Transcript Algebra Expressions and Real Numbers

Section P7
Equations
Solving Linear Equations in
One Variable
Example
Solve the equation: 2(3x-5)=-5(x-1)+x
Linear Equations with Fractions
Solving with Fractions
Example
Solve for x.
x  1 3 2x  1
 
3
2
12
Example
Solve for x.
x  3 x 1 x  2


5
2
5
Rational Equations
A rational equation is an equation containing one
or more rational expressions. Notice how the variables
appear in the denominator in rational equations and the
previous examples ( Linear Equations with Fractions)
only had variables in the numerator.
Solving Rational Equations
Example
Solve the Rational Equation.
3
4
1
 2

x2 x 4 x2
Example
5
2 1
Solve the Rational Equation.
 
x3 x 2
Example
3
2
Solve the Rational Equation.
 2
6
x 1 x 1
Solving a Formula for One of Its
Variables
A-Prt=P
A=Prt+P
A=P(rt+1)
A
 P or
rt+1
A
P=
rt+1
Example
Solve for l in the formula for the Perimeter of a rectangle.
Example
1
Solve for b in the formula for the area of a triangle A= bh.
2
Equations Involving
Absolute Value
Example
Solve: x  5  9  7
Example
Solve: 2 2x  7  14  0
Quadratic Equations and
Factoring
Example
Solve the equation (2x-5)(3x+4)=0 using the Zero-Product
Principle.
Example
Solve the equation by factoring: x2  3x  4  0
Example
Solve the equation by factoring: 2x 2  7 x  4  0
Quadratic Equations and the
Square Root Property
Example
Solve the following problem by the square root property.
(x-4)2  25
Example
Solve the following problem by the square root property.
4x 2  7  0
Quadratic Equations and
Completing the Square
Obtaining a Perfect Square Trinomial
Start
Add
1 
 b
2 
2
Result
2
x  6x
1

 g 6  9
2

x  4x
1

 g 4  4
2

x 2  20 x
1

g
20

  100
2


2
x2  6x  9
 x  3
x2  4x  4
 x  2
2
2
Factored
Form
2
x 2  20 x  100
2
2
 x  10 
2
Completing the Square
Example
Complete the square to solve the following problem.
x 2  10 x  3  0
Example
Complete the square to solve the following problem.
x 2  8 x  13  0
Example
Complete the square to solve the following problem.
x 2  5 x  10
Quadratic Equations and the
Quadratic Formula
Example
Solve the equation using the quadratic formula.
x 2  6 x  3
Example
Solve the equation using the quadratic formula.
2x 2  4 x  5
Quadratic Equations and
the Discriminant
Example
Use the discriminant to find the number and types
of solutions, but don't solve the equation.
a. x 2  5 x  6  0
b. x 2  3 x  9
c. 2x 2  4 x  9
Graphing Calculator
The real solutions of a quadratic equation
ax2+bx+c=0 correspond to the x-intercepts of
the graph. The U shaped graph shown below
has two x intercepts. When y=0, the value(s) of
x will be the solution to the equation. Since y=0
these are called the zeros of the function.
Solving Polynomial Equations using the Graphing Calculator
By pressing 2nd Trace to get Calc, then the #2,you get the
zeros. It will ask you for left and right bounds, and then a
guess. For left and right bounds move the blinking cursor
(using the arrow keys-cursor keys) to the left and press
enter. Then move the cursor to the right of the x intercept
and press enter. Press enter when asked to guess. Then you
get the zeros or solution.
Repeat this process for
each x intercept.
Determining Which
Method to Use
Example
Factor and solve. -3x 2  6 x  0
Example
Solve by any method. -3x 2  15  0
Example
Solve by any method. x 2  4x  10  0
Radical Equations
A radical equation is an equation in which the variable
occurs in a square root, cube root, or any higher root.
We solve the equation by squaring both sides.
x4
If we square both sides, we obtain
x 2  16
x   16  -4 or 4
This new equation has two solutions, -4 and 4. By
contrast, only 4 is a solution of the original
equation, x=4. For this reason, when raising both
sides of an equation to an even power, check
proposed solutions in the original equation.
Extra solutions may be introduced when you raise
both sides of a radical equation to an even power.
Such solutions, which are not solutions of the
given equation are called extraneous solutions or
extraneous roots.
Example
Solve and check your answers:
x  5  x 1
1
Solve for h in the area formula for a trapezoid. A= h ( a  b)
2
(a)
(b)
(c)
(d)
2A
ab
A
2( a  b)
A
ab
A
2a  b
Solve: 3 x  8  27  0
(a) x  3, 3
(b) x  4,10
(c) x  1, 17
(d) x  17, 1