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Quadratic Equations In One Variable with Applications

Multiply binomials.

Learning Objectives

After completing this section, you should be able to:

Multiply binomials.
Factor trinomials.
Solve quadratic equations by graphing.
Solve quadratic equations by factoring.
Solve quadratic equations using square root method.
Solve quadratic equations using the quadratic formula.
Solve real world applications modeled by quadratic equations.

Identify Polynomials, Monomials, Binomials and Trinomials

You have learned that a term is a constant, or the product of a constant and one or more variables. When it is of the form \(a{x}^{m}\), where \(a\) is a constant and \({x}^{m}\) is a positive whole number, it is called a monomial. Some examples of monomial are 8, \(-2{x}^{2}\), \(4{y}^{3}\), and \(11z\).

A monomial or two or more monomials combined by addition or subtraction is a polynomial. Some examples include: \(b+11\), \(4{y}^{2}-7y+2\), and \(4{x}^{4}+{x}^{3}+8{x}^{2}-9x+1\). Some polynomials have special names, based on the number of terms. A monomial is a polynomial with exactly one term (examples: 14, \(8{y}^{2}\), \(-9{x}^{3}{y}^{5}\), and \(-13\)). A binomial has exactly two terms (examples: \(a+\ 7\), \(4b-5\), \({y}^{2}-16\), and \(3{x}^{3}-9{x}^{2}\)), and a trinomial has exactly three terms (examples: \({x}^{2}-7x+12\), \(9{y}^{2}+2y-8\), \(6{m}^{4}-{m}^{3}+8m\), and \({x}^{4}+3{x}^{2}-1\)).

Notice that every monomial, binomial, and trinomial is also a polynomial. They are just special members of the “family” of polynomials and so they have special names. We use the words monomial, binomial, and trinomial when referring to these special polynomials and just call all the rest polynomials.

Multiply Binomials

Recall multiplying algebraic expressions from Algebraic Expressions. In this section, we will continue that work and multiply binomials as well. We can use an area model to do multiplication.

Multiply Binomials

Try it.

Multiply \((x+2)(x+3)\).

Solution

Step 1: Use the distributive property: \[(x)(x)+(x)(3)+(2)(x)+(2)(3)\]

In the area model () multiply each term on the side by each term on the top (think of it as a multiplication table).

Step 2: After we multiply, we get the following equation: \[{x}^{2}+3x+2x+6\]

Step 3: Combine the like terms to arrive at: \[{x}^{2}+5x+6\]

Multiplying More Complex Binomials

Try it.

Multiply \((2x+7)(3x-5)\).

Solution

Step 1: Use the Distributive Property: \[(2x)(3x)-(2x)(5)+(7)(3x)-(7)(5)\]

Step 2: After multiplying, get the following equation: \[6{x}^{2}-10x+21x-35\]

Step 3: Combine the like terms to arrive at: \[6{x}^{2}+11x-35\]

Factoring Trinomials

We’ve just covered how to multiply binomials. Now you will need to “undo” this multiplication—to start with the product and end up with the factors. Let us review an example of multiplying binomials to refresh your memory.

\[(x+2)(x+3)={x}^{2}+5x+6\]

To factor the trinomial means to start with the product, \({x}^{2}+5x+6\), and end with the factors, \((x+2)(x+3)\). You need to think about where each of the terms in the trinomial came from. The first term came from multiplying the first term in each binomial. So, to get \({x}^{2}\) in the product, each binomial must start with an \(x\).

\[\begin{array}{l}{x}^{2}+5x+6 \\ (x)(x)\end{array}\]

The last term in the trinomial came from multiplying the last term in each binomial. So, the last terms must multiply to 6. What two numbers multiply to 6? The factors of 6 could be 1 and 6, or 2 and 3. How do you know which pair to use? Consider the middle term. It came from adding the outer and inner terms. So the numbers that must have a product of 6 will need a sum of 5.

