1. Quadratic Equations and Problem Solving
3.2
Understand basic concepts about quadratic equations
Use factoring, the square root property, completing the square, and the quadratic formula to solve quadratic equations
Understand the discriminant
Solve problems involving quadratic equations
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2. Quadratic Equation
A quadratic equation in one variable is an equation that can be written in the form
ax2 + bx + c
where a, b, and c are constants with a ≠ 0.
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3. Solving Quadratic Equations
Quadratic equations can have no real solutions, one real solution, or two real solutions.
The are four basic symbolic strategies in which quadratic equations can be solved.
Factoring
Square root property
Completing the square
Quadratic formula
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4. Factoring
Factoring is a common technique used to solve equations. It is based on the zero-product property, which states that if
ab = 0, then a = 0 or b = 0 or both. It is important to remember that this property works only for 0. For example, if ab = 1, then this equation does not imply that either a = 1 or b = 1. For example, a = 1/2 and b = 2 also satisfies ab = 1 and neither a nor b is 1.
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5. Example 1b
Solve the quadratic equation 12t2 = t + 1. Check your results Check:
Solution
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6. Example 1b Solution continued Check:
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7. The Square Root Property
Let k be a nonnegative number. Then the solutions to the equation
x2 = k
are given by
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8. Example 4
If a metal ball is dropped 100 feet from a water tower, its height h in feet above the ground after t seconds is given by h(t) = 100 – 16t2. Determine how long it takes the ball to hit the ground.
Solution
The ball strikes the ground when the equation 100 – 16t2 = 0 is satisfied.
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9. Example 4 Solution continued
The ball strikes the ground after 10/4, or 2.5, seconds.
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10. Completing the Square
Another technique that can be used to solve a quadratic equation is completing the square. If a quadratic equation is written in the form x2 + kx =d, where k and d are constants, then the equation can be solved using
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11. Example Solve 2x2 – 8x = 7. Solution Divide each side by 2:
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12. Symbolic, Numerical and Graphical Solutions
Quadratic equations can be solved symbolically, numerically, and graphically. The following example illustrates each technique for the equation x(x – 2) = 3.
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13. Symbolic Solution
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14. Numerical Solution
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15. Graphical Solution
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16. Quadratic Formula The solutions to the quadratic equation
ax2 + bx + c = 0, where a ≠ 0, are given by
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17. Example 7 Solve the equation 3x2 – 6x + 2 = 0. Solution
Let a = 3, b = 6, and c = 2.
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18. The Discriminant
If the quadratic equation ax2 + bx + c = 0 is solved graphically, the parabola y = ax2 + bx + c can intersect the x-axis zero, one, or two times. Each x-intercept is a real solution to the quadratic equation.
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19. Quadratic Equations and Discriminant
To determine the number of real solutions to ax2 + bx + c = 0 with a ≠ 0, evaluate the discriminant b2 – 4ac.
1. If b2 – 4ac > 0, there are two real solutions.
2. If b2 – 4ac = 0, there is one real solution.
3. If b2 – 4ac < 0, there are no real solutions.
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20. Example 9
Use the discriminant to find the number of solutions to 9x2 – 12.6x = 0. Then solve the equation by using the quadratic formula. Support your answer graphically.
Solution
Let a = 9, b = –12.6, and c = 4.41
b2 – 4ac = (–12.6)2 – 4(9)(4.41) = 0
Discriminant is 0, there is one solution.
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21. Example 9 Solution continued The only solution is 0.7.
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22. Example 9 Solution continued
The graph suggests there is only one intercept 0.7.
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23. Problem Solving
Many types of applications involve quadratic equations. To solve these problems, we use the steps for “Solving Application Problems” from Section 2.3 on page 122.
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24. Example 11
A box is being constructed by cutting 2-inch squares from the corners of a rectangular piece of cardboard that is 6 inches longer than it is wide. If the box is to have a volume of 224 cubic inches, find the dimensions of the piece of cardboard.
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25. Example 11
Solution
Step 1: Let x be the width and x + 6 be the length.
Step 2: Draw a picture.
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26. Solution continued
Since the height times the width times the length must equal the volume, or 224 cubic inches, the following can be written 2(x – 4)(x + 2) = 224
Step 3: Write the quadratic equation in the form ax2 + bx + c = 0 and factor.
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27. Solution continued
The dimensions can not be negative, so the width is 12 inches and the length is 6 inches more, or 18 inches.
Step 4: After the 2-inch-square corners are cut out, the dimensions of the bottom of the box are 12 – 4 = 8 inches by 18 – 4 = 14 inches. The volume of the box is then 2•8•14 = 224 cubic inches, which checks.
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