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Learning Objectives
By the end of this section, you will be able to:
 Understand what it means to solve a quadratic equation.
 Solve quadratic equations with one unknown, i.e., onevariable by using the zeroproduct property.
Be Prepared
Before you get started, take this readiness quiz.
 Simplify \(3(2+x)(3x2)\).
 Solve: \(5y−3=0\).
 Factor completely: \(n^3−9n^2−22n\).
 Simplify \(1/3+9/4\cdot 5/7(32/5)\).
 Why is the following expression quadratic? \(9 x^{2}12 x+4\).
Quadratic Equations
Just like a linear equation in one variable is an equation that is equivalent to (a linear expression\(=0\)), a quadratic equation is an equation in one variable, is an equation that is equivalent to (\(\text{a quadratic expression }=0\)), and the equation isn't equivalent to a linear equation. For example, suppose that the area of a rectangular region with the length 3 feet more than twice its width (in feet) is equal to a square region of the twice the width of the other. We don't know what the width is (let's call it \(w\)) but we can write down this assertion about the width.
The area of a square region with width \(w\) is  \((2w)^2\) or, \(4x^2\) 
If the length of the rectangular region is 3 more than twice its width, then its length is  \(2w+3\) 
So the area of the rectangular region is length times width, which in terms of the unknown \(w\) is  \((2w+3)w\), or \(2w^2+3w\) 
The assertion that these two areas are the same can be written  \(4w^2=2w^2+3w.\) 
Now, this looks like a quadratic equation, at least on the surface, but we need to make sure it isn't a linear equation in disguise so we subtract \(4w^2\) from both sides of the equation to get
\(0=2w^2+3w.\)
We will not show this in detail, but an equation of the form \(0=\)(something quadratic) is not a linear equation.
This is the assertion about the width in the problem! Whatever \(w\) is, it satisfies the equation
\(0=2w^2+3w.\)
For example, \(w\not= 1\) since if it were, \(2w^2+3w=2 (1)^2+3(1)=1\) which is not \(0\) as asserted.
Our question is then, how do you find the width (so that if you substitute that width in for \(w\) the assertion is true!
In general we will be interested in finding the solutions to quadratic equations.
Definition
A solution to an equation in a variable (lets call it \(x\)) is a number that we can substitute in for \(x\) that makes the equation true.
Two equations are equivalent if they have the same solutions.
A quadratic equation with one variable, say, \(x\),has polynomialson both sides of the equation andcan be written in standard form: \(Ax^2+Bx+C=0\) where \(A\) is not zero, i.e., there is an equivalent equation of the form \(Ax^2+Bx+C=0\).
It is important to understand the difference between expressions and equations. We give here examples to demonstrate the two concepts.
ExpressionsvsEquations
A quadratic expressionand a quadraticequation are not the same types of objects. Here are some examples.
Expressions  Equation 
\(2x+1\), \(0\)  \(2x+1=0\) (linear equation) 
\(x^2\), \(x2\)  \(x^2=x2\) (quadratic equation) 
\(3x^2+5x1\), \(0\)  \(3x^2+5x1=0\) (quadratic equation) 
Note that \(x^2+5x=x(x+5)\) is not a polynomial equation that we aim to solve. This is an identity of polynomials that was developed in the process of factoring out the GCF.
It does not make sense to "solve a polynomial." We can only solve equations. For example, we cannot solve \(2x+1\) as there is no statement to assess. Whereas \(2x+1=0\) is either true or false for a particular value of \(x\). Below we check whether or not some values of \(x\) are solutions to the equation \(2x+1=0\).
\(\bf{x}\)  True or False  Solution or Not a Solution 
\(5\)  \(2\cdot 5 + 1= 0\) is false  Not a Solution 
\(0\)  \(2\cdot 0 +1 = 0\) is false  Not a Solution 
\(\dfrac{1}{2}\)  \(2\cdot \left(\dfrac{1}{2}\right) +1 = 0\) is true  Solution 
