − = 6) 2 1 0x x− − =


1) 24 4 1 0x x− + = 2) 23 2 1 0x x− − = 3) 23 2 0x x+ − =

4) 24 11 3 0x x+ − = 5) 28 8 6 0x x− + + = 6) 215 17 4 0x x− − + =

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2.3 Solving quadratic equation by quadratic formula The method of completion of the square is a useful tool when solving quadratic equations, but it is not the most efficient method. The next way on the Quadratic Formula is a more standard tool than completing the square. • The Quadratic Formula

The solutions of the equation ( 2 0ax bx c+ + = ) can be found in a more direct way than

the method of factorization or completing the square by using the so-called Quadratic Formula. Let’s state and prove this formula. The quadratic formula should be memorized and used to solve quadratic equations when other methods fail or are more difficult to apply.

Theorem Notes: The expression 2 4b ac− is called the discriminant for the quadratic

equation. The discriminant is denoted by ( )∆ delta or by D.

It can be used, at casual glance, to determine whether a given equation has one, two, or no solutions.

Example14: Solve the equation 23 4 1 0x x+ + =

Solution: In given equation 3a = , 4b = , and 1c = 24 4 3 1 4D = − ⋅ ⋅ =

Since 0D > there are 2 distinct real roots

1,2

4 4

2 3x

−=

⋅

∓ , 1

4 21

6x

− −= = − , and 2

4 2 1

6 3x

− += = −

Answer: -1 and1

3− .

Example15: Solve 2 5 8 0x x+ + = .

Solution: For this equation, 1a = , 5b = , and 8c = . Now, apply the quadratic formula: 2 24 5 4 1 8 17D b ac= − = − ⋅ ⋅ = −

Since 0D < no real roots.

Answer: no roots

Theorem: Consider the quadratic equation 2 0ax bx c+ + = where 0a ≠

2

b Dx

a

−=

∓ , where 2 4D b ac= −

If 0D > then there are two roots.

If 0D = then there is one root.

If 0D < then there is no root.

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Example 16: 2 10

4x x− + =

Solution: Here, 1a = , 1b = − , and 1

4c =

Now, if we apply the quadratic formula: ( )22 14 1 4 1 1 1 0

4D b ac= − = − − ⋅ ⋅ = − =

Since 0D = there is a unique root (a double root)

1 0 1

2 2 2

b Dx

a

− ± ±= = =

Answer: 1

2.

Matched Problems 5: Find solution of given equations by quadratic formula

1) 23 24 0x − = 2) 2 4 0x x− = 3) 2 2 15 0x x− − = 4) ( )22 3 2 8x− − = −

Exercise Set 2.3 1. Find the discriminants of following equations:

1) 2 5 3 0x x− + = 2) 2 4 1 0x x− + + = 3) 23 5 0x x+ =

4) 22 3 0x x− + − = 5) 22 6 5 0x x− + = 6) 25 13 2 0x x− − =

2. Solve given quadratic equations by quadratic formula:

1) 23 4 1 0x x+ + = 2) 22 5 2 0x x+ + = 3) 23 0x x+ =

4) 22 1 0x x− + + = 5) 22 7 3 0x x− + = 6) 23 11 6 0x x+ + =

3. Determine the number of roots of given quadratic equations without solution.

1) 29 6 1 0x x− + = 2) 23 5 2 0x x+ + = 3) 22 4 0x + =

4) 2 3 4 0x x− + − = 5) 23 7 6 0x x− + = 6) 23 2 3 1 0x x+ + =

4. Solve the given quadratic equations by quadratic formula:

1) 2 1 0x x+ + = 2) 225 40 16 0x x+ + = 3) ( )22 3 11 19x x− = −

4) 23 1 10

4 4 2x x− − = 5) 28 8 6 0x x− + + = 6)

2 111( 1)

2

xx

−= +

5. Solve the given equations:

1) 22 5 3x x= + 2) 25 6 1x x= − 3) ( )22 1 3 11x x− = −

4) 23 7

14

x x−= − 5) (2 1) 28x x − = 6) 23 ( 2) 1 11( 1) 2x x x x− − = + −

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2.4 Solving word problems by means of quadratic equation A great amount of problems in algebra, physics, chemistry and geometry can be solved by means of quadratic equations We now consider several applications that make use of quadratic equations. First, the strategy for solving word problems, presented below.

