Download 1.1 Solving a Linear Equation ax + b = 0 To solve an equation ax + b

1.1 Solving a Linear Equation ax + b = 0
To solve an equation ax + b = 0 :
(i) move b to the other side (subtract b from both sides)
(ii) divide both sides by a
Example: Solve 2x – 3 = 0
(i)
2x-3 = 0
2x = 3
(ii)
2x 3

2 2
3
x
2
Sometimes certain operations on an equation have to be performed before the steps above can be
applied. Those operations are:
- Removing parenthesis by using distributive property
- Clearing fractions by multiplying both sides of the equation by the LCD of all fractions
- Grouping like terms
1
1
Example: Solve ( x  5)  4  (2 x  1)
2
3
(i)
The LCD of fractions ½ and 1/3 is 6. Multiply each term by 6 and simplify
(ii)
Remove the parenthesis by using the distributive property
3x+ 15 – 24 = 4x – 2
(iii) Combine the like terms (variable on the left, constants on the right)
-x=7
(iv) Divide both sides by the coefficient of x
x = -7
1.2 Solving quadratic equation
ax2 + bx + c = 0
There are three ways to solve a quadratic equation.
(a) Solving by factoring
(i) write the equation in the standard from: ax2 + bx + c = 0
(ii) factor, if possible, the right hand side of the equation
(iii) Use the Zero-Product Property: If a b = 0, then a = 0 or b = 0;
equate each factor to zero and solve corresponding equation.
Example: Solve x(x+4) = 12
(i)
Write in the standard form (remove parenthesis and bring all terms to the left)
x(x+4) = 12
x2 + 4x – 12 =0
(ii)
Factor the right hand side
(x+6)(x-2) = 0
(iii) Use the Zero-Product Property
x+ 6=0
x=-6
or
or
x–2=0
x=2
Solutions: -6, 2. The solution set is {-6, 2}
(b) Solving by the square root method
This method can be applied to equations of the form ax2 + c = 0
(i)
Solve the equation for x2
ax2 = -c
x2 = -c/a
(ii)
The solutions are (by taking the square root of both sides)
c
x
a
If –c/a is negative, there is no real solution
If –c/a is positive, there are two solutions
Example: Solve 2x2 + 6 = 0
2x2 + 6 = 0
2x2 = -6
x2 = -3
no (real) solution (can’t have a negative number inside the square
x   6
root)
2
Example: Solve 3x – 15 = 0
3x2 – 15 = 0
3x2 = 15
x2 = 5
x 5
Solutions:
5, - 5
Example: Solve (2x+ 1)2 = 3
This equation can be solved using the square root method, since by substituting 2x+1 by u, this
equation can be rewritten as u2 = 3.
(2x+1)2 = 3
2x+1 =  3
So, we have
2x+ 1 = 3 or
2x + 1 =  3
2 x  1  3
2 x  1  3
1 3
1  3
x
2
2
 1 3
 1 3
Solutions: x 
, x
2
2
(c) Solving using the quadratic formula ( must be memorized)
(i) write the equation in the standard form: ax2 + bx + c = 0
(ii) Identify a, b, c
(iii)Use the Quadratic Formula:
x
 b  b 2  4ac
x
2a
If b 2  4ac  0 , there are two different real solutions
If b 2  4ac  0 , there is one real solution
If b 2  4ac  0 , there are no real solutions
The expression b2 – 4ac is called the discriminant of the equation. It determines the number of
real solutions.
Remark: Any quadratic equation can be solved using quadratic formula.
Example: Solve 4x2 - 1 = - x
(i)
Write the equation in the standard form
4x2- 1 = - x
4x2 + x - 1 = 0
(ii)
Identify a, b, c
a = 4, b = 1, c = - 1
(iii)
Use the formula
 1  12  4  4  (1)  1  1  16  1  17


24
8
8
 1 17
 1 17
Solutions are x 
,x
8
8
x
1.4 Solving radical equations and equations leading to quadratic
Radical equation: An equation in which a variable appears inside a radical.
To solve a radical equation:
(i) Isolate a radical
(ii) Raise both sides to a power equal to the index of the radical
(iii) Simplify and solve the resulting equation
(iv) Check
Remark: Checking the apparent solutions of a radical equation is a necessary step as raising both
sides of an equation might produce numbers that are not the solutions of the original equation
Example: Solve x 2  x  4  x  2  0
(i) Isolate the radical ( Leave the radical on the left and move all other terms to the right)
x2  x  4  x  2
(ii) square both sides (since index = 2)
x2  x  4  x  2
x

