Practice Makes Perfect Algebra - Carolyn Wheater [22]
If b2 − 4ac is positive, the equation has two real solutions. You have a positive number under the radical, so you get the positive and the negative square root and produce two solutions. If b2 − 4ac equals 0, the positive and negative square roots are both 0, so you end up with only one solution, . This is sometimes called a double root, since it really is two solutions that are exactly the same. If b2 − 4ac is negative, you know it’s impossible to find the square root of a negative number in the real numbers.(Later in your math career, you’ll learn about a set of numbers larger than the reals, where negative numbers do have square roots.) So if the discriminant is negative, there are no real solutions.
You can get one more bit of information from the discriminant. If the discriminant is positive and it’s a perfect square, the two solutions will be rational numbers, but if the discriminant is positive and not a perfect square, the two solutions will be irrational.
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EXERCISE 9.3
Solve each equation by the quadratic formula. If necessary, leave answers in simplest radical form.
1. x2 + 4x − 21 = 0
2. t2 = 10 − 3t
3. y2 − 4y = 32
4. x2 = 6 + x
5. 6x + x2 = 9
6. t2 + 6t − 15 = 0
7. 4x2 − 3 = x
8. 3x2 − 1 = 2x
9. x + 5 = 3x2 − x
10. 6x2 − 2 = x
Use the discriminant to tell whether each equation has two rational solutions, two irrational solutions, one rational solution, or no real solutions.
11. x2 + 5x − 9 = 0
12. x2 − 3x + 5 = 0
13. x2 + 3x − 4 = 0
14. x2 + 6x + 9 = 0
15. 3x2 + 6x + 1 = 0
16. 2x2 + 5x + 3 = 0
17. 5x2 + 12x + 5 = 0
18. 2x2 + 3x + 5 = 0
19. 4x2 + 4x + 1 = 0
20. −3x2 + 3x + 5 = 0
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Solving by factoring
The quadratic formula will give you solutions for any quadratic equation, but it may require complicated calculations. Sometimes there’s no way around that. Completing the square is complicated, the quadratic formula can be complicated, and irrational solutions aren’t easy. In other cases, however, the quadratic polynomial can be factored and the equation can be solved by applying the zero product property. If the product of two factors is 0, then at least one of the factors is 0. If the quadratic can be factored, that’s probably the easiest way to solve.
First, put the equation in standard form. With all terms on one side of the equation equal to 0, factor the quadratic expression. Set each factor equal to 0 and solve the resulting equations.
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EXERCISE 9.4
Solve each equation by factoring.
1. x2 + 5x + 6 = 0
2. x2 + 12 = 7x
3. y2 + 3y = 8 + y
4. a2 − 3a − 4 = 6
5. 20 = x2 + x
6. x2 + 5 = 6x
7. x2 + 3x = 0
8. x2 = 5x
9. x2 − 1 = 3x2 − 3x
10. 2x2 − x − 15 = 0
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Graphing quadratic functions
Graphing an equation of the form y = ax2 + bx + c produces a cup-shaped graph, called a parabola. Some information about the graph can be gathered from the equation without much effort and can help you construct a table of values and plot the graph. If a is a positive number, the parabola opens upward; if a is negative, the parabola opens downward(see Figure 9.1). The x-intercepts are the solutions of the equation ax2 + bx + c = 0 and the y-intercept is(0, c).
Figure 9.1 The coefficient of the square term tells you whether the parabola opens up or down.
Finding the axis of symmetry and vertex
The axis of symmetry of a parabola is an imaginary line through the center of the parabola. The parabola is symmetric about this line. If you were to fold the graph along the axis of symmetry, the two sides of the parabola would match. The vertex, or turning point, of the parabola sits right on the axis of symmetry. The equation of the vertical line that is the axis of symmetry is . Once the axis of symmetry is known, plugging that x-value into the equation of the parabola will give you the y-coordinate of the vertex.
To graph y = x2 + 4x + 1, notice that the x2-term is positive,