1. Objectives Solving Linear Equations Solving Quadratic Equations
Other Types of Equations

2. Equations
The values of the unknown that make the equation true are called the solutions or roots of the equation, and the process of finding the solutions is called solving the equation.
Two equations with exactly the same solutions are called equivalent equations. To solve an equation, we try to find a simpler, equivalent equation in which the variable stands alone on one side of the “equal” sign.

3. Equations

4. Solving Linear Equations
the difference between linear and nonlinear equations.

5. Example 1 – Solving a Linear Equation
Solve the equation 7x – 4 = 3x + 8.
Solution: We solve this by changing it to an equivalent equation with all terms that have the variable x on one side and all constant terms on the other.
7x – 4 = 3x + 8
(7x – 4) + 4 = (3x + 8) + 4
Given equation
Add 4

6. Example 1 – Solution 7x = 3x + 12 7x – 3x = (3x + 12) – 3x 4x = 12
cont’d
7x = 3x + 12
7x – 3x = (3x + 12) – 3x
4x = 12
x = 3
Simplify
Subtract 3x
Simplify
Multiply by
Simplify

7. Example 1 – Solution Check Your Answer:
cont’d
Check Your Answer:
x = 3: LHS = 7(3) – RHS = 3(3) + 8
= = 17
LHS = RHS

8. Solving Quadratic Equations

9. Solving Quadratic Equations
Linear equations are first-degree equations like 2x + 1 = 5 or 4 – 3x = 2. Quadratic equations are second-degree equations like x2 + 2x – 3 = 0 or 2x2 + 3 = 5x.

10. Solving Quadratic Equations
Set equation = to 0 Then Factor & Use…..
This means that if we can factor the left-hand side of a,
Then we can solve it by setting each factor equal to 0 in turn.
This method works only when the right-hand side of the equation is 0.

11. Example 4 – Solving a Quadratic Equation by Factoring
Solve the equation x2 + 5x = 24.
Solution: rewrite the equation so the right-hand side is 0.
x2 + 5x – 24 = 0
(x – 3)(x + 8) = 0
x – 3 = or x + 8 = Zero-Product Property
x = x = – Solve
And Check your answers!!!
Subtract 24
Factor

12. Example 5 – Solving Simple Quadratics
(a) x2 = (b) (x – 4)2 = 5
Solution: (a)
(b)Take the square root of each side of the equation.
(x – 4)2 = 5
Take the square root

13. Solving Quadratic Equations
if a quadratic equation is of the form (x  a)2 = c we can take the square root of each side….
If a quad equation does not factor , complete the square.
Add a constant to make it a perfect square.
To make x2 – 6x a perfect square,
add 9, since x2 – 6x + 9 = (x – 3)2.

14. Example 6 – Solving Quadratic Equations by Completing the Square
Solve each equation.
(a) x2 – 8x + 13 = (b) 3x2 – 12x + 6 = 0
Solution:
(a) x2 – 8x + 13 = 0
x2 – 8x = –13
x2 – 8x + 16 = –
Perfect square
Take square root
Add 4
Given equation
Subtract 13
Complete the square: add = 16

15. Example 6 – Solution 3x2 – 12x + 6 = 0 3x2 – 12x = –6 3(x2 – 4x) = –6
cont’d
3x2 – 12x + 6 = 0
3x2 – 12x = –6
3(x2 – 4x) = –6
we complete the square by adding (–2)2 = 4 inside the parentheses. Since everything inside the parentheses is multiplied by 3, this means that we are actually adding 3  4 = 12 to the left side of the equation.
Given equation
Subtract 6
Factor 3 from LHS

16. Example 6 – Solution Thus, we must add 12 to the right side as well.
cont’d
Thus, we must add 12 to the right side as well.
3(x2 – 4x + 4) = –6 + 3  4
3(x – 2)2 = 6
(x – 2)2 = 2
Complete the square: add 4
Perfect square
Divide by 3
Take square root
Add 2

17. Solving Quadratic Equations
We can use the technique of completing the square to derive a formula for the roots of the general quadratic equation ax2 + bx + c = 0.