To solve quadratic equations we need methods different from the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.
Example \(\PageIndex{1}\)
Is \(x^32x+1=x^3+2x^2+x7\) a quadratic equation? If so, is \(x=2\) a solution?
Solution
The left and the right sides of the equality are not quadratic, so the first impulse may be to say that this is not a quadratic equation. But let's be patient and see if we can put it in the required form:
\(x^32x+1=x^3+2x^2+x7\)  
We do not change the solutions to the equation if we subtract the same quantity on both sides of the equality. Let's subtract \(x^32x+1\), which is the same as adding \(x^3+2x1\).  \(x^32x+1x^3+2x1=x^3+2x^2+x7x^3+2x1\) 
And upon simplifying (which also doesn't change the solutions) we see  \(0=2x^2+3x8\), or, \(2x^2+3x8=0\) 
This is in the correct form \(Ax^2+Bx+C=0\),  \(A=2, B=3\), and \(C=8\). 
We see that \(A\) is not zero, so this is indeed a quadratic equation.
To see whether \(2\) is a solution, we can substitute it into any of the equivalent equations to check it's truth. We'll check to see if \(2\) is a solution to \(2x^2+3x8=0\).
The left hand side is \(2(2)^2+3(2)8=6\) which is not equal to the right hand side (\(0\)). So \(2\) is not a solution to the equation!
We could've checked the original equation instead and arrived at the same conclusion.
Finding solutions to a quadratic equation is, in general, a little more difficult than finding solution to a linear equation. But we know that we can sometimes write a quadratic equation as a product of two linear factors (or two linear expressions and a number). As we will see this can be vary handy!!
The zeroproduct property and solutions to quadratic equations
We know that \(a\cdot 0=0\) and \(0\cdot a=0\) no matter what \(a\) is. Suppose though that \(a\cdot b =0\). What does this tell us about \(a\) and \(b\)?
\(a\cdot b =0\)  
Either \(a=0\) or it is not. If \(a\not= 0\), then we can divide both sides of the equality by \(a\).  so, \(\dfrac{a\cdot b}{a} =\dfrac{0}{a}\), or, simplifying, \(b=0\)! 
So, either \(a=0\) or \(b=0\), or both \(a=0\) and \(b=0\). 
This is called the zero product property:
The zero product property
If \(ab=0\) then \(a=0\) or \(b=0\) or both.
This property will allow us to sometimes solve quadratic equations by reducing the problem to something we already know how to do, namely, solve linear equations.
We have seen that it is sometimes the case that we can write a quadratic expression as a product of two linear expressions. If we know this product is zero, then we can use the zeroproduct property!
We start with an example,
Example \(\PageIndex{1}\)
Lets solve \((2x3)\cdot (3x+5)=0\).
Solution
If the equation is true, then this asserts something about the unknown which we call \(x\). This equation is a product of numbers (that we don't know) and that product is zero. So we can use the zeroproduct property to conclude that
either \((2x3)=0\) or \((3x+5)=0\) (or both).
In other words, if \((2x3)\cdot (3x+5)=0\) is true about \(x\), then either \((2x3)=0\) is true or \(3x+5)=0\) is true (or both).
We have experience solving these linear equation from the last section!
If \((2x3)=0\) then \(2x3+3=0+3\), or, \(2x=3\). \(x\) is still being multiplied by \(2\) so we must divide both sides of the equal sign by \(2\)
and we see
\(2x/2=3/2\) or, simplifying, \(x=3/2\).
Similarly, if \(3x+5)=0\) then \(x=5/3.\)
To conclude, if \((2x3)\cdot (3x+5)=0\) then it must be that either \(x=3/2\) or \(x=5/3\).
So, we have solved the equation by finding the values that when substituted in for \(x\) yield a true statement, namely, \(3/2\) and \(5/3\).
In the context of an application, we may have more information about \(x\) that will help us pin down our unknown to one single answer.
Try It \(\PageIndex{4}\)
Solve \((3m−2)(2m+1)=0\).
 Answer