We’ll test both possibilities and summarize the results in the following table, which will be very helpful when you work with numbers that can be factored in many different ways.

Factors of 6	Sum of Factors
1, 6	\[1+6=7\]
2, 3	\[2+3=5\]

We see that 2 and 3 are the numbers that multiply to 6 and add to 5. We have the factors of \({x}^{2}+5x+6\). They are \((x+2)(x+3)\).

\[\begin{array}{ll}{x}^{2}+5x+6 & \text{product} \\ (x+2)(x+3) & \text{factors}\end{array}\]

You can check if the factors are correct by multiplying. Looking back, we started with \({x}^{2}+5x+6\), which is of the form \({x}^{2}+bx+c\), where \(b=5\) and \(c=6\). We factored it into two binomials of the form \((x+m)\) and \((x+n)\).

\[\begin{array}{ll}{x}^{2}+5x+6 & {x}^{2}+bx+c \\ (x+2)(x+3) & (x+m)(x+n)\end{array}\]

To get the correct factors, we found two number \(m\) and \(n\) whose product is \(c\) and sum is \(b\). With the area model (), start with an empty box and then put in the \({x}^{2}\) term and \(c\).

Factoring Trinomials

Try it.

Factor \({x}^{2}+7x+12\).

Solution

The numbers that must have a product of 12 will need a sum of 7. We will summarize the results in a table below.

Factors of 12	Sum of Factors
1, 12	\(1+12=13\)
2, 6	\(2+6=8\)
3, 4	\(3+4=7\)

We see that 3 and 4 are the numbers that multiply to 12 and add to 7. The factors of \({x}^{2}+7x+12\) are \((x+3)(x+4)\).

Condensed — the full section is in OpenStax Contemporary Mathematics.

Solving Quadratic Equations by Graphing

We have already solved and graphed linear equations in Graphing Linear Equations and Inequalities, equations of the form \(Ax+By=C\). In linear equations, the variables have no exponents. Quadratic equations are equations in which the variable is squared. The following are some examples of quadratic equations:

\[\begin{array}{llll}{x}^{2}+5x+6=0 & 3{y}^{2}+4y=106 & 4{u}^{2}-81=0 & n(n+1)=42\end{array}\]

The last equation does not appear to have the variable squared, but when we simplify the expression on the left, we will get \({n}^{2}+n\). The general form of a quadratic equation is \(a{x}^{2}+bx+c=0\), where \(a,b,\) and \(c\) are real numbers, with \(a\ne 0\). Remember that a solution of an equation is a value of a variable that makes a true statement when substituted into the equation. The solutions of quadratic equations are the values of the variables that make the quadratic equation \(a{x}^{2}+bx+c=0\) true.

To solve quadratic equations, we need methods different than the ones we used in solving linear equations. We will start by solving a quadratic equation from its graph. Just like we started graphing linear equations by plotting points, we will do the same for quadratic equations. Let us look first at graphing the quadratic equation \(y={x}^{2}\). We will choose integer values of \(x\) between \(-2\) and 2 and find their \(y\) values, as shown in the table below.

\(\begin{array}{l}y={x}^{2} \\ \end{array}\)
\(x\)	\(y\)
0	0
1	1
\(-1\)	1
2	4
\(-2\)	4

Notice when we let \(x=1\) and \(x=-1\), we got the same value for \(y\).

\[\begin{array}{lll}y={x}^{2} & y={1}^{2} & y=1 \\ y={x}^{2} & y={(-1)}^{2} & y=1\end{array}\]

The same thing happened when we let \(x=2\) and \(x=-2\). Now, we will plot the points to show the graph of \(y={x}^{2}\). See .

The graph is not a line. This figure is called a parabola. Every quadratic equation has a graph that looks like this. When \(y=0\) the solution to the quadratic \(y={x}^{2}\) is 0 because \({x}^{2}=0\) at \(x=0\).

Graphing a Quadratic Equation

Try it.

Graph \(y={x}^{2}-1\) and list the solutions to the quadratic equation.

