2.5 Quadratic Equations
Figure 1
The computer monitor on the left in Figure 1 is a 23.6-inch model and the one on the right is a 27-inch model. Proportionally, the monitors appear very similar. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods.
2.5.1 Solving Quadratic Equations by Factoring
An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as \(2x^{2} + 3x - 1 = 0\) and \(x^{2} - 4 = 0\) are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.
Often the easiest method of solving a quadratic equation is factoring. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation.
If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property, which states that if \(a \cdot b = 0,\) then \(a = 0\) or \(b = 0,\) where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.
Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression \(\left( {x - 2} \right)\left( {x + 3} \right)\) by multiplying the two factors together.
\[\begin{array}{ccl} {\left( {x - 2} \right)\left( {x + 3} \right)} & = & {x^{2} + 3x - 2x - 6} \\ & = & {x^{2} + x - 6} \end{array}\]
The product is a quadratic expression. Set equal to zero, \(x^{2} + x - 6 = 0\) is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.
The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, \(ax^{2} + bx + c = 0,\) where \(a\), \(b\), and \(c\) are real numbers, and \(a \neq 0.\) The equation \(x^{2} + x - 6 = 0\) is in standard form.
We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.
Solving Quadratics with a Leading Coefficient of 1
In the quadratic equation \(x^{2} + x - 6 = 0,\) the leading coefficient, or the coefficient of \(x^{2},\) is 1. We have one method of factoring quadratic equations in this form.
Factor and solve the equation: \(x^{2} + x - 6 = 0.\)
Solution (click to reveal)
To factor \(x^{2} + x - 6 = 0,\) we look for two numbers whose product equals \(-6\) and whose sum equals 1. Begin by looking at the possible factors of \(-6.\)
\[\begin{matrix} {1 \cdot (-6)} \\ {(-6) \cdot 1} \\ {2 \cdot (-3)} \\ {3 \cdot (-2)} \end{matrix}\]
The last pair, \(3 \cdot (-2)\) sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.
\[\left( {x - 2} \right)\left( {x + 3} \right) = 0\]
To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.
\[\begin{array}{rcl} {\left( {x - 2} \right)\left( {x + 3} \right)} & = & 0 \\ \left( {x - 2} \right) & = & 0 \\ x & = & 2 \\ \left( {x + 3} \right) & = & 0 \\ x & = & -3 \end{array}\]
The two solutions are \(2\) and \(-3.\) We can see how the solutions relate to the graph in Figure 2. The solutions are the \(x\)-intercepts of \(y = x^{2} + x - 6 = 0.\)
Figure 2
Solving a Quadratic Equation by Factoring when the Leading Coefficient is not 1
When the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures:
- With the quadratic in standard form, \(ax^{2} + bx + c = 0,\) multiply \(a \cdot c.\)
- Find two numbers whose product equals \(ac\) and whose sum equals \(b.\)
- Rewrite the equation replacing the \(bx\) term with two terms using the numbers found in step 2 as coefficients of \(x\).
- Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
- Factor out the expression in parentheses.
- Set the expressions equal to zero and solve for the variable.
Use grouping to factor and solve the quadratic equation: \(4x^{2} + 15x + 9 = 0.\)
Solution (click to reveal)
First, multiply \(ac:4(9) = 36.\) Then list the factors of \(36.\)
\[\begin{array}{l} {1 \cdot 36} \\ {2 \cdot 18} \\ {3 \cdot 12} \\ {4 \cdot 9} \\ {6 \cdot 6} \end{array}\]
The only pair of factors that sums to \(15\) is \(3 + 12.\) Rewrite the equation replacing the b term, \(15x,\) with two terms using 3 and 12 as coefficients of \(x\). Factor the first two terms, and then factor the last two terms.
\[\begin{array}{rcl} {4x^{2} + 3x + 12x + 9} & = & 0 \\ {x(4x + 3) + 3(4x + 3)} & = & 0 \\ {(4x + 3)(x + 3)} & = & 0 \end{array}\]
Solve using the zero-product property.
\[\begin{array}{rcl} {(4x + 3)(x + 3)} & = & 0 \\ {(4x + 3)} & = & 0 \\ x & = & {- \frac{3}{4}} \\ {(x + 3)} & = & 0 \\ x & = & {- 3} \end{array}\]
The solutions are \(- \frac{3}{4},\) and \(-3.\) See Figure 3.
