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Quadratic equations
Factoring, the discriminant and the quadratic formula.
A quadratic ax² + bx + c = 0 has up to two solutions — the points where the parabola crosses the x-axis. First try to factor; if that fails, the discriminant Δ = b² − 4ac tells you how many real roots there are before the quadratic formula tells you what they are.
Solving Quadratic Equations by Factoring
An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as \(2{x}^{2}+3x-1=0\) 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 \(a⋅b=0,\) 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 \((x-2)(x+3)\) by multiplying the two factors together.
\[\begin{array}{lll}(x-2)(x+3) & = & {x}^{2}+3x-2x-6 \\ & = & {x}^{2}+x-6\end{array}\]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, \(a{x}^{2}+bx+c=0,\) where a, b, and c are real numbers, and \(a\ne 0.\) 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.
Condensed — the full section is in OpenStax College Algebra 2e.
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.
Example
Try it.
Solve the quadratic using the square root property: \({x}^{2}=8.\)
Solution
Take the square root of both sides, and then simplify the radical. Remember to use a \(\pm\) sign before the radical symbol.
\[\begin{array}{lll}{x}^{2} & = & 8 \\ x & = & \pm \sqrt{8} \\ & = & \pm 2\sqrt{2}\end{array}\]The solutions are \(2\sqrt{2},\) \(-2\sqrt{2}.\)
Example
Try it.
Solve the quadratic equation: \(4{x}^{2}+1=\text{7.}\)
Solution
First, isolate the \({x}^{2}\) term. Then take the square root of both sides.
\[\begin{array}{lll}4{x}^{2}+1 & = & 7 \\ 4{x}^{2} & = & 6 \\ {x}^{2} & = & \frac{6}{4} \\ x & = & \pm \frac{\sqrt{6}}{2}\end{array}\]The solutions are \(\frac{\sqrt{6}}{2},\) and \(-\frac{\sqrt{6}}{2}.\)
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.
\[\begin{array}{lll}\frac{1}{2}(4) & = & 2 \\ {2}^{2} & = & 4\end{array}\]-
Add \({(\frac{1}{2}b)}^{2}\) to both sides of the equal sign and simplify the right side. We have
\[\begin{array}{lll}{x}^{2}+4x+4 & = & -1+4 \\ {x}^{2}+4x+4 & = & 3\end{array}\] The left side of the equation can now be factored as a perfect square.
\[\begin{array}{lll}{x}^{2}+4x+4 & = & 3 \\ {(x+2)}^{2} & = & 3\end{array}\]Use the square root property and solve.
\[\begin{array}{lll}\sqrt{{(x+2)}^{2}} & = & \pm \sqrt{3} \\ x+2 & = & \pm \sqrt{3} \\ x & = & -2\pm \sqrt{3}\end{array}\]The solutions are \(-2+\sqrt{3},\) and \(-2-\sqrt{3}.\)
Example
Try it.
Solve the quadratic equation by completing the square: \({x}^{2}-3x-5=0.\)
Solution
First, move the constant term to the right side of the equal sign.
\[{x}^{2}-3x=5\]Then, take \(\frac{1}{2}\) of the b term and square it.
\[\begin{array}{lll}\frac{1}{2}(-3) & = & -\frac{3}{2} \\ {(-\frac{3}{2})}^{2} & = & \frac{9}{4}\end{array}\]Add the result to both sides of the equal sign.
\[\begin{array}{lll}{x}^{2}-3x+{(-\frac{3}{2})}^{2} & = & 5+{(-\frac{3}{2})}^{2} \\ {x}^{2}-3x+\frac{9}{4} & = & 5+\frac{9}{4}\end{array}\]Factor the left side as a perfect square and simplify the right side.
\[{(x-\frac{3}{2})}^{2}=\frac{29}{4}\]Use the square root property and solve.
\[\begin{array}{lll}\sqrt{{(x-\frac{3}{2})}^{2}} & = & \pm \sqrt{\frac{29}{4}} \\ (x-\frac{3}{2}) & = & \pm \frac{\sqrt{29}}{2} \\ x & = & \frac{3}{2}\pm \frac{\sqrt{29}}{2}\end{array}\]The solutions are \(\frac{3+\sqrt{29}}{2}\) and \(\frac{3-\sqrt{29}}{2}\) .
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 \(a{x}^{2}+bx+c=0,\) \(a\ne 0,\) we will complete the square as follows:
First, move the constant term to the right side of the equal sign:
\[a{x}^{2}+bx=-c\]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 \({(\frac{1}{2}\frac{b}{a})}^{2}=\frac{{b}^{2}}{4{a}^{2}}\) to both sides of the equal sign:
\[{x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}\]Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:
\[{(x+\frac{b}{2a})}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}\]Now, use the square root property, which gives
\[\begin{array}{lll}x+\frac{b}{2a} & = & \pm \sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}} \\ x+\frac{b}{2a} & = & \frac{\pm \sqrt{{b}^{2}-4ac}}{2a}\end{array}\]Finally, add \(-\frac{b}{2a}\) to both sides of the equation and combine the terms on the right side. Thus,
\[x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\]
Example
Try it.