\(m=\dfrac{2}{3},\space m=−\dfrac{1}{2}\)
Try It \(\PageIndex{5}\)
Solve \((4p+3)(4p−3)=0\).
 Answer

\(p=−\dfrac{3}{4},\space p=\dfrac{3}{4}\)
Solve Quadratic Equations by Factoring
The Zero Product Property works very nicely to solve quadratic equations. The quadratic equation must be factored, with zero isolated on one side. So be sure to start with the quadratic equation in standard form, \(ax^2+bx+c=0\). Then factor the expression on the left.
Example \(\PageIndex{6}\)
Solve \(2y^2=13y+45\).
 Solution
Try It \(\PageIndex{7}\)
Solve \(3c^2=10c−8\).
 Answer

\(c=2,\space c=\dfrac{4}{3}\)
Try It \(\PageIndex{8}\)
Solve \(2d^2−5d=3\).
 Answer

\(d=3,\space d=−\dfrac{1}{2}\)
Solve a Quadratic Equation by Factoring
 Write the quadratic equation in standard form, \(ax^2+bx+c=0\).
 Factor the quadratic expression.
 Use the Zero Product Property.
 Solve the linear equations.
 Optional: Check answer by substituting each solution separately into the original equation to see if a mistake has been made.
Before we factor, we must make sure the quadratic equation is in standard form.
Solving quadratic equations by factoring will make use of all the factoring techniques we have learned in this chapter! Do we recognize the special product pattern in the next example?
Example \(\PageIndex{9}\)
Solve \(169q^2=49\).
 Solution

\(\begin{array} {ll} &169q^2=49 \\ \text{Write the quadratic equation in standard form.} &169q^2−49=0 \\ \text{Factor. It is a difference of squares.} &(13q−7)(13q+7)=0 \\ \text{Use the Zero Product Property to set each factor to }0. & \\ \text{Solve each equation.} &\begin{array} {ll} 13q−7=0 &13q+7=0 \\ 13q=7 &13q=−7 \\ q=\dfrac{7}{13} &q=−\dfrac{7}{13} \end{array} \\ \text{Check the answers.} &\text{The check is left as an exercise.} \end{array}\)
Try It \(\PageIndex{10}\)
Solve \(25p^2=49\).
 Answer

\(p=\dfrac{7}{5},p=−\dfrac{7}{5}\)
Try It \(\PageIndex{11}\)
Solve \(36x^2=121\).
 Answer

\(x=\dfrac{11}{6}\), \(x=−\dfrac{11}{6}\)
In the next example, the left side of the equation is factored, but the right side is not zero. In order to use the Zero Product Property, one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.
Example \(\PageIndex{12}\)
Solve \((3x−8)(x−1)=3x\).
 Solution

\(\begin{array} {ll} &(3x−8)(x−1)=3x \\ \text{Multiply the binomials.} &3x^2−11x+8=3x \\ \text{Write the quadratic equation in standard form.} &3x^2−14x+8=0 \\ \text{Factor the trinomial.} &(3x−2)(x−4)=0 \\ \begin{array} {l} \text{Use the Zero Product Property to set each factor to 0.} \\ \text{Solve each equation.} \end{array} &\begin{array} {ll} 3x−2=0 &x−4=0 \\ 3x=2 &x=4 \\ x=\dfrac{2}{3} & \end{array} \\ \text{Check the answers.} &\text{The check is left as an exercise.} \end{array}\)
Try It \(\PageIndex{13}\)
Solve \((2m+1)(m+3)=12m\).
 Answer

\(m=1\), \(m=\dfrac{3}{2}\)
Try It \(\PageIndex{14}\)
Solve \((k+1)(k−1)=8\).
 Answer

\(k=3\),\(k=−3\)
In the next example, when we factor the quadratic equation we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.
Example \(\PageIndex{15}\)
Solve \(3x^2=12x+63\).
 Solution