Example 17: Find all numbers with the property that when the number is added to itself , the sum is the same as when the number is multiplied by itself. Solution: Let the number be x ,

The equation will be: x x x x⋅ = + 2 2 0x x− =

( )2 0x x⋅ − =

1 0x = and 1 2x =

Answer: 0 and 2.

Example 18: The sum of two numbers is 33 and their product is 242. Find the two numbers. Solution: Let one number be x , and then the other number is 33 x− .

The equation will be: ( )33 242x x⋅ − =

2 33 242 0x x− + =

( ) ( )22 11 0x x− ⋅ − =

1 22x = and 2 11x =

Answer: 11 and 22.

Strategy for Solving Word Problems

1. Read the problem carefully several times if necessary that is, until you understand the problem, know what is to be found, and know what is given. 2. Let one of the unknown quantities be represented by a variable, say x , and try to

represent all other unknown quantities in terms of x . This is an important step and

must be done carefully. 3. If appropriate, draw figures or diagrams and label known and unknown parts. 4. Look for formulas connecting the known quantities to the unknown quantities. 5. Form an equation relating the unknown quantities to the known quantities. 6. Solve the equation and write answers to all questions asked in the problem. 7. Check and interpret all solutions in terms of the original problem—not just the equation found in step 5—since a mistake may have been made in setting up the equation in step 5.

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Example 19: Find the lengths of sides a right triangle with a hypotenuse of 15cm if one right side is 3cm shorter than the other one. Solution: Let short side’s length be x

Then long side’s length must be 3x +

Use Pythagoras formula 2 2 2( 3) 15x x+ + =

2 2 6 9 225x x x+ + + =

22 6 216 0x x+ − =

2 3 108 0x x+ − =

( )( )12 9 0x x+ − =

1 12x = − or 2 9x =

Length must be positive, so 9x =

Answer: short side is 9 cm and long side is 12 cm. Exercise Set 2.4 1. Find five consecutive integers such that the sum of the squares of the first three is

equal to the sum of the squares of the last two.

2. Find tree consecutive even integers such that the sum of the squares of the first two

is equal to the square of the last.

3. Find the dimensions of a rectangle, if its area 1800 2cm and its length exceed its

width by 5cm.

4. The square of the sum of the two consecutive natural numbers is 144 more than the

sum of the squares of these numbers. Find these numbers.

5. During the school championship in football 36 matches were played. Each team

played only once with the other. How many teams did participate in the school

championship?

6. In the tournament of chess 45 parties were played. Each competitor played only

once with others. How many participants were in the tournament?

7. The sum of two numbers is 10. The sum of their squares is 68. Find two numbers

8. In the difference of two integers is 2 and the difference of their squares is 24, find

the integers

9. Find two consecutive integers whose product is 210

10. Find two consecutive even integers whose product is 168

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2.5 Vieta’s theorem We know that the roots of a quadratic equation depend on its coefficients. There exist certain relations between the coefficients and the roots of an equation. The most important relations between the roots and the coefficients a, b, and c of a quadratic equation are coming up now.

Proof: By means of quadratic formula we know that:

2

1

4

2

b b acx

a

− + −= and

2

2

4

2

b b acx

a

− − −=

So, their sum is: 2

1 2

4b b acx x

− + −+ =

2 4

2

b b ac

a

− − −+

2

b

a a= −

and their product is: 2 2 2 2

1 2 2 2

4 4 4 4

2 2 4 4

b b ac b b ac b b ac ac cx x

a a aa a

− + − − − − − +⋅ = ⋅ = = = .

Note: The sum or product of the roots of a quadratic equation can be found immediately by Vieta’s Theorem, but be sure that quadratic equation has roots, hence, we may check its discriminant.