Solve the quadratic equation: \({x}^{2}+5x+1=0.\)
Solution
Identify the coefficients: \(a=1,b=5,c=1.\) Then use the quadratic formula.
\[\begin{array}{lll}x & = & \frac{-(5)\pm \sqrt{{(5)}^{2}-4(1)(1)}}{2(1)} \\ & = & \frac{-5\pm \sqrt{25-4}}{2} \\ & = & \frac{-5\pm \sqrt{21}}{2}\end{array}\]Condensed — the full section is in OpenStax College Algebra 2e.
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. 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 |
Example
Try it.
Use the discriminant to find the nature of the solutions to the following quadratic equations:
- ⓐ \({x}^{2}+4x+4=0\)
- ⓑ \(8{x}^{2}+14x+3=0\)
- ⓒ \(3{x}^{2}-5x-2=0\)
- ⓓ \(3{x}^{2}-10x+15=0\)
Solution
Calculate the discriminant \({b}^{2}-4ac\) for each equation and state the expected type of solutions.
- ⓐ
\({x}^{2}+4x+4=0\)
\({b}^{2}-4ac={(4)}^{2}-4(1)(4)=0.\) There will be one rational double solution.
- ⓑ
\(8{x}^{2}+14x+3=0\)
\({b}^{2}-4ac={(14)}^{2}-4(8)(3)=100.\) As \(100\) is a perfect square, there will be two rational solutions.
- ⓒ
\(3{x}^{2}-5x-2=0\)
\({b}^{2}-4ac={(-5)}^{2}-4(3)(-2)=49.\) As \(49\) is a perfect square, there will be two rational solutions.
- ⓓ
\(3{x}^{2}-10x+15=0\)
\({b}^{2}-4ac={(-10)}^{2}-4(3)(15)=-80.\) There will be two complex solutions.
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 \({90}^{∘}\) angle, and \(c\) refers to the hypotenuse, as shown in .
Example
Try it.
Find the length of the missing side of the right triangle in .
Solution
As we have measurements for side b and the hypotenuse, the missing side is a.
\[\begin{array}{lll}{a}^{2}+{b}^{2} & = & {c}^{2} \\ {a}^{2}+{(4)}^{2} & = & {(12)}^{2} \\ {a}^{2}+16 & = & 144 \\ {a}^{2} & = & 128 \\ a & = & \sqrt{128} \\ & = & 8\sqrt{2}\end{array}\]Key Equations
| quadratic formula | \(x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\) |
| Distance | \(d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\) |
| Slope of a line | \(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\) |
| Point-Slope equation of a line | \(y-{y}_{1}=m\left(x-{x}_{1}\right)\) |
| Adding Complex Numbers | \((a+bi)+(c+di)=(a+c)+(b+d)i\) |
| Subtracting Complex Numbers | \((a+bi)-(c+di)=(a-c)+(b-d)i\) |
| Square Root Property | \(\text{ If }{x}^{2}=k\text{ then }x=\pm \sqrt{k}\) |
| Quadratic Formula | \(\text{ If }a{x}^{2}+bx+c=0\text{ then }x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\) |
| Rational Exponents | \({a}^{\frac{m}{n}}={\left({a}^{\frac{1}{n}}\right)}^{m}={\left({a}^{m}\right)}^{\frac{1}{n}}=\sqrt[n]{{a}^{m}}=(\sqrt[n]{a}{)}^{m}\) |
| Absolute Value | \(\text{ If }x\ge 0,|x|=x\text{; If }x<0,|x|=-x\) |
| Addition Property of Inequalities | \(\text{ If }a |
| Multiplication Property of Inequalities | \(\begin{array}{l}\text{If }a |
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-product property is then used to find solutions. See , , and .
- Many quadratic equations with a leading coefficient other than 1 can be solved by factoring using the grouping method. See and .
- 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. See and .
- Completing the square is a method of solving quadratic equations when the equation cannot be factored. See .
- A highly dependable method for solving quadratic equations is the quadratic formula, based on the coefficients and the constant term in the equation. See and .
- The discriminant is used to indicate the nature of the roots that the quadratic equation will yield: real or complex, rational or irrational, and how many of each. See .
- 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. See .
Solve Quadratic Equations Using the Zero Product Property
We will first solve some quadratic equations by using the Zero Product Property. The Zero Product Property 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.