Solution

We will graph the equation by plotting points.

Step 1: Choose integer values for \(x\), substitute them into the equation, and solve for \(y\).

Step 2: Record the values of the ordered pairs in the chart.

\(y={x}^{2}-1\)
\(x\)	\(y\)
0	\(-1\)
1	0
\(-1\)	0
2	3
\(-2\)	3

Step 3: Plot the points and then connect them with a smooth curve. The result will be the graph of the equation \(y={x}^{2}-1\) (). The solutions are \(x=1\) and \(x=-1\).

Condensed — the full section is in OpenStax Contemporary Mathematics.

Solving Quadratic Equations by Factoring

Another way of solving quadratic equations is by factoring. We will use the Zero Product Property that says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

Solving a Quadratic Equation by Factoring

Try it.

Solve \((x+1)(x-4)=0\).

Solution

Step 1. Set each factor equal to zero.	The product equals zero, so at least one factor must equal zero.	\(\begin{array}{l}(x+1)(x-4)=0 \\ x+1=0\ \text{or}\ x-4=0\end{array}\)
Step 2. Solve the linear equations.	Solve each equation.	\(x=-1\ \text{or}\ x=4\)
Step 3. Check.	Substitute each solution separately into the original equation.	\(\begin{array}{lll}x & = & -1 \\ (x+1)(x-4) & = & 0 \\ (-1+1)(-1-4) & \overset{?}{=} & 0 \\ (0)(-5) & \overset{?}{=} & 0 \\ 0 & = & 0✓\end{array}\)
		\(\begin{array}{lll}x & = & 4 \\ (x+1)(x-4) & = & 0 \\ (4+1)(4-4) & \overset{?}{=} & 0 \\ (5)(0) & \overset{?}{=} & 0 \\ 0 & = & 0✓\end{array}\)

Solve Another Quadratic Equation by Factoring

Try it.

Solve \({x}^{2}+2x-8=0\).

Solution

Step 1. Write the quadratic equation in standard form, \({\text{ax}}^{2}+\text{bx}+c=0\).	The equation is already in standard form.	\({x}^{2}+2x-8=0\)
Step 2. Factor the quadratic expression.	Factor \(\begin{array}{l}{x}^{2}+2x-8 \\ (x+4)(x-2)\end{array}\)	\((x+4)(x-2)=0\)
Step 3. Use the Zero Product Property.	Set each factor equal to zero.	\(x+4=0\ \text{or}\ x-2=0\)
Step 4. Solve the linear equations.	We have two linear equations.	\(x=-4\ \text{or}\ x=2\)
Step 5. Check.	Substitute each solution separately into the original equation.	\(\begin{array}{lll}{x}^{2}+2x-8 & = & 0 \\ x & = & -4 \\ {(-4)}^{2}-2(-4)-8 & \overset{?}{=} & 0 \\ 16+(-8)-8 & \overset{?}{=} & 0 \\ 0 & = & 0✓\end{array}\)
		\(\begin{array}{lll}{x}^{2}+2x-8 & = & 0 \\ x & = & 2 \\ {2}^{2}-2(2)-8 & \overset{?}{=} & 0 \\ 4+4-8 & \overset{?}{=} & 0 \\ 0 & = & 0✓\end{array}\)

Condensed — the full section is in OpenStax Contemporary Mathematics.

Solving Quadratic Equations Using the Square Root Property

We just solved some quadratic equations by factoring. Let us use factoring to solve the quadratic equation \({x}^{2}=9\).

Step 1: Put the equation in standard form.	\({x}^{2}-9=0\)
Step 2: Factor the left side.	\((x+3)(x-3)=0\)
Step 3: Use the Zero Product Property.	\(x+3\ =0\)	\(x-3=0\)
Step 4: Solve each equation.	\(x=-3\)	\(x=3\)
Step 5: Combine the two solutions into \(\pm\)	\(x=\pm 3\)

The solution is read as “\(x\) is equal to positive or negative 3.”