Example 20: Find the sum and product of roots of given quadratic equations.

1) 2 5 6 0x x− + = 2) 23 5 2 0x x− + = 3) 2 102 6 0

3x x− − =

4) 25 2 1 0x x− + = 5) ( )25 1 2 1 6x x x+ − + = +

Solution: 1) Sum of roots is 5

51

b

a

−− = − = , Product of roots is

66

1

c

a= =

2) Sum of roots is 5 5

3 3

b

a

−− = − = , Product of roots is

2

3

c

a=

3) Sum of roots is 6

2, Product of roots is

1053

2 3

c

a

−

= = −

Theorem of Vieta: For any quadratic equation 2 0ax bx c+ + = , where 0a ≠ and

the roots are 1x and 2x .

1) Summation of roots: 1 2

bx x

a+ = −

2) Product of roots: 1 2

cx x

a⋅ =

18

4) 4 20 16D = − = − . Discriminant is negative, so, there is no real roots.

5) Firstly we have to write the equation in the standard form. ( )25 1 2 1 6x x x+ − + = +

25 10 5 2 1 6 0x x x x⇒ + + − + − − = 25 7 0x x⇒ + =

Sum of roots is 7

5− , Product of roots is

00

5

c

a= =

Matched Problems 6: Find the sum and product of roots of given quadratic equations.

1) 2 3 2 0x x− + = 2) 22 7 3 0x x− + = 3) 2 43 6 0

3x x− − =

4) 23 7 6 0x x− − = 5) ( )21 2 1 3 5x x x+ − + = +

Exercise Set 2.5

1. Without solving, find the sum and the product of the roots of the following equations.

1) 22 5 1 0x x+ + = 2) 2 7 2 0x x− + + = 3) ( ) ( ) ( )3 2 1 4 2 1x x x+ − = − − 4)

( )4 3 4 1x x x− = − 5) ( ) ( )2 24 3 3 1x x+ = + 6) ( )( )2 3 4 13x x− − =

2. If the sum of the roots of the equation ( ) 23 2 5 8 0a x ax a+ + − = is -5, then find the

product of the roots.

3. If the product of the roots of the equation ( ) ( )2 25 2 1 2 1 0m x m x m− + + + − = is 5,

then find m and sum of the roots.

4. 1 2x and x are the roots of the equation 25 2 1 0x mx+ + = , 1 2,x x+∈ℝ . If 1 25x x=

then find the 1 2,x x and m.

5. One of the roots of the equation 22 3 0x x k− + = is two times greater than the

other one. Find k and the roots of the equation.

6. One of the roots of the equation 22 3 0x x k− + = is 1 less than the other one. Find

k and the roots of the equation.

7. For which value of m, the equation ( ) ( )25 6 3 0m x m x+ − + + = has one double

root.

8. If -0.3 is one of the roots of the equation 25 2.5 0− + =x x t , then find t and the

second root.

9. 1 2x and x are roots of the equation 2 3 0− + =x x m . If 1 23 4 37x x− = , find m,

1 2x and x .

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3. Rational equations

3.1 Solving rational equations

Rational equations are equations that contain rational expressions. Some examples

of rational equations are 3 7 12

5 1 10

x

x

+=

+,

2 1 3

5 4

x += ,

3 104

4 2 3

x

x+ =

+.

When solving these equations, we will multiply both sides by a quantity containing a variable. When we do this, we could inadvertently multiply both sides of an equation by 0 and obtain a solution that makes the denominator of a fraction 0. In this case, we have found a false solution, called an extraneous solution. These solutions do not satisfy the equation and must be discarded. Comment: Be sure to exclude from the solution set of an equation any value that makes the denominator of a fraction equal to 0.

Example21: Solve 2

2

8 72

8 8 64

x x x x

x x x

+ + ++ =

+ − −

Solution: Multiply both sides of the equation by LCD ( 8x + ; 8x − ) which is 2 64x − ,

but we have to mention that 8, 8.x x≠ ≠ − Since denominator cannot be zero.