We will now use the Zero Product Property, to solve a quadratic equation.
How to Use the Zero Product Property to Solve a Quadratic Equation
Try it.
Solve: \((x+1)(x-4)=0\).
Solution
We usually will do a little more work than we did in this last example to solve the linear equations that result from using the Zero Product Property.
Example
Try it.
Solve: \((5n-2)(6n-1)=0\).
Solution
| \((5n-2)(6n-1)=0\) | ||
| Use the Zero Product Property to set each factor to 0. | \(5n-2=0\) | \(6n-1=0\) |
| Solve the equations. | \(n=\frac{2}{5}\) | \(n=\frac{1}{6}\) |
| Check your answers. | ||
Notice when we checked the solutions that each of them made just one factor equal to zero. But the product was zero for both solutions.
Example
Try it.
Solve: \(3p(10p+7)=0\).
Solution
| \(\ 3p(10p+7)=0\) | ||
| Use the Zero Product Property to set each factor to 0. | \(3p=0\) | \(10p+7=0\\) |
| Solve the equations. | \(p=0\) | \(10p=-7\\) |
| \(p=-\frac{7}{10}\) | ||
| Check your answers. | ||
It may appear that there is only one factor in the next example. Remember, however, that \({(y-8)}^{2}\) means \((y-8)(y-8)\).
Example
Try it.
Solve: \({(y-8)}^{2}=0\).
Solution
| \({(y-8)}^{2}=0\) | ||
| Rewrite the left side as a product. | \((y-8)(y-8)=0\) | |
| Use the Zero Product Property and set each factor to 0. | \(y-8=0\) | \(y-8=0\) |
| Solve the equations. | \(y=8\) | \(y=8\) |
| When a solution repeats, we call it a double root. | ||
| Check your answer. | ||
Solve Quadratic Equations by Factoring
Each of the equations we have solved in this section so far had one side in factored form. In order to use the Zero Product Property, the quadratic equation must be factored, with zero on one side. So we must be sure to start with the quadratic equation in standard form, \(a{x}^{2}+bx+c=0\). Then we can factor the expression on the left.
How to Solve a Quadratic Equation by Factoring
Try it.
Solve: \({x}^{2}+2x-8=0\).
Solution
Before we factor, we must make sure the quadratic equation is in standard form.
Example
Try it.
Solve: \(2{y}^{2}=13y+45\).
Solution
| \(\ 2{y}^{2}=13y+45\) | ||
| Write the quadratic equation in standard form. | \(2{y}^{2}-13y-45=0\) | |
| Factor the quadratic expression. | \(\ (2y+5)(y-9)=0\) | |
| Use the Zero Product Property to set each factor to 0. | \(\ 2y+5=0\) | \(y-9=0\) |
| Solve each equation. | \(\ y=-\frac{5}{2}\) | \(y=9\) |
| Check your answers. | ||
Example
Try it.
Solve: \(5{x}^{2}-13x=7x\).
Solution
| \(5{x}^{2}-13x=7x\) | ||
| Write the quadratic equation in standard form. | \(5{x}^{2}-20x=0\\) | |
| Factor the left side of the equation. | \(5x(x-4)=0\\) | |
| Use the Zero Product Property to set each factor to 0. | \(5x=0\\) | \(x-4=0\) |
| Solve each equation. | \(x=0\\) | \(x=4\) |
| Check your answers. | ||
Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?
Example
Try it.
Solve: \(144{q}^{2}=25\).
Solution
| Write the quadratic equation in standard form. | ||
| Factor. It is a difference of squares. | ||
| Use the Zero Product Property to set each factor to 0. | \(\begin{array}{l}12q-5=0\end{array}\) | \(\begin{array}{l}12q+5=0\end{array}\) |
| Solve each equation. | \(\begin{array}{lll}12q & = & 5 \\ q & = & \frac{5}{12}\end{array}\) | \(\begin{array}{lll}12q & = & -5 \\ q & = & -\frac{5}{12}\end{array}\) |
| Check your answers. |
The left side in the next example 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.
Condensed — the full section is in OpenStax Elementary Algebra 2e.
Solve Applications Modeled by Quadratic Equations
The problem solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to quadratic equations. We will copy the problem solving strategy here so we can use it for reference.
We will start with a number problem to get practice translating words into a quadratic equation.