What happens when we have an equation like \({x}^{2}=7\)? Since 7 is not a perfect square, we cannot solve the equation by factoring. These equations are all of the form \({x}^{2}=k\). We define the square root of a number in this way: If \({n}^{2}=m\), then \(n\) is a square root of \(m\). This leads to the Square Root Property.

Using the Square Root Property to Solve a Quadratic Equation

Try it.

Solve using the square Root Property: \({x}^{2}=169\).

Solution

Step 1: Use the Square Root Property.	\(x=\pm \sqrt{169}\)
Step 2: Simplify the radical.	\(x=\pm 13\)
Step 3: Rewrite to show the two solutions.	\(x=13\)	\(x=-13\)

Using the Square Root Property to Solve Another Quadratic Equation

Try it.

Solve using the Square Root Property: \(144{q}^{2}=25\).

Solution

Step 1: Solve for \(q\).

\({q}^{2}=\frac{25}{144}\)

Step 2: Use the Square Root Property.

\(q=\pm \sqrt{\frac{25}{144}}\)

Step 3: Simplify the radical. \[q=\pm \frac{5}{12}\]

Step 4: Rewrite to show the two solutions.
\(q=\frac{5}{12}\), \(q=-\frac{5}{12}\)

Solving Quadratic Equations Using the Quadratic Formula

This last method we will look at for solving quadratic equations is the quadratic formula. This method works for all quadratic equations, even the quadratic equations we could not factor! To use the quadratic formula, we substitute the values of \(a\), \(b\), and \(c\) into the expression on the right side of the formula. Then, we do all the math to simplify the expression. The result gives the solution(s) to the quadratic equation.

Solving a Quadratic Equation Using the Quadratic Formula

Try it.

Solve using the quadratic formula: \({x}^{2}-6x+5=0\).

Solution

\({x}^{2}-6x+5=0\)

This equation is in standard form.

\(\begin{array}{lll}a{x}^{2}+bx+c & = & 0 \\ {x}^{2}-6x+5 & = & 0\end{array}\)

Step 1: Identify the \(a\), \(b\), and \(c\) values.

\(a=1,b=-6,c=5\)

Step 2: Write the quadratic formula.

\(x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\)

Step 3: Substitute in the values of \(a\), \(b\), \(c\).

\(x=\frac{(-6)\pm \sqrt{{(-6)}^{2}-4(1)(5)}}{2.1}\)

Step 4: Simplify.

\(x=\frac{6\pm \sqrt{36-20}}{2}\)

\(x=\frac{6\pm \sqrt{16}}{2}\)

\(x=\frac{6\pm 4}{2}\)

Step 5: Rewrite to show two solutions.

\(x=\frac{6+4}{2},x=\frac{6-4}{2}\)

Step 6: Simplify.

\(x=\frac{10}{2},x=\frac{2}{2}\)

\(x=5,x=1\)

Step 7: Check.

\(\begin{array}{llllll}{x}^{2}-6x+5 & = & 0 & {x}^{2}-6x+5 & = & 0 \\ {5}^{2}-6⋅5+5 & \overset{?}{=} & 0 & {1}^{2}-6⋅1+5 & \overset{?}{=} & 0 \\ 25-30+5 & \overset{?}{=} & 0 & 1-6+5 & \overset{?}{=} & 0 \\ 0 & = & 0✓ & 0 & = & 0✓\end{array}\)

Condensed — the full section is in OpenStax Contemporary Mathematics.

Solving Real-World Applications Modeled by Quadratic Equations

There are problem solving strategies that will work well for applications that translate to quadratic equations. Here’s a problem-solving strategy to solve word problems:

Step 1: Read the problem. Make sure all the words and ideas are understood.

Step 2: Identify what we are looking for.

Step 3: Name what we are looking for. Choose a variable to represent that quantity.

Step 4: 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 algebra equation.

Step 5: Solve the equation using good algebra techniques.

