The left computer monitor in the image below 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.

Factoring and the Square Root Property

An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as [latex]2{x}^{2}+3x - 1=0[/latex] and [latex]{x}^{2}-4=0[/latex] 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 [latex]a\cdot b=0[/latex], then [latex]a=0[/latex] or [latex]b=0[/latex], 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 [latex]\left(x - 2\right)\left(x+3\right)[/latex] by multiplying the two factors together.

The product is a quadratic expression. Set equal to zero, [latex]{x}^{2}+x - 6=0[/latex] 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, [latex]a{x}^{2}+bx+c=0[/latex], where a, b, and c are real numbers and [latex]a\ne 0[/latex]. The equation [latex]{x}^{2}+x - 6=0[/latex] 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 [latex]{x}^{2}+x - 6=0[/latex], the leading coefficient, or the coefficient of [latex]{x}^{2}[/latex], is 1. We have one method of factoring quadratic equations in this form.

Example 1: Factoring and Solving a Quadratic with Leading Coefficient of 1

Factor and solve the equation: [latex]{x}^{2}+x - 6=0[/latex].

Show Solution

To factor [latex]{x}^{2}+x - 6=0[/latex], we look for two numbers whose product equals [latex]-6[/latex] and whose sum equals 1. Begin by looking at the possible factors of [latex]-6[/latex].

[latex]\begin{array}{l}1\cdot \left(-6\right)\hfill \\ \left(-6\right)\cdot 1\hfill \\ 2\cdot \left(-3\right)\hfill \\ 3\cdot \left(-2\right)\hfill \end{array}[/latex]

The last pair, [latex]3\cdot \left(-2\right)[/latex] sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.

[latex]\left(x - 2\right)\left(x+3\right)=0[/latex]

To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.

[latex]\begin{array}{l}\left(x - 2\right)\left(x+3\right)=0\hfill \\ \left(x - 2\right)=0\hfill \\ x=2\hfill \\ \left(x+3\right)=0\hfill \\ x=-3\hfill \end{array}[/latex]

The two solutions are [latex]x=2[/latex] and [latex]x=-3[/latex]. We can see how the solutions relate to the graph below. The solutions are the x-intercepts of the graph of [latex]{x}^{2}+x - 6[/latex].

Factoring by Grouping

Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[/latex] can be rewritten as [latex]\left(2x+3\right)\left(x+1\right)[/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[/latex] and then factor each portion of the expression to obtain [latex]2x\left(x+1\right)+3\left(x+1\right)[/latex]. We then pull out the GCF of [latex]\left(x+1\right)[/latex] to find the factored expression.

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 [latex]{x}^{2}[/latex] term and take the square root of the number on the other side of the equal sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the [latex]{x}^{2}[/latex] term so that the square root property can be used.

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 [latex]{a}^{2}+{b}^{2}={c}^{2}[/latex], where [latex]a[/latex] and [latex]b[/latex] refer to the legs of a right triangle adjacent to the [latex]90^\circ[/latex] angle, and [latex]c[/latex] 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

where [latex]a[/latex] and [latex]b[/latex] refer to the legs of a right triangle adjacent to the [latex]{90}^{\circ }[/latex] angle, and [latex]c[/latex] refers to the hypotenuse.

Example 5: Finding the Length of the Missing Side of a Right Triangle

Find the length of the missing side of the right triangle.

Show Solution

As we have measurements for side b and the hypotenuse, the missing side is a.

[latex]\begin{array}{l}{a}^{2}+{b}^{2}={c}^{2}\hfill \\ {a}^{2}+{\left(4\right)}^{2}={\left(12\right)}^{2}\hfill \\ {a}^{2}+16=144\hfill \\ {a}^{2}=128\hfill \\ a=\sqrt{128}\hfill \\ a=8\sqrt{2}\hfill \end{array}[/latex]

Completing the Square and the Quadratic Formula

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 [latex]{x}^{2}+4x+1=0[/latex] to illustrate each step.

1. Given a quadratic equation that cannot be factored and with [latex]a=1[/latex], first add or subtract the constant term to the right sign of the equal sign.
   [latex]{x}^{2}+4x=-1[/latex]
2. Multiply the b term by [latex]\frac{1}{2}[/latex] and square it.
   [latex]\begin{array}{l}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}[/latex]
3. Add [latex]{\left(\frac{1}{2}b\right)}^{2}[/latex] to both sides of the equal sign and simplify the right side. We have
   [latex]\begin{array}{l}{x}^{2}+4x+4=-1+4\hfill \\ {x}^{2}+4x+4=3\hfill \end{array}[/latex]
4. The left side of the equation can now be factored as a perfect square.
   [latex]\begin{array}{l}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}[/latex]
5. Use the square root property and solve.
   [latex]\begin{array}{l}\sqrt{{\left(x+2\right)}^{2}}=\pm \sqrt{3}\hfill \\ x+2=\pm \sqrt{3}\hfill \\ x=-2\pm \sqrt{3}\hfill \end{array}[/latex]
6. The solutions are [latex]x=-2+\sqrt{3}[/latex], [latex]x=-2-\sqrt{3}[/latex].