( ) ( ) ( )2 22

2

264 68 72

8 84 64

64

x x x xx x

x x xx

+ + ++ =

+ −− ⋅ − ⋅ −

−⋅

( ) ( )( ) 28 8 8 72x x x x x x− + + + = + + (Factor and simplify)

2 7 8 0x x+ − = (Write the quadratic equation in standard form)

( )( )8 1 0x x+ − = (Factor and solve)

1 8x = − , 2 1x =

Answer: -8; 1.

Example22: Solve 2 2 2

2 4 1

3 9 3

x

x x x x x

−− =

− − +

Solution: Multiply both sides of the equation by LCD (x; 3x + ; 3x − ) which is 3 9x x− , but we have to mention that 3, 3, 0.x x x≠ ≠ − ≠

( ) ( ) ( )3 3

2

3

2 2

2 4 19 9

39

9 3x x x

x

x x x xx x x

x⋅ − ⋅ −

−

+⋅−

−−=

−( ) ( )( )2 3 4 1 3x x x x+ − = − −

(Factor and simplify) 2 2 3 0x x− − = (Write quadratic equation in standard form)

( )( )3 1 0x x− + = (Factor and solve)

1 3x = , 2 1x = − (We know that 3x ≠ )

Answer: 1− .

20

Matched Problems 7: Solve the following equations.

1) x x

2

15

5

3

2

3+ = 2)

2

2 2

45 3

25 25

x x x

x x

+ −=

− − 3)

2

4 3 1 2 3

1

x x

x x x x

− +− =

− −

Exercise Set 3.1

1. Solve the following rational equations.

1) 2

2

13 360

20

x x

x x

+ +=

− + 2)

2

2

2 180

3 14 15

x

x x

−=

− +

3)1 1 5

1 4 4x x+ =

− − 4)

( )( )5 5 20

2 5 5 2

x x

x x x x

+ −− =

− − − −


1) 2

2 1 2 1

x x

x x=

− − 2)

2 9 36

3 3

x x

x x

−=

+ + 3)

2 8 20

10 10

x x

x x

+=

+ +

4)2

2 2

2 3 2 2

4 4

x x x

x x

− −=

− − 5)

2 3 2

1 1

x x

x x

−=

− − 6)

22

3 1 1 3

x x

x x=

− −


1) 1 1

3

x x

x x

− += 2)

5 8 14 12

1 3 5

x x

x x

− +=

− + 3)

x

x

x

x

−

+=

+

+

1

2 3

1

3 4

4) x

x

x

x

+

−=

+

+

3

2 1

2 3

3 5)

5 3 3 1

5 2

x x

x x

+ +=

+ + 6)

3 1 2 10

2 1

x x

x x

+ −=

− +


1) 2

1 24 0

x x− − = 2)

2

1 1 2 8

2 2 4

x x x

x x x

− + ++ =

+ − −

3) 2 2

5 4 1

24 4 4 xx x x− =

+− + − 4)

( )( )5 5 20

2 5 5 2

x x

x x x x

+ −− =

− − − −

5) 2 2

10 40 1

5 5 xx x x x− =

+ − 6)

2 2

14 21 5

2 2 xx x x x− =

− +

Algorithm to Solve Quadratic Equations in Rational Form 1) Factorize the denominators of all expressions 2) Find the least common denominator (LCD) 3) Multiply both sides by the LCD and solve the equation as a system which excludes the values that make the common denominator equal to zero.

21

3.2 Solving word problems by means of rational equations We now consider several applications that make use of rational equations. The strategy for solving word problems, presented earlier (above). The same strategy can be used here.

Example23: The sum of a number and its reciprocal is41

20. Find all such numbers.

Solution: Let the number is x then its reciprocal is 1

x :

1 41

20x

x+ =

2 1 41

20

x

x

+⇒ = (Equate denominators)

( )220 1 41x x⇒ ⋅ + = ⋅ (Multiply both sides by 20 x⋅ .)