Step 6: Check the answer in the problem and make sure it makes sense.

Finding Consecutive Integers

Try it.

The product of two consecutive integers is 132. Find the integers.

Solution

Step 1: Read the problem.

Step 2: Identify what we are looking for.

We are looking for two consecutive integers.

Step 3: Name what we are looking for.

Let \(n=\) the first integer

Let \(n+1=\) the next consecutive integer.

Step 4: Translate into an equation. Restate the problem in a sentence.

\(n(n+1)=132\)

The product of the two consecutive integers is 132. The first integer times the next integer is 132.

Step 5: Solve the equation.

\({n}^{2}+n=132\)

Bring all the terms to one side.

\({n}^{2}+n-132=0\)

Factor the trinomial.

\((n-11)(n+12)=0\)

Use the zero product property.

\(n-11=0\ \text{or}\ n+12=0\)

Solve the equations.

\(n=11,n=-12\)

There are two values for \(n\) that are solutions to this problem. So, there are two sets of consecutive integers that will work.

If the first integer is \(n=11\), then the next integer is 12. If the first integer is \(n=-12\), then the next integer is \(-11\).

Step 6: Check the answer.

The consecutive integers are 11, 12 and \(-11\), \(-12\). The product of 11 and \(12=132\) and the product of \(-11(-12)=132\). Both pairs of consecutive integers are solutions.

Step 7: Answer the question.

The consecutive integers are 11, 12, and \(-11\), \(-12\).

Condensed — the full section is in OpenStax Contemporary Mathematics.

Key Concepts

A quadratic equation is an algebraic equation where the highest power (degree) of the equation is two.
To solve a quadratic equation is to find the value(s) that when substituted in for the variables, will make the equation equal to zero.
There can be two, one, or no solutions to any quadratic equation.
There are several methods to solve a quadratic equation. These methods include factoring quadratic equations, graphic quadratic equations, using the square root method, and using the quadratic formula.

Videos

Factoring with the Box Method (Area Model)
Solving Quadratics with the Zero Property
Solving Quadratics with the Quadratic Formula

Practice (14)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Multiply \((x+2)(x+3)\).

Atskleisti atsakymą

Step 1: Use the distributive property: \[(x)(x)+(x)(3)+(2)(x)+(2)(3)\]
In the area model () multiply each term on the side by each term on the top (think of it as a multiplication table).
Step 2: After we multiply, we get the following equation: \[{x}^{2}+3x+2x+6\]
Step 3: Combine the like terms to arrive at: \[{x}^{2}+5x+6\]
Multiply \((2x+7)(3x-5)\).

Atskleisti atsakymą

Step 1: Use the Distributive Property: \[(2x)(3x)-(2x)(5)+(7)(3x)-(7)(5)\]
Step 2: After multiplying, get the following equation: \[6{x}^{2}-10x+21x-35\]
Step 3: Combine the like terms to arrive at: \[6{x}^{2}+11x-35\]
Factor \({x}^{2}+7x+12\).

Atskleisti atsakymą

The numbers that must have a product of 12 will need a sum of 7. We will summarize the results in a table below.
Factors of 12 Sum of Factors
1, 12 \(1+12=13\)
2, 6 \(2+6=8\)
3, 4 \(3+4=7\)
We see that 3 and 4 are the numbers that multiply to 12 and add to 7. The factors of \({x}^{2}+7x+12\) are \((x+3)(x+4)\).
Factor \({x}^{2}-11x+28\).

Atskleisti atsakymą

The numbers that must have a product of 28 will need a sum of \(-11\). We will summarize the results in a table.
Factors of 28 Sum of Factors
\(1,28\) \(1+28=29\)
\(2,14\) \[2+14=16\]
\(4,7\) \[4+7=11\]
We see that 4 and 7 are the numbers that multiply to 28 and add to 11. But we needed \(-11\), so we will need to use \(-4\) and \(-7\) because \((-4)(-7)=28\) and \((-4)+(-7)=-11\). The factors of \({x}^{2}-11x+28\) are \((x-4)(x-7)\).
Graph \(y={x}^{2}-1\) and list the solutions to the quadratic equation.