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 [latex]-1[/latex] and obtain a positive a. Given [latex]a{x}^{2}+bx+c=0[/latex], [latex]a\ne 0[/latex], we will complete the square as follows:

1. First, move the constant term to the right side of the equal sign:
   [latex]a{x}^{2}+bx=-c[/latex]
2. As we want the leading coefficient to equal 1, divide through by a:
   [latex]{x}^{2}+\frac{b}{a}x=-\frac{c}{a}[/latex]
3. Then, find [latex]\frac{1}{2}[/latex] of the middle term, and add [latex]{\left(\frac{1}{2}\frac{b}{a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}[/latex] to both sides of the equal sign:
   [latex]{x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}[/latex]
4. Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:
   [latex]{\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}[/latex]
5. Now, use the square root property, which gives
   [latex]\begin{array}{l}x+\frac{b}{2a}=\pm \sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \\ x+\frac{b}{2a}=\frac{\pm \sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}[/latex]
6. Finally, add [latex]-\frac{b}{2a}[/latex] to both sides of the equation and combine the terms on the right side. Thus,
   [latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex]

Notice they are written in standard form of a complex number. When a solution is a complex number, you must separate the real part from the imaginary part and write it in standard form.

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, [latex]{b}^{2}-4ac[/latex]. The discriminant tells us whether the solutions are real numbers or complex numbers as well as how many solutions of each type to expect. The table below relates the value of the discriminant to the solutions of a quadratic equation.

Key Concepts

• Many quadratic equations can be solved by factoring when the equation has a leading coefficient of 1 or if the equation is a difference of squares. The zero-factor property is then used to find solutions.
• Many quadratic equations with a leading coefficient other than 1 can be solved by factoring using the grouping method.
• Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the square root of both sides of the equation. The solution will yield a positive and negative solution.
• Completing the square is a method of solving quadratic equations when the equation cannot be factored.
• A highly dependable method for solving quadratic equations is the quadratic formula based on the coefficients and the constant term in the equation.
• The discriminant is used to indicate the nature of the solutions that the quadratic equation will yield: real or complex, rational or irrational, and how many of each.
• The Pythagorean Theorem, among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation.

Glossary

completing the square

a process for solving quadratic equations in which terms are added to or subtracted from both sides of the equation in order to make one side a perfect square

discriminant

the expression under the radical in the quadratic formula that indicates the nature of the solutions, real or complex, rational or irrational, single or double roots.

Pythagorean Theorem

a theorem that states the relationship among the lengths of the sides of a right triangle, used to solve right triangle problems

quadratic equation

an equation containing a second-degree polynomial; can be solved using multiple methods

quadratic formula

a formula that will solve all quadratic equations

square root property

one of the methods used to solve a quadratic equation in which the [latex]{x}^{2}[/latex] term is isolated so that the square root of both sides of the equation can be taken to solve for x

zero-product property

the property that formally states that multiplication by zero is zero so that each factor of a quadratic equation can be set equal to zero to solve equations

Section 2.1 Homework Exercises

1. How do we recognize when an equation is quadratic?

2. When we solve a quadratic equation, how many solutions should we always start seeking out? Explain why when solving a quadratic equation in the form [latex]ax^{2}+bx+c=0[/latex] we may graph the equation [latex]\text{ }y=ax^{2}+bx+c[/latex] and have no zeroes ([latex]x[/latex]-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?

4. In the quadratic formula, what is the name of the expression under the radical sign [latex]\text{ }b^{2}-4ac[/latex], 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.

For the following exercises, solve the quadratic equation by factoring.

6. [latex]\text{ }x^{2}-4x-21=0[/latex]

7. [latex]\text{ }x^{2}-9x+18=0[/latex]

8. [latex]\text{ }2x^{2}+9x-5=0[/latex]

9. [latex]\text{ }6x^{2}+17x+5=0[/latex]

10. [latex]\text{ }4x^{2}-12x+8=0[/latex]

11. [latex]\text{ }3x^{2}-75=0[/latex]

12. [latex]\text{ }8x^{2}+6x-9=0[/latex]

13. [latex]\text{ }4x^{2}=9[/latex]

14. [latex]\text{ }2x^{2}+14x=36[/latex]

15. [latex]\text{ }5x^{2}=5x+30[/latex]

16. [latex]\text{ }4x^{2}=5x[/latex]

17. [latex]\text{ }7x^{2}+3x=0[/latex]

18. [latex]\text{ }\dfrac{x}{3}-\dfrac{9}{x}=2[/latex]

For the following exercises, solve the quadratic equation by using the square root property.