220 41 20 0x x− + = (Write quadratic equation in standard form)

( )( )5 4 4 5 0x x− − = (Factor and solve)

1

4

5x = 2

5

4x = (We know that 0x ≠ )

Answer: 4 5

; .5 4

Example24: An excursion boat takes 14,4 hours longer to go 36 miles up a river than to return. If the rate of the current is 4 miles per hour, what is the rate of the boat in still water? Solution: Let

x = Rate of boat in still water

4x + = Rate downstream

4x − = Rate upstream

The equation will be as follow: 14,4time time

upstream downstream

− =

36 36

14, 44 4x x

− =− +

( ) ( ) ( )2 2 236 3614, 4

4 416 16 16x x

xx

x⇒ − =⋅

−− ⋅ − ⋅ −

+

( ) ( ) ( )236 4 36 4 14,4 16x x x+ − − = ⋅ − (Divide both sides by 36)

( ) ( ) ( )24 4 0, 4 16x x x+ − − = − (Simplify the equation)

214,4 0,4x= 236 x⇒ = (Solve)

1 6x = 2 6x = − (We know that rate cannot be negative)

Answer: 6miles/h. Matched Problems 8:

22

1) The sum of a number and its reciprocal is 2,5 . Find all such numbers.

2) A ship moves 16km in the direction of a river and 30km in the opposite direction. Trip took 1,5h. If the speed of river is 1km/h, find speed of the ship. 3) Two workers can do a job in 16h together. One of them can do this job 24h faster than the other one. Find time that requires each worker to do this job alone.

Exercise Set 3.2

1. One of the sides of a rectangle is 9cm longer than the other one and area of the rectangle is 36cm2. Find perimeter of the rectangle.

2. The sum of two numbers is 10 and sum of their squares is 68. Find these numbers.

3. From each corner of a square piece of a sheet metal, remove a square of side 3 cm. Turn up the edges to form an open box. If the box is to hold 48 cm3

what should be the dimensions of the piece of sheet metal 4. A wire that is 32cm long is cut into two pieces, and each piece is bent to form

a square. The total area enclosed is 34cm2. Find the length of each piece of wire.

5. Two cars move from one city to another in the same time. The distance between two cities is 300km. Speed of one of the cars is 10km/h more than the other one and faster car arrived the city 1h earlier. Find the speeds of each car.

6. A car moved from one city to another and turned back by using another way. First way is 21km and the second way is 16km. In the return way, the speed of the car is increased by 3km/h. Trip in the first way took 1h more time. Find speed of the car in first way.

7. A man did a job $156. It took him 7 hours longer than he expected and so he earned $14 for an hour less than he anticipated. How long time did he expect?

8. It takes two workers 12 days to do a job. If each worked alone, one of them could do the job in ten days less time than the other. How long would it take the faster one the complete job alone?

9. A Chemist wishes to mix some alloy of %25 silver with another alloy of %40 silver. How much of each should be used to produce 60kg of an alloy which is %30 silver

10. It takes two workers 4 days to complete the job. If each worked alone, one of them could do the job in six days less time than the other. After the faster one works 3 days alone, second one joins. How long would it take together to complete remaining work?

11. Two workers working together can paint the front of a house in 60 hours. One of the painters alone can finish this job in 50 hours less time than the other painter alone. How much time does each worker need to do this job alone?

23

3.3 Solving rational equations by graphical method

Let’s analyze the equation 2 4x

x= . When we multiply both sides by x , we get

3 4x = which is a third degree equation; we haven’t learnt to solve a third degree

equation. Such equations can be easily solved by graphical method:

Firstly, we sketch the graphs of both 2y x= and

4y

x=

Secondly, we find the intersection points of the graphs which is ( )1,5874;2,5198

Finally, the abscissa(s) of intersection points will be solution of the equation which is1,5874 .

As another example, let’s solve 3 10 0x x+ − = . This equation also can be solved by

graphical method easily:

Firstly we write equation in the form 3 10x x= − + sketch graphs of both 3y x= and

10y x= − + .

