In Algebra 1, it was shown that equations fall into three "truth" categories:

1. Equations that are TRUE under certain conditions.        Consider: x - 1 = 9     Only true when x = 10. Such an equation is called a "conditional equation". 2. Equations that are ALWAYS TRUE.        Consider x + 7 = 7 + x    True for all values of x. Such an equation is called an "identity". 3. Equations that are always FALSE.        Consider: x + 7 = x     Never true. Such an equation is called a "contradiction".

In this lesson, we are going to be examining "identities" associated quadratic equations.

Equations that are identities tend to be statements involving "properties" or "rules", such as a property of the real numbers (commutative property, distributive property, etc.), an arithmetic operation on the variable (addition, subtraction, etc), a rule for factoring, and so on. Both sides of the equation represent the same algebraic expression, just written in a different manner.

An equation that is true for every value of the variable is called an identity.

To show that an equation is an identity:
Start with either side of the equation and show that it can algebraically be changed into the other side. Or start with both sides of the equation and show that they both can be changed into the same algebraic expression.

You may be asked to verify (show, prove) that a known "rule" is an identity (always true).
You may, or may not, be asked to supply "justification" for your work.

1. Show, with justification, that (a - b)(a + b) = a2 - b2 is an identity. (a - b)(a + b) = a2 + ab - ab - b2    by the distributive property.                       = a2 - b2    by combining like terms. (a - b)(a + b) = a2 - b2 Identity! (always true)

You may be asked to verify (show, prove) that an algebraic equation is an identity.

NOTE: If you are asked to show that an equation "is" an identity, you may assume that the equation IS an identity. If you are asked to determine "whether" an equation is an identity, be careful as the answer may be that it is not an identity.

2. Show that x2 + 8(x + 2) = (x + 4)2 is an identity. (show left side creates right side) • x2 + 8(x + 2) = x2 + 8x + 16 (distribute)                      = (x + 4)(x + 4) (factor)                      = (x + 4)2 (multiply) • x2 + 8(x + 2) = (x + 4)2 (Is an identity.)

3. Determine whether b(2b + a + 1) + a = (a + b)(2b + 1) is an identity. (show both sides don't desolve to same term) • b(2b + a + 1) + a = (a + b)(2b + 1) • 2b2 + ab + b + a = 2ab + a + 2b2 + b • 2b2 + ab + b + a = 2b2 + 2ab + b + a • ab = 2ab • 1 = 2 (FALSE) This is not an identity.

4. Determine whether x2 + y2 = ( x + yi )( x - yi ) is an identity. ("i " is the imaginary unit.) (show right side creates the left side) • ( x + yi )( x - yi ) = x2 - xyi + xyi - (yi)2 multiplication (distribution)                     = x2 - (yi)2 combine like terms                     = x2 - y2• i2 property of exponents                     = x2 - y2• (-1) i2 = -1                     = x2 + y2 multiplication by (-1) • x2 + y2 = ( x + yi )( x - yi ) is an identity.

You may be asked to "check" someone's work to see if they properly showed the existence of an identity.

5. (3x - 2)(x - 3)2 Left side of equation. Step 1: = (3x - 2)(x - 3)(x - 3) Interpretation of squaring. Step2: = (3x - 2)(x2 - 3x - 3x + 9) Multiplication (distribution). Step 3: = (3x - 2)(x2 - 6x + 9) Combining like terms. Step 4: = 3x3 - 18x2 + 27x - 2x2 - 12x - 18 Multiplication (distribution). Step 5: = 3x3 - 20x2 + 15x - 18 Combining like terms. (3x - 2)(x - 3)2 = 3x3 - 20x2 + 15x - 18     Makes it an identity.

Keith made an error in Step 4. His fifth term should be +12x (not -12x).
(3x - 2)(x - 3)² does not equal 3x³ - 20x² + 15x - 18 for all values of x and is NOT an identity.
[Yes, Keith's equation is TRUE when x = 0,
but zero alone is insufficient to show the equation to be an "identity".]