19. [latex]\text{ }x^{2}=36[/latex]

20. [latex]\text{ }x^{2}=49[/latex]

21. [latex]\text{ }(x-1)^{2}=25[/latex]

22. [latex]\text{ }(x-3)^{2}=7[/latex]

23. [latex]\text{ }(2x+1)^{2}=9[/latex]

24. [latex]\text{ }(x-5)^{2}=4[/latex]

For the following exercises, solve the quadratic equation by completing the square. Show each step.

25. [latex]\text{ }x^{2}-9x-22=0[/latex]

26. [latex]\text{ }2x^{2}-8x-5=0[/latex]

27. [latex]\text{ }x^{2}-6x=13[/latex]

28. [latex]\text{ }x^{2}+\dfrac{2}{3}x-\dfrac{1}{3}=0[/latex]

29. [latex]\text{ }2+z=6z^{2}[/latex]

30. [latex]\text{ }6p^{2}+7p-20=0[/latex]

31. [latex]\text{ }2x^{2}-3x-1=0[/latex]

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. [latex]\text{ }2x^{2}-6x+7=0[/latex]

33. [latex]\text{ }x^{2}+4x+7=0[/latex]

34. [latex]\text{ }3x^{2}+5x-8=0[/latex]

35. [latex]\text{ }9x^{2}-30x+25=0[/latex]

36. [latex]\text{ }2x^{2}-3x-7=0[/latex]

37. [latex]\text{ }6x^{2}-x-2=0[/latex]

For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution.

38. [latex]\text{ }2x^{2}+5x+3=0[/latex]

39. [latex]\text{ }x^{2}+x=4[/latex]

40. [latex]\text{ }2x^{2}-8x-5=0[/latex]

41. [latex]\text{ }3x^{2}-5x+1=0[/latex]

42. [latex]\text{ }x^{2}+4x+2=0[/latex]

43. [latex]\text{ }4+\dfrac{1}{x}-\dfrac{1}{x^{2}}=0[/latex]

44. [latex]\text{ }x+\dfrac{4}{x}+6=0[/latex]

45. Beginning with the general form of the quadratic equation [latex]\text{ }ax^{2}+bx+c=0[/latex], solve for x by using the completing the square method, thus deriving the quadratic formula.

46. Show that the sum of the two solutions to the quadratic formula is [latex]-\dfrac{b}{a}[/latex].

47. 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 [latex]119 ft^{2}[/latex]. Solve the quadratic equation to find the length and width.

48. Abercrombie and Fitch stock had a price given as [latex]P=0.2t^{2}-5.6t+50.2[/latex], where [latex]t[/latex] is the time in months from 1999 to 2001. ([latex]t=1[/latex] is January 1999). Find the two months in which the price of the stock was $30.

49. Suppose that an equation is given [latex]p=-2x^{2}+280x-1000[/latex], where [latex]x[/latex] represents the number of items sold at an auction and [latex]p[/latex] is the profit made by the business that ran the auction. How many items solve 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 [latex]x[/latex] [0,200] and [latex]y[/latex] [0,10000].

50. A formula for the normal systolic blood pressure for a man age A, measured in mmHg, is given as [latex]P=0.006A^{2}-0.02A+120[/latex]. Find the age to the nearest year of a man whose normal blood pressure measures 125 mmHg.

51. The cost function for a certain company is [latex]C=60x+300[/latex] and the revenue is given by [latex]R=100x-0.5x^{2}[/latex]. Recall the the profit is revenue minus cost. Set up a quadratic equation and find two values of [latex]x[/latex] (production level) that will create a profit of $300.

52. A falling object travels a distance given by the formula [latex]d=5t+16t^{2}[/latex] ft, where [latex]t[/latex] is measured in seconds. How long will it take for the object to travel 74 ft?

53. A vacanot lot is being converted into a community garden. The garden and walkway around its perimeter have an area of [latex]378\text{ }ft^{2}[/latex]. Find the width of the walkway if the garden is [latex]12ft[/latex] wide by [latex]15ft[/latex] long.

54. 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 [latex]t[/latex] days after it broke out is given by the model [latex]P=-t^{2}+13t+130[/latex], where [latex]1<=t<=6[/latex]. Find the day that 160 students had the flu. Recall that the restriction on [latex]t[/latex] is at most 6.