Figure 3
2.5.2 Using the Square Root Property
When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property, in which we isolate the \(x^{2}\) term and take the square root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the \(x^{2}\) term so that the square root property can be used.
2.5.3 Completing the Square
Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a, must equal 1. If it does not, then divide the entire equation by a. Then, we can use the following procedures to solve a quadratic equation by completing the square.
We will use the example \(x^{2} + 4x + 1 = 0\) to illustrate each step.
- Given a quadratic equation that cannot be factored, and with \(a = 1,\) first add or subtract the constant term to the right side of the equal sign.
\[x^{2} + 4x = -1\]
- Multiply the b term by \(\frac{1}{2}\) and square it.
\[\begin{array}{rcl} {\frac{1}{2}(4)} & = & 2 \\ 2^{2} & = & 4 \end{array}\]
- Add \(\left( {\frac{1}{2}b} \right)^{2}\) to both sides of the equal sign and simplify the right side. We have
\[\begin{array}{rcl} {x^{2} + 4x + 4} & = & {- 1 + 4} \\ {x^{2} + 4x + 4} & = & 3 \end{array}\]
- The left side of the equation can now be factored as a perfect square.
\[\begin{array}{rcl} {x^{2} + 4x + 4} & = & 3 \\ {(x + 2)}^{2} & = & 3 \end{array}\]
- Use the square root property and solve.
\[\begin{array}{rcl} \sqrt{{(x + 2)}^{2}} & = & {\pm \sqrt{3}} \\ {x + 2} & = & {\pm \sqrt{3}} \\ x & = & {-2 \pm \sqrt{3}} \end{array}\]
- The solutions are \(-2 + \sqrt{3},\operatorname{}\) and \(-2 - \sqrt{3}.\)
2.5.4 Using the Quadratic Formula
The fourth method of solving a quadratic equation is by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.
We can derive the quadratic formula by completing the square. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by \(-1\) and obtain a positive \(a\). Given \(ax^{2} + bx + c = 0,\) \(a \neq 0,\) we will complete the square as follows:
- First, move the constant term to the right side of the equal sign:
\[ax^{2} + bx = - c\]
- As we want the leading coefficient to equal 1, divide through by \(a\):
\[x^{2} + \frac{b}{a}x = - \frac{c}{a}\]
- Then, find \(\frac{1}{2}\) of the middle term, and add \(\left( {\frac{1}{2}\frac{b}{a}} \right)^{2} = \frac{b^{2}}{4a^{2}}\) to both sides of the equal sign:
\[x^{2} + \frac{b}{a}x + \frac{b^{2}}{4a^{2}} = \frac{b^{2}}{4a^{2}} - \frac{c}{a}\]
- Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:
\[\left( {x + \frac{b}{2a}} \right)^{2} = \frac{b^{2} - 4ac}{4a^{2}}\]
- Now, use the square root property, which gives
\[\begin{array}{rcl} {x + \frac{b}{2a}} & = & {\pm \sqrt{\frac{b^{2} - 4ac}{4a^{2}}}} \\ {x + \frac{b}{2a}} & = & \frac{\pm \sqrt{b^{2} - 4ac}}{2a} \end{array}\]
- Finally, add \(- \frac{b}{2a}\) to both sides of the equation and combine the terms on the right side. Thus,
\[x = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}\]
2.5.5 The Discriminant
The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant, or the expression under the radical, \(b^{2} - 4ac.\) The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. Table 1 relates the value of the discriminant to the solutions of a quadratic equation.