18. Example 7 – Using the Quadratic Formula
Find all solutions of each equation.
(a) 3x2 – 5x – 1 = (b) 4x2 + 12x + 9 = 0
(c) x2 + 2x + 2 = 0
Solution: (a) In this quadratic equation a = 3, b = –5, and c = –1.
3x2 – 5x – 1 = 0

19. Example 7 – Solution By the Quadratic Formula,
cont’d
By the Quadratic Formula,
If approximations are desired, we can use a calculator to obtain
and

20. Example 7 – Solution
cont’d
(b) Using the Quadratic Formula with a = 4, b = 12, and c = 9 gives
This equation has only one solution,

21. Example 7 – Solution
cont’d
(c) Using the Quadratic Formula with a = 1, b = 2, and c = gives

22. Solving Quadratic Equations

23. Example 8 – Using the Discriminant
Use the discriminant to determine how many real solution each equation has.
(a) x2 + 4x – 1 = (b) 4x2 – 12x + 9 = 0
(c) x2 – 2x + 4 = 0
Solution:
(a) The discriminant is D = 42 – 4(1)(–1) = 20 > 0, so the equation has two distinct real solutions.
(b) The discriminant is D = (–12)2 – 4  4  9 = 0, so the equation has exactly one real solution.

24. Example 8 – Solution
cont’d
(c) The discriminant is D = (–2)2 – so the equation has no real solution.

25. Example 10 – An Equation Involving Fractional Expressions
Solve the equation
Solution: We eliminate the denominators by multiplying each side by the lowest common denominator.
x(x + 2) = 2x(x + 2)
3(x + 2) + 5x = 2x2 + 4x
Multiply by LCD x(x + 2)
Expand

26. Example 10 – Solution 8x + 6 = 2x2 + 4x 0 = 2x2 – 4x – 6
cont’d
8x + 6 = 2x2 + 4x
0 = 2x2 – 4x – 6
0 = x2 – 2x – 3
0 = (x – 3)(x + 1)
x – 3 = or x + 1 = 0
x = x = –1
Expand LHS
Subtract 8x + 6
Divide both sides by 2
Factor
Zero-Product Property
Solve

27. Example 10 – Solution
cont’d
We must check our answers because multiplying by an expression that contains the variable can introduce extraneous solutions. From Check Your Answers we see that the solutions are x = 3 and –1.
Check Your Answers:
x = 3:
LHS = = = RHS = 2
LHS = RHS

28. Example 10 – Solution x = –1: LHS = = –3 + 5 = 2 RHS = 2 LHS = RHS
cont’d
x = –1:
LHS = = –3 + 5 = 2
RHS = 2
LHS = RHS

29. Example 11 – An Equation Involving a Radical
Solve the equation
Solution: To eliminate the square root, we first isolate it on one side of the equal sign, then square:
(2x – 1)2 = 2 – x
4x2 – 4x + 1 = 2 – x
Subtract 1
Square each side
Expand LHS

30. Example 11 – Solution 4x2 – 3x – 1 = 0 (4x + 1)(x – 1) = 0
cont’d
4x2 – 3x – 1 = 0
(4x + 1)(x – 1) = 0
4x + 1 = 0 or x – 1 = 0
x = x = 1
The values x = and x = 1 are only potential solutions. We must check them to see if they satisfy the original equation. From Check Your Answers we see that x = is a solution but x = 1 is not. The only solution is x =
Add –2 + x
Factor
Zero-Product Property
Solve

31. Example 11 – Solution
cont’d
Check Your Answers:
LHS =
RHS =
LHS = RHS

32. Example 11 – Solution x = 1: LHS = 2(1) = 2 RHS = = 1 – 1 = 0
cont’d
x = 1:
LHS = 2(1)
= 2
RHS =
= 1 – 1
= 0
LHS ≠ RHS

33. Other Types of Equations
When we solve an equation, we may end up with one or more extraneous solutions, that is, potential solutions that do not satisfy the original equation.
In Example 11 the value x = 1 is an extraneous solution. Extraneous solutions may be introduced when we square each side of an equation because the operation of squaring can turn a false equation into a true one.
For example, –1 ≠ 1, but (–1)2 = Thus, the squared equation may be true for more values of the variable than the original equation.

34. Other Types of Equations
That is why you must always check your answers to make sure that each satisfies the original equation.
An equation of the form aW2 + bW + c = 0, where W is an algebraic expression, is an equation of quadratic type. We solve equations of quadratic type by substituting for the algebraic expression, as we see in the next example.

35. Example 12 – A Fourth-Degree Equation of Quadratic Type
Find all solutions of the equation x4 – 8x2 + 8 = 0.
Solution: If we set W = x2, then we get a quadratic equation in the new variable W:
(x2)2 – 8x2 + 8 = 0
W2 – 8W + 8 = 0
Write x4 as (x2)2
Let W = x2

36. Example 12 – Solution cont’d Quadratic Formula W = x2
Take square roots

37. Example 12 – Solution So, there are four solutions:
cont’d
So, there are four solutions:
Using a calculator, we obtain the approximations x  2.61, 1.08, –2.61, –1.08.