\(\begin{array} {ll} &3x^2=12x+63 \\ \text{Write the quadratic equation in standard form.} &3x^2−12x−63=0 \\ \text{Factor the greatest common factor first.} &3(x^2−4x−21)=0 \\ \text{Factor the trinomial.} &3(x−7)(x+3)=0 \\ \begin{array} {l} \text{Use the Zero Product Property to set each factor to 0.} \\ \text{Solve each equation.} \end{array} &\begin{array} {lll} 3\neq 0 &x−7=0 &x+3=0 \\ 3\neq 0 &x=7 &x=−3 \end{array} \\ \text{Check the answers.} &\text{The check is left as an exercise.} \end{array}\)
Try It \(\PageIndex{16}\)
Solve \(18a^2−30=−33a\).
 Answer

\(a=−\dfrac{5}{2}\), \(a=\dfrac{2}{3}\)
Try It \(\PageIndex{17}\)
Solve \(123b=−6−60b^2\).
 Answer

\(b=−2\),\(b=−\dfrac{1}{20}\)
The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree greater than two by using the Zero Product Property, just like we solved quadratic equations.
Example \(\PageIndex{18}\)
Solve \(9m^3+100m=60m^2\).
 Solution

\(\begin{array} {ll} & 9m^3+100m=60m^2 \\ \text{Bring all the terms to one side so that the other side is zero.} &9m^3−60m^2+100m=0 \\ \text{Factor the greatest common factor first.} &m(9m^2−60m+100)=0 \\ \text{Factor the trinomial.} &m(3m−10)^2=0 \end{array}\\ \begin{array} {l} \text{Use the Zero Product Property to set each factor to 0.} \\ \text{Solve each equation.} &\begin{array} {lll} m=0 &3m−10=0 &{}\\ m=0 &m=\dfrac{10}{3} & {} \end{array}\\ \text{Check your answers.} &\text{The check is left to you.} \end{array}\)
Try It \(\PageIndex{19}\)
Solve \(8x^3=24x^2−18x\).
 Answer

\(x=0\),\(x=\dfrac{3}{2}\)
Try It \(\PageIndex{20}\)
Solve \(16y^2=32y^3+2y\).
 Answer

\(y=0\), \(y=14\)
We will now continue with our application:
Example \(\PageIndex{1}\)
Suppose that the area of a rectangular region with the length 3 feet more than twice its width (in feet) is equal to a square region twice the width of the other. Find the dimensions of the two regions.
Solution
We saw in the introduction that if we call the width \(w\), that the length is then \(2w+3\) and the assertion is
\((2w)^2=(2w+3)w\)
which is equivalent to
\(2w^2+3w=0.\)
So, let's factor this so that it looks more like our previous example (which we now know how to go about solving).
We have experience factoring from Unit 1, but we will remind you to factor out the GCF which is \(w\).
An equivalent equation is therefore
\(w(2w+3)=0.\)
Again, this is an assertion about the width, i.e., the width \(w\) that we are looking for is a solution to this equation. Using the zeroproduct property we see that
if \(w(2w+3)=0\) then \(w=0\) or \(2w+3=0\)
or, equivalently,
if \(w(2w+3)=0\) then \(w=0\) or \(w=3/2\).
Since \(w\) is a width, we see that it doesn't make sense for \(w\) to be zero. So, \(w\) must be 3/2.
The length of the rectangle is then \(2(3/2)+3=6\) feet. So the rectangle is 6 ft by 3/2 ft and the square is 3 ft by 3 ft. (we can see that the areas of both regions is 9 sq ft).
Example \(\PageIndex{21}\)
The product of two consecutive odd integers is 323. Find the integers.
 Solution