Atskleisti atsakymą

We will graph the equation by plotting points.
Step 1: Choose integer values for \(x\), substitute them into the equation, and solve for \(y\).
Step 2: Record the values of the ordered pairs in the chart.
\(y={x}^{2}-1\)
\(x\) \(y\)
0 \(-1\)
1 0
\(-1\) 0
2 3
\(-2\) 3
Step 3: Plot the points and then connect them with a smooth curve. The result will be the graph of the equation \(y={x}^{2}-1\) (). The solutions are \(x=1\) and \(x=-1\).
Find the solutions of \(y={x}^{2}+5x+4\) from its graph ().

Atskleisti atsakymą

The solutions of a quadratic equations are the values of \(x\) that make the equation a true statement when set equal to zero (i.e. when \(y=0\)). \({x}^{2}+5x+4=0\) at \(x=-4\) and \(x=-1\).

Solve \((x+1)(x-4)=0\).

Atskleisti atsakymą

Step 1. Set each factor equal to zero.	The product equals zero, so at least one factor must equal zero.	\(\begin{array}{l}(x+1)(x-4)=0 \\ x+1=0\ \text{or}\ x-4=0\end{array}\)
Step 2. Solve the linear equations.	Solve each equation.	\(x=-1\ \text{or}\ x=4\)
Step 3. Check.	Substitute each solution separately into the original equation.	\(\begin{array}{lll}x & = & -1 \\ (x+1)(x-4) & = & 0 \\ (-1+1)(-1-4) & \overset{?}{=} & 0 \\ (0)(-5) & \overset{?}{=} & 0 \\ 0 & = & 0✓\end{array}\)
		\(\begin{array}{lll}x & = & 4 \\ (x+1)(x-4) & = & 0 \\ (4+1)(4-4) & \overset{?}{=} & 0 \\ (5)(0) & \overset{?}{=} & 0 \\ 0 & = & 0✓\end{array}\)

Solve \({x}^{2}+2x-8=0\).

Atskleisti atsakymą

Step 1. Write the quadratic equation in standard form, \({\text{ax}}^{2}+\text{bx}+c=0\).	The equation is already in standard form.	\({x}^{2}+2x-8=0\)
Step 2. Factor the quadratic expression.	Factor \(\begin{array}{l}{x}^{2}+2x-8 \\ (x+4)(x-2)\end{array}\)	\((x+4)(x-2)=0\)
Step 3. Use the Zero Product Property.	Set each factor equal to zero.	\(x+4=0\ \text{or}\ x-2=0\)
Step 4. Solve the linear equations.	We have two linear equations.	\(x=-4\ \text{or}\ x=2\)
Step 5. Check.	Substitute each solution separately into the original equation.	\(\begin{array}{lll}{x}^{2}+2x-8 & = & 0 \\ x & = & -4 \\ {(-4)}^{2}-2(-4)-8 & \overset{?}{=} & 0 \\ 16+(-8)-8 & \overset{?}{=} & 0 \\ 0 & = & 0✓\end{array}\)
		\(\begin{array}{lll}{x}^{2}+2x-8 & = & 0 \\ x & = & 2 \\ {2}^{2}-2(2)-8 & \overset{?}{=} & 0 \\ 4+4-8 & \overset{?}{=} & 0 \\ 0 & = & 0✓\end{array}\)

Solve using the square Root Property: \({x}^{2}=169\).

Atskleisti atsakymą

Step 1: Use the Square Root Property. \(x=\pm \sqrt{169}\)
Step 2: Simplify the radical. \(x=\pm 13\)
Step 3: Rewrite to show the two solutions. \(x=13\) \(x=-13\)
Solve using the Square Root Property: \(144{q}^{2}=25\).