Secondly, we see the intersection points of the graphs which is ( )2;8

Finally, the abscissa(s) of intersection points will be solution of the equation which is2 .

x-4 -3 -2 -1 1 2 3 4

y

-4

-3

-2

-1

1

2

3

4

4y

x=

2y x=

24

Exercise Set 3.3

1. Solve the following equations by using graph of 2y x= .

1) 2 6x x= + 2) 2 3 10 0x x− − = 3) 2 2 0x x− − =

2. Solve the following equations by graphical method.

1) 4

xx

= 2)3

4xx

− = + 3) 3x x= −


1) 2 2x

x= 2) 2 3

xx

= − 3) 3 4 3x x= −


1) 6x x= − 2) 2x x= − 3) 8

xx

=

1 2 3 4 5 6 7 8 9 10 11

-1

1

2

3

4

5

6

7

8

9

10

11

10y x= − +

3y x=

25

Review Exercises 1. Solve the given equations by factoring.

1) 23 27 0x x− = 2) 264 25 0x − = 3) 2 6 27 0x x− − =

4) 22 4 17 0x x− + + = 5) 2 2 0x x− − = 6) 2 3 2 0x x+ + =

7) 2 2 1 0x x− + = 8) 210 19 6 0x x− + = 9) 26 15 0x x+ − =

10) 23 5 2 0x x− + = 11) 26 91 15 0x x− + = 12) 28 9 9 0x x− − =

2. Solve the given equations by completing perfect square.

1) 2 2 0x x− = 2) 22 12 0x x− = 3) 2 3 0x x− =

4) 2 4 5 0x x+ − = 5) 2 3 1 0x x− − = 6) 2 5 2 0x x+ + =

7) 22 3 1 0x x− + = 8) 25 6 3 0x x− − = 9) 2 1 0x x+ − =

3. Solve the given equations by completing quadratic formula.

1) 22 3 1 0x x+ + = 2) 2 2 15 0x x− − = 3) 2 3 10 0x x− − =

4) 22 3 5 0x x+ − = 5) 21 30

2 2x x− − = 6) 2 1

02

x x+ − =

7) 224 6 0

3x x− + = 8) 2 10 3 0x x− − = 9) 2 2 5 5 0x x+ + =

4. Find two consecutive positive even integers whose product is 168. 5. If one of the legs of a right triangle is 14cm less than the other leg and its

hypotenuse is 34cm, find the lengths of the sides of the triangle. 6. Sum of the length and width of a rectangle is 79cm and its diagonal is 65cm.

Find length of each side. 7. One year ago, a father was 8 times as old as his son. Now his age is the square

of his son’s age. Find their present ages. 8. Find three consecutive integers such that the sum of the squares of the first

two is equal to the square of the last. 9. The difference of the squares of the two consecutive even natural numbers is

72 less than the sum of the squares of these numbers. Find the numbers. 10. During the school championship in football 55 matches were played. Each

team played only once with the other. How many teams did participate in the school championship?

11. In the tournament of chess 66 parties were played. Each competitor played only once with others. How many participants were in the tournament?

12. Without solving, find the sum and the product of the roots of the following

equations.

1) 23 6 1 0x x+ + = 2) 2 5 7 0x x− + + = 3) ( )( ) ( )4 2 1 3 2 1x x x+ − = − −

26

13. For which value of m, the equation ( )2 2 1 1 0mx m x m+ + + − = has two real

roots, and then find m.

14. For which value of a , the equation 2 3 10 0+ + =ax x , have

1) Two distinct real roots 2) One double root 3) No solution

15. 1 2x and x are roots of the equation 2 13 5 0− + =x x . Without solving the

equation find the values of the following expressions.