Example
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. | \(\begin{array}{l}\text{Let}\ n=\text{the first integer} \\ n+1=\ \text{the next consecutive integer}\end{array}\) |
| Step 4. Translate into an equation. Restate the problem in a sentence. | The product of the two consecutive integers is 132. |
| The first integer times the next integer is 132. | |
| Translate to an equation. | \(\ \begin{array}{lll} \\ n(n+1) & = & 132\end{array}\) |
| Step 5. Solve the equation. | \(\ \begin{array}{lll} \\ {n}^{2}+n & = & 132\end{array}\) |
| Bring all the terms to one side. | \(\ \begin{array}{lll} \\ {n}^{2}+n-132 & = & 0\end{array}\) |
| Factor the trinomial. | \(\begin{array}{lll} \\ (n-11)(n+12) & = & 0\end{array}\) |
| Use the zero product property. Solve the equations. | \(\ \begin{array}{llllllllll}\ n-11 & = & 0 & & & & & n+12 & = & 0 \\ n & = & 11 & & & & & n & = & -12\end{array}\) |
| There are two values for \(n\) that are solutions to this problem. So there are two sets of consecutive integers that will work. | |
| \(\begin{array}{llllll}\text{If the first integer is}\ n=11 & & & & & \text{If the first integer is}\ n=-12 \\ \\ \\ \text{then the next integer is}\ n+1 & & & & & \text{then the next integer is}\ n+1 \\ \\ \\ 11+1 & & & & & -12+1 \\ \\ \\ 12 & & & & & -11\end{array}\) | |
| Step 6. Check the answer. | |
| The consecutive integers are \(11,12\) and \(-11,-12\). The product \(11\cdot 12=132\) and the product \(-11(-12)=132\). 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 Elementary Algebra 2e.
Key Concepts
- Zero Product Property If \(a\cdot b=0\), then either \(a=0\) or \(b=0\) or both. See .
- Solve a quadratic equation by factoring To solve a quadratic equation by factoring: See .
- Write the quadratic equation in standard form, \(a{x}^{2}+bx+c=0\).
- Factor the quadratic expression.
- Use the Zero Product Property.
- Solve the linear equations.
- Check.
- Use a problem solving strategy to solve word problems See .
- 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 algebra equation.
- Solve the equation using good algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.
Chapter 7 Review Exercises
Find the Greatest Common Factor of Two or More Expressions
In the following exercises, find the greatest common factor.
Try it.
\(42,60\)
Solution
6
Try it.
\(450,420\)
Try it.
\(90,150,105\)
Solution
\(15\)
Try it.
\(60,294,630\)
Factor the Greatest Common Factor from a Polynomial
In the following exercises, factor the greatest common factor from each polynomial.
Try it.
\(24x-42\)
Solution
\(6(4x-7)\)
Try it.
\(35y+84\)
Try it.
\(15{m}^{4}+6{m}^{2}n\)
Solution
\(3{m}^{2}(5{m}^{2}+2n)\)
Try it.
\(24p{t}^{4}+16{t}^{7}\)
Factor by Grouping
In the following exercises, factor by grouping.
Try it.
\(ax-ay+bx-by\)
Solution
\((a+b)(x-y)\)
Try it.
\({x}^{2}y-x{y}^{2}+2x-2y\)
Try it.
\({x}^{2}+7x-3x-21\)
Solution
\((x-3)(x+7)\)
Try it.
\(4{x}^{2}-16x+3x-12\)
Try it.
\({m}^{3}+{m}^{2}+m+1\)
Solution
\(({m}^{2}+1)(m+1)\)
Try it.
\(5x-5y-y+x\)
Factor Trinomials of the Form \({x}^{2}+bx+c\)
In the following exercises, factor each trinomial of the form \({x}^{2}+bx+c\).
Try it.
\({u}^{2}+17u+72\)
Solution
\((u+8)(u+9)\)
Try it.
\({a}^{2}+14a+33\)
Try it.
\({k}^{2}-16k+60\)
Solution
\((k-6)(k-10)\)
Try it.
\({r}^{2}-11r+28\)
Try it.
\({y}^{2}+6y-7\)
Solution
\((y+7)(y-1)\)
Try it.
\({m}^{2}+3m-54\)
Try it.
\({s}^{2}-2s-8\)
Solution
\((s-4)(s+2)\)
Try it.
\({x}^{2}-3x-10\)
Factor Trinomials of the Form \({x}^{2}+bxy+c{y}^{2}\)
In the following examples, factor each trinomial of the form \({x}^{2}+bxy+c{y}^{2}\).
Try it.
\({x}^{2}+12xy+35{y}^{2}\)
Solution
\((x+5y)(x+7y)\)
Try it.
\({u}^{2}+14uv+48{v}^{2}\)
Try it.
\({a}^{2}+4ab-21{b}^{2}\)
Solution
\((a+7b)(a-3b)\)
Try it.
\({p}^{2}-5pq-36{q}^{2}\)
Recognize a Preliminary Strategy to Factor Polynomials Completely
In the following exercises, identify the best method to use to factor each polynomial.
Try it.