\(\begin{array} {ll} \textbf{Step 1. Read }\text{the problem.} & \\ \textbf{Step 2. Identify }\text{what we are looking for.} &\text{We are looking for two consecutive integers.} \\ \textbf{Step 3. Name}\text{ what we are looking for.} &\text{Let } n=\text{ the first integer.} \\ &n+2= \text{ next consecutive odd integer} \\ \begin{array} {l} \textbf{Step 4. Translate }\text{into an equation. Restate the}\hspace{20mm} \\ \text{problem in a sentence.} \end{array} &\begin{array} {l} \text{The product of the two consecutive odd} \\ \text{integers is }323. \end{array} \\ &\quad n(n+2)=323 \\ \textbf{Step 5. Solve }\text{the equation.} n^2+2n=323 \\ \text{Bring all the terms to one side.} &n^2+2n−323=0 \\ \text{Factor the trinomial.} &(n−17)(n+19)=0 \\ \begin{array} {l} \text{Use the Zero Product Property.} \\ \text{Solve the equations.} \end{array} &\begin{array} {ll} n−17=0 \hspace{10mm}&n+19=0 \\ n=17 &n=−19 \end{array} \end{array} \)
There are two values for \(n\) that are solutions to this problem. So there are two sets of consecutive odd integers that will work.\(\begin{array} {ll} \text{If the first integer is } n=17 \hspace{60mm} &\text{If the first integer is } n=19 \\ \text{then the next odd integer is} &\text{then the next odd integer is} \\ \hspace{53mm} n+2 &\hspace{53mm} n+2 \\ \hspace{51mm} 17+2 &\hspace{51mm} 19+2 \\ \hspace{55mm} 19 &\hspace{55mm} 17 \\ \hspace{51mm} 17,19 &\hspace{51mm} 17,19 \\ \textbf{Step 6. Check }\text{the answer.} & \\ \text{The results are consecutive odd integers} & \\ \begin{array} {ll} 17,\space 19\text{ and }−19,\space −17. & \\ 17·19=323\checkmark &−19(−17)=323\checkmark \end{array} & \\ \text{Both pairs of consecutive integers are solutions.} & \\ \textbf{Step 7. Answer }\text{the question} &\text{The consecutive integers are }17, 19\text{ and }−19,−17. \end{array} \)
Try It \(\PageIndex{22}\)
The product of two consecutive odd integers is 255. Find the integers.
 Answer

\(−15,−17\) and \(15, 17\)
Try It \(\PageIndex{23}\)
The product of two consecutive odd integers is 483 Find the integers.
 Answer

\(−23,−21\) and \(21, 23\)
Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give positive results.
In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.
Example \(\PageIndex{24}\)
A rectangular bedroom has an area 117 square feet. The length of the bedroom is four feet more than the width. Find the length and width of the bedroom.
 Solution

Step 1. Read the problem. In problems involving
geometric figures, a sketch can help you visualize
the situation.Step 2. Identify what you are looking for. We are looking for the length and width. Step 3. Name what you are looking for. Let \(w=\text{ the width of the bedroom}\). The length is four feet more than the width. \(w+4=\text{ the length of the garden}\) Step 4. Translate into an equation. Restate the important information in a sentence. The area of the bedroom is 117 square feet. Use the formula for the area of a rectangle. \(A=l·w\) Substitute in the variables. \(117=(w+4)w\) Step 5. Solve the equation Distribute first. \(117=w^2+4w\) Get zero on one side. \(117=w^2+4w\) Factor the trinomial. \(0=w^2+4w−117\) Use the Zero Product Property. \(0=(w^2+13)(w−9)\) Solve each equation. \(0=w+13\quad 0=w−9\) Since \(w\) is the width of the bedroom, it does not
make sense for it to be negative. We eliminate that value for \(w\).\(\cancel{w=−13}\) \(\quad w=9\) \(w=9\) The width is 9 feet. Find the value of the length. \(w+4\)
\(9+4\)
13 The length is 13 feet.Step 6. Check the answer.
Does the answer make sense?
Yes, this makes sense.Step 7. Answer the question. The width of the bedroom is 9 feet and
the length is 13 feet.
Try It \(\PageIndex{25}\)
A rectangular sign has area 30 square feet. The length of the sign is one foot more than the width. Find the length and width of the sign.
 Answer

