5. Solving Quadratic Equations by Using the Quadratic Formula
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| Roots | Factors |
|---|---|
| 2, 5 | (x-2), (x-5) |
| 1, 9 | (x-1), (x-9) |
| - 1, 3 | (x+1), (x-3) |
| 0, 6 | x, (x-6) |
| 1/2, 7 | (x-1/2), (x-7) |
| - 2/3, 4 | (x+2/3), (x-4) |
| Roots | Factors | Equation |
|---|---|---|
| 2, 5 | (x-2), (x-5) | x^2-7x+10=0 |
| 1, 9 | (x-1), (x-9) | x^2-10x+9=0 |
| - 1, 3 | (x+1), (x-3) | x^2-2x-3=0 |
| 0, 6 | x, (x-6) | x^2-6x=0 |
| 1/2, 7 | (x-1/2), (x-7) | 2x^2-15x+ 7=0 |
| - 2/3, 4 | (x+2/3), (x-4) | 3x^2-10x-8=0 |
Equation: x^3-6x^2+11x-6=0
No, see solution.
If pand mare two roots of a quadratic equation, then(x- p)and(x- m) are the factors of the equation. Now, let's complete the second column of the table.
| Roots | Factors |
|---|---|
| 2, 5 | (x- 2), (x- 5) |
| 1, 9 | (x- 1), (x- 9) |
| - 1, 3 | (x-( - 1)), (x- 3) or (x+1), (x-3) |
| 0, 6 | (x- 0), (x- 6) or x, (x-6) |
| 1/2, 7 | (x- 1/2), (x- 7) |
| - 2/3, 4 | (x-( - 2/3)), (x- 4) [0.5em] or [0.5em] (x+2/3), (x-4) |
As we can see, the coefficient of x is the sum of the roots and the constant term is their product. With this in mind, let's complete the last column.
| Roots | Factors | Equation |
|---|---|---|
| 2, 5 | (x-2), (x-5) | (x-2)(x-5)=0 x^2-7x+10=0 |
| 1, 9 | (x-1), (x-9) | (x-1)(x-9)=0 x^2-10x+9=0 |
| - 1, 3 | (x+1), (x-3) | (x+1)(x-3)=0 x^2-2x-3=0 |
| 0, 6 | x, (x-6) | x(x-6)=0 x^2-6x=0 |
| 1/2, 7 | (x-1/2), (x-7) | (x-1/2)(x-7)=0 x^2-15/2x+ 7/2=0 |
| - 2/3, 4 | (x+2/3), (x-4) | (x+2/3)(x-4)=0 x^2-10/3x- 8/3=0 |
To get rid of the fractions in the last two equations, we can multiply both sides of the equations by the denominator of the fractions.
| Roots | Factors | Equation |
|---|---|---|
| 2, 5 | (x-2), (x-5) | x^2-7x+10=0 |
| 1, 9 | (x-1), (x-9) | x^2-10x+9=0 |
| - 1, 3 | (x+1), (x-3) | x^2-2x-3=0 |
| 0, 6 | x, (x-6) | x^2-6x=0 |
| 1/2, 7 | (x-1/2), (x-7) | 2x^2-15x+ 7=0 |
| - 2/3, 4 | (x+2/3), (x-4) | 3x^2-10x-8=0 |