\({y}^{2}-17y+42\)
Solution
Undo FOIL
Try it.
\(12{r}^{2}+32r+5\)
Try it.
\(8{a}^{3}+72a\)
Solution
Factor the GCF
Try it.
\(4m-mn-3n+12\)
Factor Trinomials of the Form \(a{x}^{2}+bx+c\) with a GCF
In the following exercises, factor completely.
Try it.
\(6{x}^{2}+42x+60\)
Solution
\(6(x+2)(x+5)\)
Try it.
\(8{a}^{2}+32a+24\)
Try it.
\(3{n}^{4}-12{n}^{3}-96{n}^{2}\)
Solution
\(3{n}^{2}(n-8)(n+4)\)
Try it.
\(5{y}^{3}+25{y}^{2}-70y\)
Factor Trinomials Using the “ac” Method
In the following exercises, factor.
Try it.
\(2{x}^{2}+9x+4\)
Solution
\((x+4)(2x+1)\)
Try it.
\(3{y}^{2}+17y+10\)
Try it.
\(18{a}^{2}-9a+1\)
Solution
\((3a-1)(6a-1)\)
Try it.
\(8{u}^{2}-14u+3\)
Try it.
\(15{p}^{2}+2p-8\)
Solution
\((5p+4)(3p-2)\)
Try it.
\(15{x}^{2}+x-2\)
Try it.
\(40{s}^{2}-s-6\)
Solution
\((5s-2)(8s+3)\)
Try it.
\(20{n}^{2}-7n-3\)
Factor Trinomials with a GCF Using the “ac” Method
In the following exercises, factor.
Try it.
\(3{x}^{2}+3x-36\)
Solution
\(3(x+4)(x-3)\)
Try it.
\(4{x}^{2}+4x-8\)
Try it.
\(60{y}^{2}-85y-25\)
Solution
\(5(4y+1)(3y-5)\)
Try it.
\(18{a}^{2}-57a-21\)
Condensed — the full section is in OpenStax Elementary Algebra 2e.
דוגמה עובדת: x^2 - 5x + 6 = 0
צעד אחר צעד
- x^{2} - 5 x + 6 = 0
Start from the equation as given.
- a = 1,\quad b = -5,\quad c = 6
Read off the coefficients of the standard form ax² + bx + c = 0.
- \left(x - 3\right) \left(x - 2\right) = 0
Factor the quadratic into two linear factors.
- x - 3 = 0 \;\Rightarrow\; x = 3
A product is zero only when one of its factors is zero (zero-product property).
- x - 2 = 0 \;\Rightarrow\; x = 2
A product is zero only when one of its factors is zero (zero-product property).
גלה את התשובה
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
-
- ⓐ \(5{x}^{3}-15{x}^{2}-50x\)
- ⓑ \(4{x}^{2}+20xy+25{y}^{2}\)
- ⓒ \(9{x}^{2}-16\)
- ⓓ \(8{x}^{3}+27{y}^{3}\)
- ⓔ \(16{x}^{2}+24xy-4x-6y\)
גלה את התשובה
ⓐ \(5{x}^{3}-15{x}^{2}-50x\) First let's factor out the common factor of \(5x\) . \(5x({x}^{2}-3x-10)\) Consider the resulting trinomial with a leading coefficient of \(1\) . Are there two factors of \(-10\) that add to \(-3\) ? Yes, \(-5\) and \(2\) , multiply to \(-10\) and add to \(-3\) . Use these factors. \(5x(x-5)(x+2)\) ⓑ \(4{x}^{2}+20xy+25{y}^{2}\) There are no factors common to all three terms in the trinomial. Notice that the first term, \(4{x}^{2}\) is a perfect square \((2x{)}^{2}\) and last term, \(25{y}^{2}\) , is a perfect square \((5y{)}^{2}\) , so this may be a perfect square trinomial. \((a+b{)}^{2}={a}^{2}+2ab+{b}^{2}\) Check the middle term to see if it is equivalent to \(2ab\) , or \(2(2x)(5y)=20xy\) , and it is. \(ad(2x+5y)(2x+5y)=(2x+5y{)}^{2}\) ⓒ \(9{x}^{2}-16\) There are no factors common to both terms in the binomial. Notice that the first term, \(9{x}^{2}\) is a perfect square \((3x{)}^{2}\) and last term, \(16\) , is a perfect square \({4}^{2}\) , so this form is a difference of two squares. \({a}^{2}-{b}^{2}=(a-b)(a+b)\) \(9{x}^{2}-16=(3x-4)(3x+4)\) ⓓ \(8{x}^{3}+27{y}^{3}\) There are no factors common to both terms in the binomial. Notice that the first term, \(8{x}^{3}\) is a perfect cube \((2x{)}^{3}\) and last term, \(27{y}^{3}\) , so this form is a sum of two cubes. \({a}^{3}+{b}^{3}=(a+b)({a}^{2}-ab+{b}^{2})\) \({(2x)}^{3}+{(3y)}^{3}=(2x+3y)(4{x}^{2}-6xy+9{y}^{2})\) ⓔ \(16{x}^{2}+24xy-4x-6y\) First, let’s factor out the common factor of \(2\) . \(2(8{x}^{2}+12xy-2x-3y)\) The first two terms in the parentheses have a 4x in common, let’s factor that out. Since the second two terms have nothing in common, but are both negative, we’ll factor out a negative \(1\) . \(2[4x(2x+3y)-1(2x+3y)]\) Inside the parentheses, the four terms have become two terms with a common factor of \((2x+3y)\) Factor this from the two terms \(2[(2x+3y)(4x-1)]\) -