| Value of Discriminant | Results |
|---|---|
| \(b^{2} - 4ac = 0\) | One rational solution (double solution) |
| \(b^{2} - 4ac > 0,\) perfect square | Two rational solutions |
| \(b^{2} - 4ac > 0,\) not a perfect square | Two irrational solutions |
| \(b^{2} - 4ac < 0\) | Two complex solutions |
Table 1
2.5.6 Using the Pythagorean Theorem
One of the most famous formulas in mathematics is the Pythagorean Theorem. It is based on a right triangle, and states the relationship among the lengths of the sides as \(a^{2} + b^{2} = c^{2},\) where \(a\) and \(b\) refer to the legs of a right triangle adjacent to the \(90{^\circ}\) angle, and \(c\) refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.
We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side.
The Pythagorean Theorem is given as
\[a^{2} + b^{2} = c^{2}\]
where \(a\) and \(b\) refer to the legs of a right triangle adjacent to the \(90^{\circ}\) angle, and \(c\) refers to the hypotenuse, as shown in Figure 4.
Figure 4
Find the length of the missing side of the right triangle in Figure 5.
Figure 5
Solution (click to reveal)
As we have measurements for side \(b\) and the hypotenuse, the missing side is \(a\).
\[\begin{array}{ccl} {a^{2} + b^{2}} & = & c^{2} \\ {a^{2} + {(4)}^{2}} & = & {(12)}^{2} \\ {a^{2} + 16} & = & 144 \\ a^{2} & = & 128 \\ a & = & \sqrt{128} \\ & = & {8\sqrt{2}} \end{array}\]
Section Exercises
Verbal
1. How do we recognize when an equation is quadratic?
Solution (click to reveal)
It is a second-degree equation (the highest variable exponent is 2).
2. When we solve a quadratic equation, how many solutions should we always start out seeking? Explain why when solving a quadratic equation in the form \(ax^{2} + bx + c = 0\) we may graph the equation \(y = ax^{2} + bx + c\) and have no zeroes (\(x\)-intercepts).
3. When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side?
Solution (click to reveal)
We want to take advantage of the zero property of multiplication in the fact that if \(a \cdot b = 0\) then it must follow that each factor separately offers a solution to the product being zero: \(a = 0\mspace{9mu} or\mspace{9mu}\text{b} = 0.\)
4. In the quadratic formula, what is the name of the expression under the radical sign \(b^{2} - 4ac,\) and how does it determine the number of and nature of our solutions?
5. Describe two scenarios where using the square root property to solve a quadratic equation would be the most efficient method.
Solution (click to reveal)
One, when no linear term is present (no \(x\) term), such as \(x^{2} = 16.\) Two, when the equation is already in the form \({(ax + b)}^{2} = d.\)
Algebraic
For the following exercises, solve the quadratic equation by factoring.
6. \(x^{2} + 4x - 21 = 0\)
7. \(x^{2} - 9x + 18 = 0\)
Solution (click to reveal)
\(x = 6,\) \(x = 3\)
8. \(2x^{2} + 9x - 5 = 0\)
9. \(6x^{2} + 17x + 5 = 0\)
Solution (click to reveal)
\(x = \frac{- 5}{2},\) \(x = \frac{- 1}{3}\)
10. \(4x^{2} - 12x + 8 = 0\)
11. \(3x^{2} - 75 = 0\)
Solution (click to reveal)
\(x = 5,\) \(x = -5\)
12. \(8x^{2} + 6x - 9 = 0\)
13. \(4x^{2} = 9\)
Solution (click to reveal)
\(x = \frac{- 3}{2},\) \(x = \frac{3}{2}\)
14. \(2x^{2} + 14x = 36\)
15. \(5x^{2} = 5x + 30\)
Solution (click to reveal)
\(x = -2,3\)
16. \(4x^{2} = 5x\)
17. \(7x^{2} + 3x = 0\)
Solution (click to reveal)
\(x = 0,\) \(x = \frac{- 3}{7}\)
18. \(\frac{x}{3} - \frac{9}{x} = 2\)
For the following exercises, solve the quadratic equation by using the square root property.