 x  4  x  2
(iii) Simplify and solve the resulting equation
x2 – x – 4 = x2 + 4x + 4
-5x = 8
x = - 8/5
(iv) Check
2
2
2
2
?
 8  8
 8
      4     20
 5  5
 5
?
64 8
8
  4   20
25 5
5
4 2?
  0 yes
25 5
Solution: -8/5
Solving equations quadratic in form
An equation quadratic in form is an equation that becomes quadratic after using an appropriate
substitution
Example: Consider 3(1- y)2 + 5(1-y) + 2 = 0
Notice that if we replace (1 – y) by u (we let 1- y = u) we get the equation 3u2 + 5u + 2 = 0.
Example: Solve z1/2 – 4z1/4 + 4 = 0
Since ½ = 2( 1 4 ) we can re-write the equations as
(z1/4)2 – 4z1/4 + 4 = 0
Now substituting z1/4 by u (u = z1/4) gives
u2 – 4u + 4 = 0
This is a quadratic equation. There is only one solution u = 2.
Going back to the z variable we get z1/4 = 2 or 4 z  2
After raising both sides to the fourth power, we get z = 24 = 16
?
Check: (16)1/2 – 4 (16)1/4 + 4  0
4 – 4(2) + 4 = 0, yes
Solution: z = 16
Solving Polynomial Equations: polynomial = 0 , where the polynomial has the degree greater
than 2
(i)
Write the polynomial in the standard form: polynomial = 0
(ii)
Factor the polynomial
(iii) Use the Zero-Product Property
Example: Solve x3 + 2x2 = 3x
(i)
x3 + 2x2 = 3x
x3 + 2x2 – 3x = 0
(ii)
x(x2 + 2x – 3) = 0
x(x+3)(x-1) = 0
(iii) x = 0 or x + 3 = 0 or x – 1 = 0
x = -3
x=1
Solutions: 0, -3, 1
1.5 & 5.4 Solving polynomial and rational inequalities
Intervals
An interval is a subset of the set of real numbers that can be visualized as a segment (finite or
infinite) on the number line
If a < b, then
- The closed interval [a,b] consists of all real numbers x such that a < x < b. We can graph
this set on the number line as follows (the endpoints of the segment are included)
-
The open interval (a, b) consists of all real numbers x such that a < x < b. Below is the
graph of this set (the endpoints are not included)
-
The half open interval (a, b] is the set of all x for which a < x < b
-
The half open interval [a, b) is the set of all x such that a < x < b
-
The infinite interval (a, ) is the set of all x such that x > a
-
The infinite interval [a, ) is the set of all x such that x > a
-
The infinite interval (- , a) is the set of all x such that x< a
-
The infinite interval (- , a] is the set of all x such that x < a
-
The infinite interval (- , ) is the set of all real numbers
Solving a Linear Inequality: ax + b > (>, <, < ) 0
(i)
(ii)
Move b to the right hand side
Divide both sides by a, remembering that when dividing or multiplying an inequality by
a negative number, we MUST reverse the inequality
Example: Solve 2 – 3x > 0 and write the solution in the interval notation
(i)
2- 3x > 0
-3x > -2
(ii)
Divide both sides by (-3). The inequality will be reversed.
2
x
3
2
x
3
Solution: (, 2/3)
Solving a Polynomial Inequality: polynomial > (>, <, <) 0
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Write the inequality in the standard form (0 on the right hand side)
Solve the equation : polynomial = 0
Plot the solution on the number line
The solutions divide the number line into a finite number of intervals. Choose a number in
each interval and evaluate the value of the polynomial at each number.
If the value of the polynomial at the chosen number is positive (> 0), then the polynomial is
positive on the whole interval
If the value of the polynomial is negative (< 0), then the polynomial is negative on the whole
interval
Choose, as the solution, the intervals on which the polynomial has a desired sign. Use
interval notation. Include the endpoints only when the original inequality is < or > .
If there are two separate intervals on which the polynomial has a desired sign, use the union
sign, , between the intervals.
Example: Solve x3 > x
(i)
x3 > x
x3 – x > 0
(ii)
x3 – x = 0
x(x2- 1)= 0
x = 0 or x2 – 1 =0
x2 = 1
x =  1  1
(iii)
(iv)
x
-2
- 0.5
0.5
2
(v)
x3-x
(-2)3 –(-2) = -8 + 2= -6
(-0.5)2- (-0.5)= -.125+.5=.375
(.5)3-(.5)= -.375
23- 2= 6
negative
positive
negative
positive
(vi)
Since the inequality is x3 – x  0, we choose the intervals on which the polynomial is
positive and include the endpoints. There are two intervals, so we use the union
symbol .
Solution: [-1,0]  [1,)
P
 (, , )0 , P, Q are polynomials
Q
P
(i)
Write the inequality in the standard form  (, , )0
Q
(ii)
Solve the equations: P = 0 and Q = 0
(iii) Plot the solutions on the number line. Place open circle at each solution of Q = 0; those
numbers cannot ever be included in the solution set (they make the denominator zero!)
(iv)
The solutions divide the number line into a finite number of intervals. Choose a number
P
in each interval and evaluate the value of the expression at each number.
Q
P
P
(v)
If the value of is positive (> 0), then
is positive on the whole interval
Q
Q
P
P
If the value of is negative (< 0), then
is negative on the whole interval
Q
Q
P
(vi)
Choose, as the solution, the intervals on which
has a desired sign. Use interval
Q
notation. Include the endpoints only when the original inequality is < or > . Remember
to never include the endpoint with an open circle!
If there are two or more such intervals, use the union sign, .
x2
2
Example: Solve
x4
(i)
x2
2
x4
x2
20
x4
Solving a Rational Inequality:
(ii)
x  2 2( x  4)

0
x4
x4
x  2  2x  8
0
x4
 x  10
0
x4
Numerator = 0
–x+10 = 0
x = 10
denominator = 0
x- 4 = 0
x=4
(iii)
(iv)
x
0
5
11
 x  10
x4
 0  10 10

04
4
 5  10 5

54
1
 11  10  1

11  4
7
negative
positive
negative
(v)
 x  10
 x  10
> 0, we choose the intervals on which
is
x4
x4
positive and include endpoints that do not have an open circle .
Solution: (4, 10]
(vi)
Since the inequality is