\({x}^{2}+10x+24\)
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\({y}^{2}-20y+36\)
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\(2{x}^{2}+16x+30\)
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\(9{x}^{2}+42x+49\)
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\(25{n}^{2}-90n+81\)
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\(10{x}^{3}y+65{x}^{2}y-35xy\)
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\(121{q}^{2}-100\)
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\(50{m}^{2}+72\)
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\(125{z}^{3}+27\)
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\(12ab-6a+10b-5\)
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\(18{x}^{2}-12xy+2{y}^{2}-98\)
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\({(3x-5)}^{2}-7(3x-5)+12\)
-
Factor and solve the equation: \({x}^{2}+x-6=0.\)
גלה את התשובה
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 \(-6.\)
\[\begin{array}{l}1⋅(-6) \\ (-6)⋅1 \\ 2⋅(-3) \\ 3⋅(-2)\end{array}\]The last pair, \(3⋅(-2)\) sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.
\[(x-2)(x+3)=0\]To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.
\[\begin{array}{lll}(x-2)(x+3) & = & 0 \\ (x-2) & = & 0 \\ x & = & 2 \\ (x+3) & = & 0 \\ x & = & -3\end{array}\]The two solutions are \(2\) and \(-3.\) We can see how the solutions relate to the graph in . The solutions are the x-intercepts of \(y={x}^{2}+x-6=0.\)
-
Factor and solve the quadratic equation: \({x}^{2}-5x-6=0.\)
גלה את התשובה
\((x-6)(x+1)=0;x=6,x=-1\)
-
Solve the quadratic equation by factoring: \({x}^{2}+8x+15=0.\)
גלה את התשובה
Find two numbers whose product equals \(15\) and whose sum equals \(8.\) List the factors of \(15.\)
\[\begin{array}{l}1⋅15 \\ 3⋅5 \\ (-1)⋅(-15) \\ (-3)⋅(-5)\end{array}\]The numbers that add to 8 are 3 and 5. Then, write the factors, set each factor equal to zero, and solve.
\[\begin{array}{lll}(x+3)(x+5) & = & 0 \\ (x+3) & = & 0 \\ x & = & -3 \\ (x+5) & = & 0 \\ x & = & -5\end{array}\]The solutions are \(-3\) and \(-5.\)
-
Solve the quadratic equation by factoring: \({x}^{2}-4x-21=0.\)
גלה את התשובה
\((x-7)(x+3)=0,\) \(x=7,\) \(x=-3.\)
-
Solve the difference of squares equation using the zero-product property: \({x}^{2}-9=0.\)
גלה את התשובה
Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property.
\[\begin{array}{lll}{x}^{2}-9 & = & 0 \\ (x-3)(x+3) & = & 0 \\ (x-3) & = & 0 \\ x & = & 3 \\ (x+3) & = & 0 \\ x & = & -3\end{array}\]The solutions are \(3\) and \(-3.\)
-
Solve by factoring: \({x}^{2}-25=0.\)
גלה את התשובה
\((x+5)(x-5)=0,\) \(x=-5,\) \(x=5.\)
-
Use grouping to factor and solve the quadratic equation: \(4{x}^{2}+15x+9=0.\)
גלה את התשובה
First, multiply \(ac:4(9)=36.\) Then list the factors of \(36.\)
\[\begin{array}{l}1⋅36 \\ 2⋅18 \\ 3⋅12 \\ 4⋅9 \\ 6⋅6\end{array}\]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.
\[\begin{array}{lll}4{x}^{2}+3x+12x+9 & = & 0 \\ x(4x+3)+3(4x+3) & = & 0 \\ (4x+3)(x+3) & = & 0\end{array}\]Solve using the zero-product property.
\[\begin{array}{lll}(4x+3)(x+3) & = & 0 \\ (4x+3) & = & 0 \\ x & = & -\frac{3}{4} \\ (x+3) & = & 0 \\ x & = & -3\end{array}\]The solutions are \(-\frac{3}{4},\) and \(-3.\) See .