19. \(x^{2} = 36\)
Solution (click to reveal)
\(x = -6,\) \(x = 6\)
20. \(x^{2} = 49\)
21. \(\left( {x - 1} \right)^{2} = 25\)
Solution (click to reveal)
\(x = 6,\) \(x = -4\)
22. \(\left( {x - 3} \right)^{2} = 7\)
23. \(\left( {2x + 1} \right)^{2} = 9\)
Solution (click to reveal)
\(x = 1,\) \(x = -2\)
24. \(\left( {x - 5} \right)^{2} = 4\)
For the following exercises, solve the quadratic equation by completing the square. Show each step.
25. \(x^{2} - 9x - 22 = 0\)
Solution (click to reveal)
\(x = -2,\) \(x = 11\)
26. \(2x^{2} - 8x - 5 = 0\)
27. \(x^{2} - 6x = 13\)
Solution (click to reveal)
\(x = 3 \pm \sqrt{22}\)
28. \(x^{2} + \frac{2}{3}x - \frac{1}{3} = 0\)
29. \(2 + z = 6z^{2}\)
Solution (click to reveal)
\(z = \frac{2}{3},\) \(z = - \frac{1}{2}\)
30. \(6p^{2} + 7p - 20 = 0\)
31. \(2x^{2} - 3x - 1 = 0\)
Solution (click to reveal)
\(x = \frac{3 \pm \sqrt{17}}{4}\)
For the following exercises, determine the discriminant, and then state how many solutions there are and the nature of the solutions. Do not solve.
32. \(2x^{2} - 6x + 7 = 0\)
33. \(x^{2} + 4x + 7 = 0\)
Solution (click to reveal)
Not real
34. \(3x^{2} + 5x - 8 = 0\)
35. \(9x^{2} - 30x + 25 = 0\)
Solution (click to reveal)
One rational
36. \(2x^{2} - 3x - 7 = 0\)
37. \(6x^{2} - x - 2 = 0\)
Solution (click to reveal)
Two real; rational
For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution.
38. \(2x^{2} + 5x + 3 = 0\)
39. \(x^{2} + x = 4\)
Solution (click to reveal)
\(x = \frac{- 1 \pm \sqrt{17}}{2}\)
40. \(2x^{2} - 8x - 5 = 0\)
41. \(3x^{2} - 5x + 1 = 0\)
Solution (click to reveal)
\(x = \frac{5 \pm \sqrt{13}}{6}\)
42. \(x^{2} + 4x + 2 = 0\)
43. \(4 + \frac{1}{x} - \frac{1}{x^{2}} = 0\)
Solution (click to reveal)
\(x = \frac{- 1 \pm \sqrt{17}}{8}\)
Technology
For the following exercises, enter the expressions into your graphing utility and find the zeroes to the equation (the \(x\)-intercepts) by using 2nd CALC 2:zero. Recall finding zeroes will ask left bound (move your cursor to the left of the zero,enter), then right bound (move your cursor to the right of the zero,enter), then guess (move your cursor between the bounds near the zero, enter). Round your answers to the nearest thousandth.
44. \(\text{Y}_{1} = 4x^{2} + 3x - 2\)
45. \(\text{Y}_{1} = -3x^{2} + 8x - 1\)
Solution (click to reveal)
\(x \approx 0.131\) and \(x \approx 2.535\)
46. \(\text{Y}_{1} = 0.5x^{2} + x - 7\)
47. To solve the quadratic equation \(x^{2} + 5x - 7 = 4,\) we can graph these two equations
\(\begin{array}{l} \\ \begin{array}{l} {\text{Y}_{1} = x^{2} + 5x - 7} \\ {\text{Y}_{2} = 4} \end{array} \end{array}\)
and find the points of intersection. Recall 2nd CALC 5:intersection. Do this and find the solutions to the nearest tenth.
