37
QUADRATIC EQUATIONS
Proof of the quadratic formula
A QUADRATIC is a polynomial whose highest exponent is 2.
Question 1. What is the standard form of a quadratic equation?
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ax² + bx + c = 0
The quadratic is on the left. 0 is on the right.
Question 2. What do we mean by a root of a quadratic?
A solution to the quadratic equation.
For example, this quadratic
x² + 2x − 8
can be factored as
(x + 4)(x − 2).
Now, if x = −4, then the first factor will be 0. While if x = 2, the second factor will be 0. But if any factor is 0, then the entire product will be 0. That is, if x = −4 or 2, then
x² + 2x − 8 = 0.
Therefore, −4 and 2 are the roots of that quadratic. They are the solutions to the quadratic equation.
A root of a quadratic is also called a zero. Because, as we will see, at those values of x, the graph has the value 0.
Question 3. How many roots has a quadratic?
Always two.
Question 4. What do we mean by a double root?
The two roots are equal.
For example, this quadratic
x² − 10x + 25
can be factored as
(x − 5)(x − 5).
If x = 5, then each factor will be 0, and therefore the quadratic will be 0. 5 is called a double root.
A quadratic will have a double root if the quadratic is a perfect square trinomial.
Problem 1. If either a = 0 or b = 0, then what can you conclude about ab ?
ab = 0
Solution by factoring
Problem 2. Find the roots of each quadratic by factoring.
| a) | x² − 3x + 2 | b) | x² + 7x + 12 | |
| (x − 1)(x − 2) | (x + 3)(x + 4) | |||
| x = 1 or 2. | x = −3 or −4. | |||
Notice that we use the conjunction "or," because x takes on only one value at a time.
| c) | x² + 3x − 10 | d) | x² − x − 30 | |
| (x + 5)(x − 2) | (x + 5)(x − 6) | |||
| x = −5 or 2. | x = −5 or 6. | |||
| e) | 2x² + 7x + 3 | f) | 3x² + x − 2 | |||||
| (2x + 1)(x + 3) | (3x − 2)(x + 1) | |||||||
| x = − | 1 2 |
or −3. | x = | 2 3 |
or −1. | |||
| g) | x² + 12x + 36 | h) | x² − 2x + 1 | |
| (x + 6)² | (x − 1)² | |||
| x = −6, −6. | x = 1, 1. | |||
| A double root. | A double root. | |||
Example 1. c = 0. Solve this quadratic equation:
ax² + bx = 0
Solution. Since there is no constant term -- c = 0 -- x is a common factor:
| x(ax + b) | = | 0. | ||||
| This implies: | ||||||
| x | = | 0 | ||||
| or | ||||||
| x | = | − | b a |
. | ||
Those are the two roots.
Problem 3. Find the roots of each quadratic.
| a) | x² − 5x | b) | x² + x | |
| x(x − 5) | x(x + 1) | |||
| x = 0 or 5. | x = 0 or −1. | |||
| c) | 3x² + 4x | d) | 2x² − x | |||
| x(3x + 4) | x(2x − 1) | |||||
| x = 0 or − | 4 3 |
x = 0 or ½ | ||||
Example 2. b = 0. Solve this quadratic equation:
ax² − c = 0.
Solution. In the case where there is no middle term, we can write:
| ax² | = | c. | |
| This implies: | |||
| x² | = | c a |
|
| x | = |  , according to Lesson 26. |
|
However, if the form is the difference of two squares --
x² − 16
-- then we can factor:
(x + 4)(x −4)
The roots are ±4.
In fact, if the quadratic is
x² − c,
then we could factor:
(x + 
)(x − )
so that the roots are ±.
Problem 4. Find the roots of each quadratic.
| a) | x² − 3 | b) | x² − 25 | c) | x² − 10 | ||
| x² = 3 | (x + 5)(x − 5) | (x +  )(x − ) |
|||||
x = ± . |
x = ±5. | x = ±. |
|||||
Example 3. Solve this quadratic equation:
| x² | = | x + 20 |
| Solution. First, rewrite the equation in the standard form, by transposing all the terms to the left: | ||
| x² − x − 20 | = | 0 |
| (x + 4)(x − 5) | = | 0 |
| x | = | −4 or 5. |
Thus, an equation is solved when x is isolated on the left.
x = ±
is not a solution.
Problem 5. Solve each equation for x.
| a) | x² = 5x − 6 | b) | x² + 12 = 8x | |
| x² − 5x + 6 = 0 | x² − 8x + 12 = 0 | |||
| (x − 2)(x − 3) = 0 | (x − 2)(x − 6) = 0 | |||
| x = 2 or 3. | x = 2 or 6. | |||
| c) | 3x² + x = 10 | d) | 2x² = x | |
| 3x² + x − 10 = 0 | 2x² − x = 0 | |||
| (3x − 5)(x + 2) = 0 | x(2x − 1) = 0 | |||
| x = 5/3 or − 2. | x = 0 or 1/2. | |||
Example 4. Solve this equation
| 3 − | 5 2 |
x − 3x² | = | 0 |
Solution. We can put this equation in the standard form by changing all the signs on both sides. 0 will not change. We have the standard form:
| 3x² + | 5 2 |
x − 3 | = | 0 |
Next, we can get rid of the fraction by multiplying both sides by 2. Again, 0 will not change.
| 6x² + 5x − 6 | = | 0 |
| (3x − 2)(2x + 3) | = | 0 |
| The roots are | 2 3 |
and − | 3 2 |
. |
Problem 6. Solve for x.
| a) | 3 − | 11 2 |
x − 5x² | = 0 | b) | 4 + | 11 3 |
x − 5x² | = 0 |
| 5x² + | 11 2 |
x − 3 | = 0 | 5x² − | 11 3 |
x − 4 | = 0 | |
| 10x² + 11x − 6 | = 0 | 15x² − 11x − 12 | = 0 | |||||
| (5x − 2 )(2x + 3) | = 0 | (3x − 4)(5x + 3 ) | = 0 | |||||
| The roots are | 2 5 |
and − | 3 2 |
. | The roots are | 4 3 |
and − | 3 5 |
. |
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Copyright © 2001-2007 Lawrence Spector
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