-
Solve using factoring by grouping: \(12{x}^{2}+11x+2=0.\)
גלה את התשובה
\((3x+2)(4x+1)=0,\) \(x=-\frac{2}{3},\) \(x=-\frac{1}{4}\)
-
Solve the equation by factoring: \(-3{x}^{3}-5{x}^{2}-2x=0.\)
גלה את התשובה
This equation does not look like a quadratic, as the highest power is 3, not 2. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out \(-x\) from all of the terms and then proceed with grouping.
\[\begin{array}{lll}-3{x}^{3}-5{x}^{2}-2x & = & 0 \\ -x(3{x}^{2}+5x+2) & = & 0\end{array}\]Use grouping on the expression in parentheses.
\[\begin{array}{lll}-x(3{x}^{2}+3x+2x+2) & = & 0 \\ -x[3x(x+1)+2(x+1)] & = & 0 \\ -x(3x+2)(x+1) & = & 0\end{array}\]Now, we use the zero-product property. Notice that we have three factors.
\[\begin{array}{lll}-x & = & 0 \\ x & = & 0 \\ 3x+2 & = & 0 \\ x & = & -\frac{2}{3} \\ x+1 & = & 0 \\ x & = & -1\end{array}\]The solutions are \(0,\) \(-\frac{2}{3},\) and \(-1.\)
-
Solve by factoring: \({x}^{3}+11{x}^{2}+10x=0.\)
גלה את התשובה
\(x=0,x=-10,x=-1\)
-
Solve the quadratic using the square root property: \({x}^{2}=8.\)
גלה את התשובה
Take the square root of both sides, and then simplify the radical. Remember to use a \(\pm\) sign before the radical symbol.
\[\begin{array}{lll}{x}^{2} & = & 8 \\ x & = & \pm \sqrt{8} \\ & = & \pm 2\sqrt{2}\end{array}\]The solutions are \(2\sqrt{2},\) \(-2\sqrt{2}.\)
-
Solve the quadratic equation: \(4{x}^{2}+1=\text{7.}\)
גלה את התשובה
First, isolate the \({x}^{2}\) term. Then take the square root of both sides.
\[\begin{array}{lll}4{x}^{2}+1 & = & 7 \\ 4{x}^{2} & = & 6 \\ {x}^{2} & = & \frac{6}{4} \\ x & = & \pm \frac{\sqrt{6}}{2}\end{array}\]The solutions are \(\frac{\sqrt{6}}{2},\) and \(-\frac{\sqrt{6}}{2}.\)
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Solve the quadratic equation using the square root property: \(3{(x-4)}^{2}=15.\)
גלה את התשובה
\(x=4\pm \sqrt{5}\)
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Solve the quadratic equation by completing the square: \({x}^{2}-3x-5=0.\)
גלה את התשובה
First, move the constant term to the right side of the equal sign.
\[{x}^{2}-3x=5\]Then, take \(\frac{1}{2}\) of the b term and square it.
\[\begin{array}{lll}\frac{1}{2}(-3) & = & -\frac{3}{2} \\ {(-\frac{3}{2})}^{2} & = & \frac{9}{4}\end{array}\]Add the result to both sides of the equal sign.
\[\begin{array}{lll}{x}^{2}-3x+{(-\frac{3}{2})}^{2} & = & 5+{(-\frac{3}{2})}^{2} \\ {x}^{2}-3x+\frac{9}{4} & = & 5+\frac{9}{4}\end{array}\]Factor the left side as a perfect square and simplify the right side.
\[{(x-\frac{3}{2})}^{2}=\frac{29}{4}\]Use the square root property and solve.
\[\begin{array}{lll}\sqrt{{(x-\frac{3}{2})}^{2}} & = & \pm \sqrt{\frac{29}{4}} \\ (x-\frac{3}{2}) & = & \pm \frac{\sqrt{29}}{2} \\ x & = & \frac{3}{2}\pm \frac{\sqrt{29}}{2}\end{array}\]The solutions are \(\frac{3+\sqrt{29}}{2}\) and \(\frac{3-\sqrt{29}}{2}\) .