Solution (click to reveal)
\(x \approx - 6.7\) and \(x \approx 1.7\)
48. To solve the quadratic equation \(0.3x^{2} + 2x - 4 = 2,\) we can graph these two equations
\(\begin{array}{l} \\ \begin{array}{l} {\text{Y}_{1} = 0.3x^{2} + 2x - 4} \\ {\text{Y}_{2} = 2} \end{array} \end{array}\)
and find the points of intersection. Recall 2nd CALC 5:intersection. Do this and find the solutions to the nearest tenth.
Extensions
49. Beginning with the general form of a quadratic equation, \(ax^{2} + bx + c = 0,\) solve for \(x\) by using the completing the square method, thus deriving the quadratic formula.
Solution (click to reveal)
\(\begin{array}{rcl} {ax^{2} + bx + c} & = & 0 \\ {x^{2} + \frac{b}{a}x} & = & \frac{- c}{a} \\ {x^{2} + \frac{b}{a}x + \frac{b^{2}}{4a^{2}}} & = & {\frac{- c}{a} + \frac{b}{4a^{2}}} \\ \left( {x + \frac{b}{2a}} \right)^{2} & = & \frac{b^{2} - 4ac}{4a^{2}} \\ {x + \frac{b}{2a}} & = & {\pm \sqrt{\frac{b^{2} - 4ac}{4a^{2}}}} \\ x & = & \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a} \end{array}\)
50. Show that the sum of the two solutions to the quadratic equation is \(- \frac{b}{a}\) .
51. A person has a garden that has a length 10 feet longer than the width. Set up a quadratic equation to find the dimensions of the garden if its area is 119 ft.2. Solve the quadratic equation to find the length and width.
Solution (click to reveal)
\(x(x + 10) = 119;\) 7 ft. and 17 ft.
52. Abercrombie and Fitch stock had a price given as \(P = 0.2t^{2} - 5.6t + 50.2,\) where \(t\) is the time in months from 1999 to 2001. ( \(t = 1\) is January 1999). Find the two months in which the price of the stock was $30.
53. Suppose that an equation is given \(p = -2x^{2} + 280x - 1000,\) where \(x\) represents the number of items sold at an auction and \(p\) is the profit made by the business that ran the auction. How many items sold would make this profit a maximum? Solve this by graphing the expression in your graphing utility and finding the maximum using 2nd CALC maximum. To obtain a good window for the curve, set \(x\) [0,200] and \(y\) [0,10000].
Solution (click to reveal)
maximum at \(x = 70\)
Real-World Applications
54. A formula for the normal systolic blood pressure for a man age \(A,\) measured in mmHg, is given as \(P = 0.006A^{2} - 0.02A + 120.\) Find the age to the nearest year of a man whose normal blood pressure measures 125 mmHg.
55. The cost function for a certain company is \(C = 60x + 300\) and the revenue is given by \(R = 100x - 0.5x^{2}.\) Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of \(x\) (production level) that will create a profit of $300.
Solution (click to reveal)
The quadratic equation would be \((100x-0.5x^{2}) - (60x + 300) = 300.\) The two values of \(x\) are 20 and 60.
56. A falling object travels a distance given by the formula \(d = 5t + 16t^{2}\) ft, where \(t\) is measured in seconds. How long will it take for the object to travel 74 ft?
57. A vacant lot is being converted into a community garden. The garden and the walkway around its perimeter have an area of 378 ft2. Find the width of the walkway if the garden is 12 ft. wide by 15 ft. long.
Solution (click to reveal)
3 feet
58. An epidemiological study of the spread of a certain influenza strain that hit a small school population found that the total number of students, \(P\), who contracted the flu \(t\) days after it broke out is given by the model \(P = - t^{2} + 13t + 130,\) where \(1 \leq t \leq 6.\) Find the day that 160 students had the flu. Recall that the restriction on \(t\) is at most 6.