Atskleisti atsakymą

Step 1: Solve for \(q\).
\({q}^{2}=\frac{25}{144}\)
Step 2: Use the Square Root Property.
\(q=\pm \sqrt{\frac{25}{144}}\)
Step 3: Simplify the radical. \[q=\pm \frac{5}{12}\]
Step 4: Rewrite to show the two solutions.
\(q=\frac{5}{12}\), \(q=-\frac{5}{12}\)
Solve using the quadratic formula: \({x}^{2}-6x+5=0\).

Atskleisti atsakymą

\({x}^{2}-6x+5=0\)
This equation is in standard form.
\(\begin{array}{lll}a{x}^{2}+bx+c & = & 0 \\ {x}^{2}-6x+5 & = & 0\end{array}\)
Step 1: Identify the \(a\), \(b\), and \(c\) values.
\(a=1,b=-6,c=5\)
Step 2: Write the quadratic formula.
\(x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\)
Step 3: Substitute in the values of \(a\), \(b\), \(c\).
\(x=\frac{(-6)\pm \sqrt{{(-6)}^{2}-4(1)(5)}}{2.1}\)
Step 4: Simplify.
\(x=\frac{6\pm \sqrt{36-20}}{2}\)
\(x=\frac{6\pm \sqrt{16}}{2}\)
\(x=\frac{6\pm 4}{2}\)
Step 5: Rewrite to show two solutions.
\(x=\frac{6+4}{2},x=\frac{6-4}{2}\)
Step 6: Simplify.
\(x=\frac{10}{2},x=\frac{2}{2}\)
\(x=5,x=1\)
Step 7: Check.
\(\begin{array}{llllll}{x}^{2}-6x+5 & = & 0 & {x}^{2}-6x+5 & = & 0 \\ {5}^{2}-6⋅5+5 & \overset{?}{=} & 0 & {1}^{2}-6⋅1+5 & \overset{?}{=} & 0 \\ 25-30+5 & \overset{?}{=} & 0 & 1-6+5 & \overset{?}{=} & 0 \\ 0 & = & 0✓ & 0 & = & 0✓\end{array}\)

Solve using the quadratic formula: \(2{x}^{2}+9x-5=0\).

Atskleisti atsakymą

Step 1. Write the quadratic equation in standard form. Identify the \(a\), \(b\), \(c\) values.	This equation is in standard form.	\(\begin{array}{l}{\text{ax}}^{2}+\text{bx}+c=0 \\ 2{x}^{2}+9x-5=0 \\ a=2,\ b=9,\ c=-5\end{array}\)
Step 2. Write the quadratic formula. Then substitute in the values of \(a\), \(b\), \(c\).	Substitute in \(a=2\), \(b=9\), \(c=-5\)	\(\begin{array}{l}x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a} \\ x=\frac{-9\pm \sqrt{{9}^{2}-4(2)(-5)}}{2⋅2}\end{array}\)
Step 3. Simplify the fraction, and solve for \(x\).		\(\begin{array}{l}x=\frac{-9\pm \sqrt{81-(-40)}}{4} \\ x=\frac{-9\pm \sqrt{121}}{4} \\ x=\frac{-9\pm 11}{4} \\ x=\frac{-9+11}{4}\ x=\frac{-9-11}{4} \\ x=\frac{2}{4}\ x=\frac{-20}{4} \\ x=\frac{1}{2}\ x=-5\end{array}\)
Step 4. Check the solutions.	Put each answer in the original equation to check. Substitute \(x=\frac{1}{2}\).	\(\begin{array}{lll}2{x}^{2}+9x-5 & = & 0 \\ 2{(\frac{1}{2})}^{2}+9⋅\frac{1}{2}-5 & \overset{?}{=} & 0 \\ 2⋅\frac{1}{4}+9⋅\frac{1}{2}-5 & \overset{?}{=} & 0 \\ \frac{1}{2}+\frac{9}{2}-5 & \overset{?}{=} & 0 \\ \frac{10}{2}-5 & \overset{?}{=} & 0 \\ 5-5 & \overset{?}{=} & 0 \\ 0 & = & 0✓\end{array}\)
	Substitute \(x=-5\).	\(\begin{array}{lll}2{x}^{2}+9x-5 & = & 0 \\ 2{(-5)}^{2}+9(-5)-5 & \overset{?}{=} & 0 \\ 2⋅25-45-5 & \overset{?}{=} & 0 \\ 50-45-5 & \overset{?}{=} & 0 \\ 0 & = & 0✓\end{array}\)

