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.
Moreover, it is standard for the leading coefficient a to be positive. If it is not, we can always make it positive by changing every sign. For example,
| −3x² + 2x + 5 | = | 0 | |
| implies | |||
| 3x² − 2x − 5 | = | 0. | |
Since −0 = 0, we again have the standard form.
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 (Lesson 6). 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 |
. |
Example 5. A rectangular field has a perimeter of 110 feet, and an area of 700 square feet. Find its length and width.
Solution. Let x be the length of the field, and let w be its width.
Then we have the following simultaneous equations:
| 1) | 2x + 2w | = | 110 (feet) |
| 2) | xw | = | 700 (square feet) |
Now equation 1) is linear but equation 2) is not, therefore we have to use the method of substitution (Lesson 35). We will solve equation 1) for w, and substitute it in equation 2).
First, on dividing both sides of equation 1) by 2, we have
| x + w | = | 55. | |
| Therefore | |||
| w | = | 55 − x. | |
Equation 2) then becomes:
| x(55 − x) | = | 700 |
| 55x − x² | = | 700 |
| x² − 55x + 700 | = | 0, |
on putting the equation in standard form.
The quadratic can be factored as follows:
(x − 35)(x − 20).
The two roots are 35 and 20. And either one of them is a value for x. When x = 35, then upon substituting that in the equations, we would find w = 20. And when x = 20, we would find w = 35. That happens in this case because the equations are symmetrical in x and w. That means that if we exchanged x and w, the equations would remain the same. Therefore, we could choose either root as the length, and the other will be the width.
Problem 7. Find two numbers such that one of them plus twice the other is 100, while their product is 450.
If you call one number x and the other y, then
| x + 2y | = | 100 |
| xy | = | 450. |
If you solve the top equation for x, and substitute that in xy, then you will eventually have the quadratic equation
y² − 50y + 225 = 0. This has roots 5 and 45. y therefore may have either value, and so there are two solutions. When y = 5, x = 90. When
y = 45, x = 10.
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Copyright © 2001-2008 Lawrence Spector
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