Quadratic equations: Section 2
Proof of the quadratic formula
IN LESSON 18, we saw a technique called completing the square. We will now see how to apply it to solving a quadratic equation.
Completing the square
If we try to solve this quadratic equation by factoring,
| x² + 6x + 2 | = | 0 |
| we cannot. Therefore, we complete the square. That means to make the quadratic into the form | ||
| a² + 2ab + b² | = | (a + b)². |
The technique is valid only when 1 is the coefficient of x².
| 1) | Transpose the constant term to the right: |
x² + 6x = −2
| 2) | Add a square number to both sides. Add the square of half the coefficient of x. In this case, add the square of 3: |
x² + 6x + 9 = −2 + 9
The left-hand side is now the perfect square of (x + 3).
(x + 3)² = 7
3 is half of the coefficient 6.
This equation has the form
| a² | = | b | |
| which implies | |||
| a | = | ± . |
|
| Therefore, | |||
| x + 3 | = | ± |
|
| x | = | −3 ±. |
|
That is, the solutions to
x² + 6x + 2 = 0
are the conjugate pair,
−3 + , −3 − .
For a method of checking these roots, see Lesson 10 of Topics in Precalculus,
In Lesson 18, there are examples and problems in which the coefficient of x is odd. Also, some of the quadratics below have complex roots, and some involve simplifying radicals.
Problem 6. Solve each quadratic equation by completing the square.
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Do the problem yourself first!
| a) | x² − 2x − 2 | = | 0 | b) | x² + 4x − 6 | = | 0 | |
| x² − 2x | = | 2 | x² + 4x | = | 6 | |||
| x² − 2x + 1 | = | 2 + 1 | x² + 4x + 4 | = | 6 + 4 | |||
| (x − 1)² | = | 3 | (x + 2)² | = | 10 | |||
| x − 1 | = | ± |
x + 2 | = | ± |
|||
| x | = | 1 ± |
x | = | −2 ± |
|||
| c) | x² − 4x + 13 | = | 0 | d) | x² + 6x + 29 | = | 0 | |
| x² − 4x | = | −13 | x² + 6x | = | −29 | |||
| x² − 4x + 4 | = | −13 + 4 | x² + 6x + 9 | = | −29+ 9 | |||
| (x − 2)² | = | −9 | (x + 3)² | = | −20 | |||
| x − 2 | = | ±3i | x + 3 | = | ± |
|||
| x | = | 2 ± 3i | x | = | −3 ± 2i |
|||
| e) | x² − 5x − 5 | = | 0 | f) | x² + 3x + 1 | = | 0 | |
| x² − 5x | = | 5 | x² + 3x | = | −1 | |||
| x² − 5x + 25/4 | = | 5 + 25/4 | x² + 3x + 9/4 | = | −1 + 9/4 | |||
| (x − 5/2)² | = | 5 + 25/4 | (x + 3/2)² | = | − 1 + 9/4 | |||
| x − 5/2 | = | ± |
x + 3/2 | = | ± |
|||
| x | = |  |
x | = |  |
|||
The quadratic formula
Here is the quadratic formula -- which is proved by completing the square! In other words, the quadratic formula completes the square for us.
Theorem. If
ax² + bx + c = 0,
Theorem. then
We will prove this below.
Example 4. Use the quadratic formula to solve this quadratic equation:
3x² + 5x − 8 = 0
Solution. We have: a = 3, b = 5, c = −8.
Therefore, according to the formula:
| x | = |  |
| = |  |
|
| = |  |
|
| = |  |
|
That is,
| x | = | −5 + 11 6 |
or | −5 − 11 6 |
| x | = | 6 6 |
or | −16 6 |
| x | = | 1 | or − | 8 3 |
. |
These are the two roots. And they are rational. This tells us that we could have solved the equation by factoring, which is always the simplest method.
| 3x² + 5x − 8 | = | (3x + 8)(x − 1) | ||
| x | = | − | 8 3 |
or 1. |
Problem 7. Use the quadratic formula to find the roots of each quadratic.
a) x² − 5x + 5
a = 1, b = −5, c = 5.
| x | = |  |
= |  |
= |  |
b) 2x² − 8x + 5
a = 2, b = −8, c = 5.
| x | = |  |
||||||
| = |  |
= |  |
= |  |
= | 2 + ½ |
|
c) 5x² − 2x + 2
a = 5, b = −2, c = 2.
| x | = |  |
||||||
| = |  |
= |  |
= | 2 ± 6i 10 |
= | 1 ± 3i 5 |
|
The discriminant
The radicand b² − 4ac is called the discriminant. If the discriminant is
| a) Positive: | The roots are real and conjugate. |
| b) Negative: | The roots are complex and conjugate. |
| c) Zero: | The roots are rational and equal -- i.e. a double root. |
Proof of the quadratic formula
To prove the quadratic formula, we complete the square. But to do that, the coefficient of x² must be 1. Therefore, we will divide both sides of the original equation by a:
on multiplying both c and a by 4a, thus making the denominators the same (Lesson 23),
This is the quadratic formula.
Section 3: The graph of y = A quadratic
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