1)1 2

1 1

x x+ 2) 2 2

1 2x x+ 3) 2 2

1 2 1 2x x x x+

16. Solve the rational equations

1) 3 4 1

32 3

x x+ −+ = 2)

1 13

2x x+ = 3)

13 4x

x+ =

4) 2 3 4

3 4 6

x x x+ ++ = 5)

1 5

2x

x+ = 6)

209 x

x= −

7) 1 5

2x

x= − 8)

1 5

2

x x

x

+ −= 9)

2 3

2

x

x x

+=

−

17. Solve the rational equations

1) 5

7

3

2

1

1=

++

− xx 2) 2

3

2

1

3=

++

+ xx 3)

10

9

2

1

1

2−

+

−=

−

−

x

x

x

x

4) ( )

15

7

5

62

−+

=+ xx

5)2

2 32

4 4 16

x

x x x− =

− + −

6)22 5 2

4 12

x xx

x

− += +

− 7)

2

1 1

1

x

x x x x− =

+ +

18. The sum of a number and its reciprocal is10

3 . Find all such numbers.

19. Denominator of a fraction is 3 more than numerator. After adding 7 to

numerator and 5 to denominator, the fraction increases by 1

2. Find this

fraction. 20. Numerator of a principal fraction is 5 less than denominator. After decreasing

numerator by 2 and increasing denominator by 16, the fraction decreases by

1

3. Find this fraction.

21. A speedboat takes 1 hour longer to go 24 miles up a river than to return. If the boat cruises at 10 miles per hour in still water, what is the rate of the current?

27

22. Flying rates A jet plane, flying 120 mph faster than a propeller-driven plane, travels 3,520 miles in 3 hours less time than the propeller plane requires to fly the same distance. How fast does each plane fly?

23. Two planes travel at right angles to each other after leaving the same airport at the same time. One hour later they are 260 miles apart. If one travels 140 miles per hour faster than the other, what is the rate of each?

24. A train moved from Kazan to Moscow. Distance between two cities is 810km. After taking 1/3 of the distance, it decreased its speed by 30 km/h. Total trip took 12h. Find initial speed of the train.

25. One pipe can fill a tank in 6 hours less than another. Together they can fill the tank in 4 hours. How long would it take each alone to fill the tank?

26. One pipe can fill a pool 4 hours faster than another one. After slower pipe filled the pool 7h, other pipe was opened and hey filled 2h more together and filled all pool. Find time that requires each pipe to fill the pool alone.


1) 2 1x

x= 2) 2 1

xx

= − 3) 2 3 2x x= −

4) 6x x= − 5)3

2

xx

−= 6) 3

x x= −

28. Find the coordinates of the intersecting point of the following functions.

1)7

2 35

y x and yx

= + =−

2)23 5

23

x xy and y x

x

−= =

+

29. Find the roots of the equations.

1)2

3 2 3 2 10

3 23 2 3 2

x x x

xx x

+ −+ =

−− + 2)

2

1 5 1 5 9

1 51 5 1 5

y y y

yy y

− ++ =

−+ −

3) 2 2

8 32 1

4 4 xx x x x− =

+ − 4)

2

3 2 2 6

1 1 1

x x x

x x x

+ + ++ =

− + −

5)

2

2

4 4 4 6

1

x x x

x xx x

− + ++ =

++ 6)

2 2

1 4 1

24 4 4 xx x x− =

+− + −

7)2 3 2

1 2 5

2 1 2 2a a a a a+ =

− + − + − 8)

2 2 2

6 3 120

36 6 6

x

x x x x x

−− + =

− − +

9)2

1 1 2 18

3 3 9

x x x

x x x

− + ++ =

+ − − 10)

2 2

3 4 1

24 4 4 xx x x+ =

−+ + −

28

Review Test 1. Which one is a quadratic equation?

