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Solving a Quadratic Equation: ax2 + bx +c = 0
| NOTE: You may not have heard of "Solving quadratic equations by Inspection". And, unfortunately, there seems to be a variety of ways to interpret this concept. So, we are going to go with a description of this strategy that we think makes the most sense. |
 To solve a quadratic equation by inspection:
Solving quadratic equations by inspection is the process of finding the roots by "observing" the structure of an equation and recognizing patterns or factors without using the usual formal methods to arrive at these values.
This technique relies on your abilities such as using simple factoring strategies, and simplifying an equation into a recognizable form. In other words, you basically solve the problem in your head based upon previous knowledge.
At the Algebra 2 level, your knowledge of working with quadratic equations will allow you to quickly recognize the roots of a variety of simple quadratic equations without any computations.
Simple Quadratic Observation:
• Example 1: Solve x2 = 25 (you can easily see that x = ±5) |
Simple Trinomial Factoring - Leading Coefficient of 1
:
| • Example 2: Solve x2 - 2x - 3 = 0 |
(you can easily see the factors are (x - 3)(x + 1).
Thus, the roots are x = 3 and x = -1.) |
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Recognize a Pattern:
| • Example 3: Solve x2 = -64 |
(you can easily see the perfect square and the indication of a complex solution. x = ±8i) |
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Recognize a Pattern:
| • Example 4: Solve x2 - 16 = 0 |
(you can easily see the difference of perfect
squares "pattern" in this problem. x = ±4) |
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Solving by inspection is usually reserved for working with quadratic equations
that are easily factored, or easily solved.
Some folks like to place factoring by
"Guess and Check" or "Trial and Error" as "Inspection".
Trial and Error
Factor: x2 - x - 12 |
Step 1: The first step when factoring is to always check for a common factor among the terms. Since the leading coefficients in this section are always "1", there is no common factor (other than 1). |
Does Not Apply to this problem
Due to leading coefficient of 1. |
Step 2: To get the leading term of x2, each first term of the factors will need to be x. |
(x )(x ) |
Step 3: The product of the last terms must be -12. Unfortunately, there are a variety of ways to arrive at this product of -12.
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+12 and -1
-12 and +1
+6 and - 2
-6 and +2
+4 and -3
-4 and +3
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ALL of these pairings
yield a product of -12.
But only ONE of them
is the correct choice!
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Possible answers?
(x + 12)(x - 1)
(x - 12)(x + 1)
(x + 6)(x - 2)
(x - 6)(x + 2)
(x + 4)(x - 3)
(x - 4)(x + 3)
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Step 4: Only one of the possible answers shown above will be the correct factoring for this problem. Which choice, when multiplied, will create the correct "middle term" needed for this example?
As we multiply through the various options, we discover that the needed middle term for this problem is "-x".
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(x + 12)(x - 1) middle term +11x
(x - 12)(x + 1) middle term -11x
(x + 6)(x - 2) middle term +4x
(x - 6)(x + 2) middle term -4x
(x + 4)(x - 3) middle term +x
(x - 4)(x + 3) middle term -x
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Step 5: Write the answer. |
x2 - x - 12 = (x - 4)(x + 3) |
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