EQUATIONS AND INEQUALITIESjfrabajante.weebly.com/uploads/1/1/5/5/11551779/13_math_17.pdf · 1. Solve linear equations in one variable 2. Solve quadratic equations in one variable

EQUATIONS

2

A mathematical statement that asserts that two quantities are equal is called an equation.

Examples:

12 9 21

2 3 11x 2 24 0x xy y

1.

2.

3.

Mathematics Division, IMSP, UPLB

3

Consider the following equations: 1. 12 + 9 = 21 2. x + 9 = 21

Equation (1) is always true while equation (2) is true for a particular value of the variable x.

An equation involving a variable is called an open equation.

Mathematics Division, IMSP, UPLB

A solution (root) of an open equation in one variable is the value of the variable for which the equation is true. The set of all solutions of an equation is called the solution set of the equation. In Math 17 we will consider complex solutions.

If the solution of an open equation is the entire set of real numbers, the equation is said to be an identity. Otherwise, the equation is called a conditional equation.

4 Mathematics Division, IMSP, UPLB

Example 1:

The solution of the equation

x + 9 = 21

is x = 12.

The solution set is { 12 }.

This is a conditional equation.


Example 2: x2 + x – 6 = 0

The solution of the equation

x2 + x – 6 = 0

are x = -3 or x = 2.

The solution set is { -3, 2 }.

This is a conditional equation.


Example 3: 3(x + 2) = 3x + 6

The equation

3(x + 2) = 3x + 6

is true for any real number x.

The solution set is R (set of real numbers).

What kind of equation is this?


Two equations are said to be equivalent if we can obtain one from the other by applying the properties of operations on the real numbers or the properties of equality.

Examples: 5x – 3 = 0 is equivalent to 5x = 3. x2 – 4 = 5 is equivalent to (x + 3)(x - 3)=0


LINEAR AND QUADRATIC

EQUATIONS

At the end of this section, you should be able to:

1. Solve linear equations in one variable 2. Solve quadratic equations in one variable by: Factoring Completing the Squares Using the Quadratic Formula

3. Determine the nature of the solutions of a quadratic equation


An equation in one variable, say x, that can be written in the form

where a and b are real numbers with

is called a linear equation in the variable x. Examples: 1. 2.

0ax b 0a

2 3 6 2 3 0x x

7 2 0 7 14 0x x


standard form

of a linear

equation.

From the linear equation

a x + b = 0,

We obtain the following equivalent equations:

(ax + b) + (-b ) = 0 + (-b) by APE

ax + [b + (-b )] = 0 + (-b) by Asso +

ax + 0 = 0 + (-b) by Inv +

ax = -b by Ident +

(1/a) · ax = (1/a) · (-b) by MPE

[(1/a) · a] x = (1/a) · (-b) by Asso ·

1 · x = (1/a) · (-b) by Inv ·

x = (1/a) · (-b) by Ident ·

x = -(b/a) by Defn of Div.


Thus, the solution of the linear equation

is given by . The solution

set of the linear equation is .

bx

a

a

b

0, 0ax b a


Example 1:

Find the solution set of the following linear equations:

1 7 5 16

2 2 3 3 10

4 2 13

5 3 2

. x

. x x

x x. x


1

3 42 1

2 3

3

. ax b cx d; a

xy. x; x

y

. rs at bt s; s

Example 2:

Solve for the indicated variable in terms of

the other variables:


Any equation in one variable, say x, that can be written in the form

where a, b and c are real numbers, ,

is called a quadratic equation in the variable x.

2 0ax bx c

0a


standard form

of a quadratic

equation.

2 2

2

1 3 7 5 2

2 5 6

3 5 2 2 1 7

. x x x

. t t

. t t t t

Examples: Write the ff. in standard form:


To solve quadratic equations, we use any of the following methods:

1. By factoring

2. By completing the squares

3. By using the Quadratic Formula.


In this method, we use the following rule:

Let u and v be real numbers or quantities,

u v = 0 if and only if u = 0 or v = 0.

Thus, if we can factor the quadratic equation

ax2 + bx + c = 0, a 0, into two linear factors say (px + q) and (sx + t), then the solutions of this quadratic equation are the solutions of the linear factors:

(px + q) = 0 or (sx + t) = 0


Example 1: Solve x2 – x – 6 = 0.

Solution: x2 – x – 6 = (x – 3)(x + 2) Hence, x2 – x – 6 = 0 will imply that (x – 3)(x + 2) = 0

x – 3 = 0 or x + 2 = 0 x = 3 or x = -2.


2,3 SS

WARNING!!! The following is NOT always true:

If ab = k (where k≠0) then a=k or b=k…

Kaya walang gagawa nito!!!


Example 2: Solve t(5t + 2) = 2t(1 – t ) + 7.

Solution:

t(5t + 2) = 2t(1 – t ) + 7 7t2 – 7 = 0 t2 – 1 = 0 why? (t – 1)(t + 1) = 0 t – 1 = 0 or t + 1 = 0 t = 1 or t = – 1.