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Solve by completing the square: \({x}^{2}-6x=13.\)
גלה את התשובה
\(x=3\pm \sqrt{22}\)
-
Solve the quadratic equation: \({x}^{2}+5x+1=0.\)
גלה את התשובה
Identify the coefficients: \(a=1,b=5,c=1.\) Then use the quadratic formula.
\[\begin{array}{lll}x & = & \frac{-(5)\pm \sqrt{{(5)}^{2}-4(1)(1)}}{2(1)} \\ & = & \frac{-5\pm \sqrt{25-4}}{2} \\ & = & \frac{-5\pm \sqrt{21}}{2}\end{array}\] -
Use the quadratic formula to solve \({x}^{2}+x+2=0.\)
גלה את התשובה
First, we identify the coefficients: \(a=1,b=1,\) and \(c=2.\)
Substitute these values into the quadratic formula.
\[\begin{array}{lll}x & = & \frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a} \\ & = & \frac{-(1)\pm \sqrt{{(1)}^{2}-(4)⋅(1)⋅(2)}}{2⋅1} \\ & = & \frac{-1\pm \sqrt{1-8}}{2} \\ & = & \frac{-1\pm \sqrt{-7}}{2} \\ & = & \frac{-1\pm i\sqrt{7}}{2}\end{array}\]The solutions to the equation are \(\frac{-1+i\sqrt{7}}{2}\) and \(\frac{-1-i\sqrt{7}}{2}\)
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Solve the quadratic equation using the quadratic formula: \(9{x}^{2}+3x-2=0.\)
גלה את התשובה
\(x=-\frac{2}{3},\) \(x=\frac{1}{3}\)
-
Use the discriminant to find the nature of the solutions to the following quadratic equations:
- ⓐ \({x}^{2}+4x+4=0\)
- ⓑ \(8{x}^{2}+14x+3=0\)
- ⓒ \(3{x}^{2}-5x-2=0\)
- ⓓ \(3{x}^{2}-10x+15=0\)
גלה את התשובה
Calculate the discriminant \({b}^{2}-4ac\) for each equation and state the expected type of solutions.
- ⓐ
\({x}^{2}+4x+4=0\)
\({b}^{2}-4ac={(4)}^{2}-4(1)(4)=0.\) There will be one rational double solution.
- ⓑ
\(8{x}^{2}+14x+3=0\)
\({b}^{2}-4ac={(14)}^{2}-4(8)(3)=100.\) As \(100\) is a perfect square, there will be two rational solutions.
- ⓒ
\(3{x}^{2}-5x-2=0\)
\({b}^{2}-4ac={(-5)}^{2}-4(3)(-2)=49.\) As \(49\) is a perfect square, there will be two rational solutions.
- ⓓ
\(3{x}^{2}-10x+15=0\)
\({b}^{2}-4ac={(-10)}^{2}-4(3)(15)=-80.\) There will be two complex solutions.
-
Find the length of the missing side of the right triangle in .
גלה את התשובה
As we have measurements for side b and the hypotenuse, the missing side is a.
\[\begin{array}{lll}{a}^{2}+{b}^{2} & = & {c}^{2} \\ {a}^{2}+{(4)}^{2} & = & {(12)}^{2} \\ {a}^{2}+16 & = & 144 \\ {a}^{2} & = & 128 \\ a & = & \sqrt{128} \\ & = & 8\sqrt{2}\end{array}\] -
Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.
גלה את התשובה
\(5\) units
-
How do we recognize when an equation is quadratic?
גלה את התשובה
It is a second-degree equation (the highest variable exponent is 2).
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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 \(a{x}^{2}+bx+c=0\) we may graph the equation \(y=a{x}^{2}+bx+c\) and have no zeroes (x-intercepts).
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When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side?
גלה את התשובה
We want to take advantage of the zero property of multiplication in the fact that if \(a⋅b=0\) then it must follow that each factor separately offers a solution to the product being zero: \(a=0\ or\ \text{b}=0.\)
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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?
-
Describe two scenarios where using the square root property to solve a quadratic equation would be the most efficient method.
גלה את התשובה
One, when no linear term is present (no x term), such as \({x}^{2}=16.\) Two, when the equation is already in the form \({(ax+b)}^{2}=d.\)
-
\({x}^{2}+4x-21=0\)
Symbols used here
The non-negative number whose square (n-th power) is x.
Logical connectives.
1/360 of a full turn. 180° = π radians.
Both signs at once: x = 3 ± 2 means 5 and 1.
Inequalities that allow equality; < and > exclude it.
The two sides are different.
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
i² = −1.
The exponent b must be raised to for x; ln uses base e.
Naturals, integers, rationals, reals, complex numbers.
How to: Quadratic equations
- Solve quadratic equations by using the Zero Product Property
- Solve quadratic equations factoring
- Solve applications modeled by quadratic equations
- Write the quadratic equation in standard form,
- Factor the quadratic expression.
- Use the Zero Product Property.
- Solve the linear equations.
- Check.
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.
נסה את שלך.
Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
יותר בפנים. Algebra
Linear equationsSystems of equationsInequalitiesFactoringExpandingSimplifying expressionsFunctions and graphsExponential and logarithmic equationsPolynomial equationsAbsolute value