The width is 5 feet and length is 6 feet.
Try It \(\PageIndex{26}\)
A rectangular patio has area 180 square feet. The width of the patio is three feet less than the length. Find the length and width of the patio.
 Answer

The length of the patio is 12 feet and the width 15 feet.
In the next example, we will use the Pythagorean Theorem \((a^2+b^2=c^2)\). This formula gives the relation between the legs and the hypotenuse of a right triangle.
We will use this formula to in the next example.
Example \(\PageIndex{27}\)
A boat’s sail is in the shape of a right triangle as shown. The hypotenuse will be 17 feet long. The length of one side will be 7 feet less than the length of the other side. Find the lengths of the sides of the sail.
 Solution

Step 1. Read the problem Step 2. Identify what you are looking for. We are looking for the lengths of the
sides of the sail.Step 3. Name what you are looking for.
One side is 7 less than the other.Let \(x=\text{ length of a side of the sail}\).
\(x−7=\text{ length of other side}\)Step 4. Translate into an equation. Since this is a
right triangle we can use the Pythagorean Theorem.\(a^2+b^2=c^2\) Substitute in the variables. \(x^2+(x−7)^2=17^2\) Step 5. Solve the equation
Simplify.\(x^2+x^2−14x+49=289\) \(2x^2−14x+49=289\) It is a quadratic equation, so get zero on one side. \(2x^2−14x−240=0\) Factor the greatest common factor. \(2(x^2−7x−120)=0\) Factor the trinomial. \(2(x−15)(x+8)=0\) Use the Zero Product Property. \(2\neq 0\quad x−15=0\quad x+8=0\) Solve. \(2\neq 0\quad x=15\quad x=−8\) Since \(x\) is a side of the triangle, \(x=−8\) does not
make sense.\(2\neq 0\quad x=15\quad \cancel{x=−8}\) Find the length of the other side. If the length of one side is
then the length of the other side is
8 is the length of the other side.Step 6. Check the answer in the problem
Do these numbers make sense?Step 7. Answer the question The sides of the sail are 8, 15 and 17 feet.
Try It \(\PageIndex{28}\)
Justine wants to put a deck in the corner of her backyard in the shape of a right triangle. The length of one side of the deck is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the deck.
 Answer

5 feet and 12 feet
Try It \(\PageIndex{29}\)
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of the other leg. Find the lengths of the hypotenuse and the other leg.
 Answer

24 feet and 25 feet
Writing Exercises \(\PageIndex{30}\)
 What is the Zero Product Property?
 Is there an analogue of the Zero Product Propertyif we replace 0 with 1? Give an example supporting your answer.
 What is the goal of solving a quadratic equation?
 Must we check our answers even if we know we didn’t make a mistake?
Exit Problem \(\PageIndex{31}\)
Solve the equation \(x(x+1)=−x+35\) by using the Zero Product Property.
Key Concepts
 Polynomial Equation: A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.
 Quadratic Equation: An equation of the form \(ax^2+bx+c=0\), where \(a,b,c,\) are real numbers and \(a\neq 0\nonumber\), is called a quadratic equation.
 Zero Product Property: If \(a·b=0\), then either \(a=0\) or \(b=0\) or both.
 How to use the Zero Product Property
 Set each factor equal to zero.
 Solve the linear equations.
 Check.
 How to solve a quadratic equation by factoring.
 Write the quadratic equation in standard form, \(ax^2+bx+c=0\).
 Factor the quadratic expression.
 Use the Zero Product Property.
 Solve the linear equations.
 Check. Substitute each solution separately into the original equation.
 How to use a problem solving strategy to solve word problems.
 Read the problem. Make sure all the words and ideas are understood.
 Identify what we are looking for.
 Name what we are looking for. Choose a variable to represent that quantity.
 Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
 Solve the equation using appropriate algebra techniques.
 Check the answer in the problem and make sure it makes sense.
 Answer the question with a complete sentence.