a) ( )2 2 2 0x x − = b) ( )( )2 2 0x x− + =

c) 3 2 0x x x+ − = d) 2

10x

x+ = e) 2 3 5 0x x

− + − =

2. Find the solutions of the quadratic equation 22 3 5 0x x+ − =

a) -5;0,5 b) 0,5;1 c) -2,5;1 d) 2,5;-1 e) 2;1

3. Find the solutions of the quadratic equation 22 0x x+ =

a) 1

;02

− b) 0;2 c) 1;2− d) 2;0− e) 2;1−

4. Find the solutions of the quadratic equation ( )22 6 2 3 0x x+ + + =

a) 2; 3− b) 3;2− c) 1; 3− d) 3

3;2

− − e) 2;3−

5. Find the sum of the roots of the equation ( ) ( )3 2 3 2 4,5 2x x x x+ = + + .

a) 0 b) 0,5 c) -0,5 d) -2 e) 2

6. Find x if 2

5 50

5 5 25

x x

x x x

++ =

+ − −

a) 2,5 b) 5− c) 5 d) 2,5− e) 2,5; 5−

7. What is the value of m if sum of the roots of the equation 22 4 0x mx+ − = is 4.

a) 2− b) 2 c) 4− d) 4 e) 8−

8. In a still water speed of a boat is 20km/h.If it travels against the river 36 km and 22km with river, this trip takes 3 hours. Find the speed of river. a)2 b)4 c)6 d)8 e)10 9. Two technicians can complete a mailing in 4 hours while working together. Alone, the first technician can complete the mailing 6 hours faster than the other technician. How long will it take each technician to complete the mailing alone? a) 2; 2 b) 6; 9 c) 6; 12 d) 8; 12 e) 10; 12

10. Find x if 26 24 0x + =

a) 2 b) 2± c) 4± d) ∅ e) 4

29

11. Find the number of the roots of the equation 2 4x

x= .

a) 0 b) 1 c) 2 d) 3 e) 4

12. Find the discriminant of the equation 2 2 3 3 0x x− + =

a) 12 b)24 c)0 d)6 e)48 13. If sum of the squares of 3 consecutive natural numbers is 869, find the greatest of them. a) 34 b) 35 c) 36 d) 37 e) 38

14. Find the product of the roots of the equation ( )22 4 6 2 3x x x+ − = +

a) 1− b) 1 c) 6 d) 6− e) 0

15. Find the roots of the equation 23 6 2 0x x− + = .

a) 3 2 3

3

± b)

3 4 3

3

± c)

3 3

3

± d)

6 3

6

± e)

3 3

6

±

16. Solve the equation 2 7 12

4 4

x x

x x

−=

− −

a) 3;4− b) 3 c) 3; 4− d) 4 e) 3;4

17. For which value of m trinomial 2 3x x m+ + is a perfect square:

a) 1

12 b) 1 c)

1

4 d)

2

15 e) 3

18. The sum of a number and its reciprocal is 29

10 . Which one of the following is

product of such numbers? a) 2− b) 1 c) 2 d) 3− e) 3

19. Which one of the following is false for the equation 22 7 5 0x x− + = ?

a) The roots have the same sign b) One of the roots is rational c) The roots are both positive d) Both roots are not integer e) Product of the roots is positive.

30

20. It is known that 1 2,x x are roots of the equation 2 ( 1) 1 0x m x− − + = , for which value

of m: 1 2x x= .

a) 2− b) 1 c) 2 d) 3− e) 3

21. Find the product of roots of the following equation 4 2

2 3

a a

a

+ −=

a) 2

7− b)

7

2 c) 6 d) 6− e)

7

2

22. Find the solution sets of the following equation 1 1 5

2 3x x+ =

+

a) 2; 3− − b) 1;5 c) 2;3 d) 5;1− e) 6;1−


1 6x x+ =

+

a) 3

2;5

− b) 3;2 c)3

2;5

d)3

;22

− e) 2; 3− −


2 2 2 4

x

x x x+ − =

− + −

a) 4− b) ∅ c) 2− d) 2 e) 3

25. If 22 3 5 0x x− + = , find the value of 2

2

254x

x+

a) 14 b) 21 c) 25 d) 29 e) 31

31

Ildar Ilyazov

Quadratic equations

Workbook

Format 60×80 Quire 1,25

Number of printed copies 50

Atlantic International School 115088, Moscow, Sharikopodshipnikovskaya Street, 30A/block 1

www.atlanticschool.ru

32

ATLANTIC INTERNATIONAL SCHOOL

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