1,1 SS

21. 15 8 26

2. 3 1 2 2

3. 2 3 1 2

y y

x x x

a a a a

2 2 2

2 2 2

1. 18 2 15 0;

2. 3 k 5 4 ;

a x abx b x

t s kst t

Exercise:

A. Solve the following quadratic equations by

factoring:

B. Solve for the indicated variable:


Note:

a2 = b2 a = b or a = – b (why?)

Recall:

(x a)2 = x2 2ax + a2

We use the above results in solving quadratic equations by completing the squares.


Solve: x2 + 4x – 2 = 0.

Solution:

Transpose the constant – 2 to the right:

x2 + 4x = 2

Complete the squares on the left side by adding 4 to both sides of the equation (why?):

x2 + 4x + 4 = 2 + 4

Factor the left side completely and simplify:

(x + 2)2 = 6

We obtain x + 2 = to get x = – 2 . 6 6

25 Mathematics Division, IMSP, UPLB 62,62 SS


Solve: x2 – 6x + 7 = 0. Solution: Transpose the constant 7 to the right:

x2 – 6x = – 7 Complete the square on the left side by adding 9

to both sides of the equation (why?): x2 – 6x + 9 = – 7 + 9

Factor the left side completely and simplify: (x – 3)2 = 2

We obtain x – 3 = , to get x = 3 2 2

23,23 SS


In general, if you have ax2 + bx + c = 0 then factor out a: a[x2 + (b/a)x + c/a] = 0.

Hence, you will have: x2 + (b/a)x + c/a = 0.

To complete the square, transpose the constant

to the right: x2 + (b/a)x = –c/a

Then, add [(b/a)/2]2 to both sides: x2 + (b/a)x + [(b/a)/2]2 = –c/a + [(b/a)/2]2

The left hand side is already a perfect square:

[x + (b/a)/2]2 = –c/a + [(b/a)/2]2

Exercise:

Solve the following by completing the square:

1. 4x2 – 8x + 7 = 0

2. x2 – 5x = 24

3. x2 – 7x + 6 = 0.


2 b cx x

a a

222

22

a

b

a

c

a

bx

a

bx

From ax2 + bx + c = 0 , a 0,

We get ax2 + bx = – c.

Dividing both sides of the equation by a 0:

Completing the squares on the left side:


RECALL:

a

acbbx

2

42

2

2

2

22

4

4

42 a

acb

a

b

a

c

a

bx

2

2

4

2 4

b b acx

a a

Simplifying:

The Quadratic

Formula


or 2

142

2

142

xx

4

1424

4

564

22

524442

x

Example 1: Solve 2x2 – 4x – 5 = 0

Solution: Here a = 2, b = -4 and c = -5

By the quadratic formula:


2

142,

2

142SS


2

3x

Example 2: Solve 4x2 – 12x + 9 = 0

Solution: Here a = 4, b = -12 and c = 9


212 12 4 4 9

2 4

12 144 144 12 0

8 8

x

2

3SS

Example 3: Solve 2x2 – 4x + 5 = 0

Solution: Here a = 2, b = -4 and c = 5



Solution. real No Hence,

number. real anot is 24-

4

244

22

524442

x

61

2SS i

2

0 two real solutions

4 0 one real solution

0 no real solution (i.e., two complex solutions)

b ac

In the quadratic formula, the radicand

b2 – 4ac

is called the discriminant of the quadratic equation.

The discriminant gives the nature of the solutions of a quadratic equation:


In each of the following, determine the nature of the solutions of the given quadratic equation:

2

2

2

1 2 4 9 0

2 5 3 20 0

3 8 16 0

. x x

. x x

. w w


2 2

1 2

4 4

2 2

b b ac b b acr or r

a a

2 2

1 2

4 4

2 2

b b ac b b acr r

a a

a

b

a

b

2

2

Note that the solutions of the quadratic equation

ax2 + bx + c = 0 , a 0, is given by

Taking the sum of these solutions give:


2 2

1 2

4 4

2 2

b b ac b b acr r

a a

22 2 4

2 2

b b ac

a a

2 2

2

4

4

b b ac c

aa

Also, the product of the two solutions is given by:


2

2

2

1 2 4 9 0

2 5 3 20 0

3 8 16 0

. x x

. x x

. w w

In each of the following, find the sum and

product of the solutions of the given quadratic

equation:


In this, you have learned that

1. the solution of the linear equation ax + b = 0 is x = -b/a;

2. the quadratic equation ax2 + bx + c = 0 can be solve by factoring, completing the squares and by the quadratic formula.


3. the quadratic formula is given by:

2 4

2

b b acx

a

1 2 1 2

b cr r and r r

a a

4. the sum and product of the solutions of a

quadratic equation is given by


Documents

EQUATIONS AND INEQUALITIESjfrabajante.weebly.com/uploads/1/1/5/5/11551779/13_math_17.pdf · 1. Solve linear equations in one variable 2. Solve quadratic equations in one variable