Download IMPORTANT INFO! Verbal Expressions and

Review
Solving Equations
Applications of Quadratic
Equations
Objectives: To write equations from given
information and use the six steps in solving an
applied problem.
IMPORTANT INFO!
Verbal Expressions and Mathematical
Expressions: twice, sum, product, less
than, difference, quotient, etc.
Don’t forget: An equation has an equal sign.
An expression does not.
1. Clear the equation of fractions. (Multiply
every term by the LCD to remove fractions.)
2. Use the Distributive Property to remove
parentheses on each side.
3. Combine like terms to get variable on one
side. (Undo addition or subtraction.)
4. Solve. (Undo multiplication or division.)
5. Check your solution by substituting what you
get into the original equation.
Steps
1. Read the problem completely.
2. Decide which unknown quantity
3.
4.
5.
6.
the variable will represent.
Write an equation .
Solve.
State the answer to the question.
Is it reasonable?
Check the solution.
Example 1 -- Solution
Example 1
The product of two consecutive locker
numbers at a health club is 132. Find the
locker numbers.
The product of two consecutive locker numbers at a
health club is 132. Find the locker numbers.
If the locker numbers are consecutive, that means
there is one number between them, or “ + 1”.
x = 1st locker
and (x + 1) = 2nd locker
Product means to multiply and the product equals
132.
x(x+1) = 132
x2 + x – 132 = 0 (quadratic so set = 0)
(x + 12)(x – 11) = 0 (factor)
x = -12 or x = 11 (set each factor & solve= 0)
Since locker numbers would not be negative, the
lockers are numbered 11 and 11+1 or 12.
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Example 2 -- Solution
Example 2
The product of two consecutive even integers is
four more than two times their sum. Find the
integers
The product of two consecutive even integers is four
more than two times their sum. Find the integers.
If the integers are consecutive evens, that means there are two numbers between them, or “ +
2”.
x = 1st integer and
(x + 2) = 2nd integer
Product means to multiply and sum means to add.
x(x+2) = 4 + 2[x + (x + 2)]
x2 + 2x = 4 + 2[2x + 2]
SIMPLIFY
x2 + 2x - 4 - 4x – 4 = 0
x2 – 2x – 8 = 0
(quadratic so set = 0)
(x – 4)(x + 2) = 0 (factor)
x = 4 or x = - 2 (set each factor = 0)
Since If x = 4 then, x + 2 = 4 + 2 = 6.
If x = -2, then x + 2 = -2 + 2 = 0.
Therefore, the integers are 4 & 6, or -2 & 0.
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