The product of two consecutive integers is 132. Find the integers.

Atskleisti atsakymą

Step 1: Read the problem.
Step 2: Identify what we are looking for.
We are looking for two consecutive integers.
Step 3: Name what we are looking for.
Let \(n=\) the first integer
Let \(n+1=\) the next consecutive integer.
Step 4: Translate into an equation. Restate the problem in a sentence.
\(n(n+1)=132\)
The product of the two consecutive integers is 132. The first integer times the next integer is 132.
Step 5: Solve the equation.
\({n}^{2}+n=132\)
Bring all the terms to one side.
\({n}^{2}+n-132=0\)
Factor the trinomial.
\((n-11)(n+12)=0\)
Use the zero product property.
\(n-11=0\ \text{or}\ n+12=0\)
Solve the equations.
\(n=11,n=-12\)
There are two values for \(n\) that are solutions to this problem. So, there are two sets of consecutive integers that will work.
If the first integer is \(n=11\), then the next integer is 12. If the first integer is \(n=-12\), then the next integer is \(-11\).
Step 6: Check the answer.
The consecutive integers are 11, 12 and \(-11\), \(-12\). The product of 11 and \(12=132\) and the product of \(-11(-12)=132\). Both pairs of consecutive integers are solutions.
Step 7: Answer the question.
The consecutive integers are 11, 12, and \(-11\), \(-12\).
A rectangular garden has an area 15 square feet. The length of the garden is 2 feet more than the width. Find the length and width of the garden.

Atskleisti atsakymą

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. The length is 2 feet more than width.
Let \(W=\) the width of the garden.
\(W+2=\) the length of the garden.
Step 4: Translate into an equation.
Restate the important information in a sentence.
The area of the rectangular garden is 15 square feet.
Use the formula for the area of a rectangle.
\(A=L\times W\)
Substitute in the variables.
\(15=(W+2)W\)
Step 5: Solve the equation. Distribute first.
\(15={W}^{2}+2W\)
Get zero on one side.
\(0={W}^{2}+2W-15\)
Factor the trinomial.
\(0=(W+5)(W-3)\)
Use the Zero Product Property.
\(0=W+5\)
\(0=W-3\)
Solve each equation.
\(-5=W\)
\(3=W\)
Since \(W\) is the width of the garden, it does not make sense for it to be negative. We eliminate that value for \(W\).
\(W=3\)
Width is 3 feet.
Find the value of the length.
\(W+2=\) length.
\(3+2\)
5
Length is 5 feet.
Step 6: Check the answer ().
Does the answer make sense?

Yes, this makes sense.
Step 7: Answer the question.
The width of the garden is 3 feet and the length is 5 feet.

Symbols used here

\neq

not equal
The two sides are different.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Quadratic Equations In One Variable with Applications

Multiply binomials.
Factor trinomials.
Solve quadratic equations by graphing.
Solve quadratic equations by factoring.
Solve quadratic equations using square root method.
Solve quadratic equations using the quadratic formula.
Solve real world applications modeled by quadratic equations.
monomial

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Pabandyk savo pačių

Parts of this page are adapted from OpenStax Contemporary Mathematics (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Daugiau informacijos Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

Factors of 28	Sum of Factors
\(1,28\)	\(1+28=29\)
\(2,14\)	\[2+14=16\]
\(4,7\)	\[4+7=11\]