We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Study Guides > ALGEBRA / TRIG I

Quadratic Equations

Learning Outcomes

  • Recognize a quadratic equation
  • Use the zero product principle to solve a quadratic equation that can be factored
When a polynomial is set equal to a value (whether an integer or another polynomial), the result is an equation. An equation that can be written in the form ax2+bx+c=0ax^{2}+bx+c=0 is called a quadratic equation.  When solving polynomials where the highest degree is degree 2, we want to confirm that the equation is written in standard form, ax2+bx+c=0a{x}^{2}+bx+c=0, where a, b, and c are real numbers and a0a\ne 0. The equation x2+x6=0{x}^{2}+x - 6=0 is in standard form.

Quadratic Equation

A quadratic equation is an equation containing a second-degree polynomial; for example,
ax2+bx+c=0a{x}^{2}+bx+c=0
where a, b, and c are real numbers, and if a0a\ne 0, it is in standard form.
Often the easiest method of solving a quadratic equation is by factoring. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation. Note that we will not spend a lot of time explaining how to factor in this section. We cover factoring in an earlier module of this course and you can sharpen your skills there. Solving by factoring depends on the Principle of Zero Products. What if we told you that we multiplied two numbers together and got an answer of zero? What could you say about the two numbers? Could they be 22 and 55? Could they be 99 and 11? No! When the result (answer) from multiplying two numbers is zero, that means that one of them had to be zero. This idea is called the zero product principle, and it is useful for solving polynomial equations that can be factored. You can further review the Principle of Zero Products here.

Principle of Zero Products

The Principle of Zero Products states that if the product of two numbers is 00, then at least one of the factors is 00. If ab=0ab=0, then either a=0a=0 or b=0b=0, or both a and b are 00.
In this section, we will show several examples of solving quadratic equations using factoring.  For each example, we will be using the same general technique:

How to solve a quadratic equation using factoring

  1. Make sure your quadratic equation is written in standard form, that is, ax2+bx+c=0a{x}^{2}+bx+c=0 where a, b, and c are real numbers, and a0a\ne 0.
  2. Factor the quadratic expression on the left-hand side of the equation.
  3. Solve using the zero-product property by setting each factor equal to zero and solving for the variable.
Let us start with a simple example. We will factor a GCF from a binomial and apply the principle of zero products to solve a polynomial equation.  (You can review how to factor out a greatest common factor here)

Example

Solve: t2+t=0-t^2+t=0

Answer: Each term has a common factor of tt, so we can factor and use the zero product principle. Rewrite each term as the product of the GCF and the remaining terms. t2=t(t)t=t(1)\begin{array}{c}-t^2=t\left(-t\right)\\t=t\left(1\right)\end{array} Rewrite the polynomial equation using the factored terms in place of the original terms.

t2+t=0t(t)+t(1)t(t+1)=0\begin{array}{c}-t^2+t=0\\t\left(-t\right)+t\left(1\right)\\t\left(-t+1\right)=0\end{array}

Now we have a product on one side and zero on the other, so we can set each factor equal to zero using the zero product principle.

t=0         OR            t+1=0                                                         1   1                                                         t=1                                                         t1=11                                                         t=1\begin{array}{c}t=0\,\,\,\,\,\,\,\,\text{ OR }\,\,\,\,\,\,\,\,\,\,\,-t+1=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-1}\,\,\,\underline{-1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-t=-1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{-t}{-1}=\frac{-1}{-1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=1\end{array}

Therefore, t=0 OR t=1t=0\text{ OR }t=1.

In the following video, we show two more examples of using both factoring and the principle of zero products to solve a polynomial equation. https://youtu.be/gIwMkTAclw8   In our next example, we will solve a quadratic equation that has a leading coefficient of 1.  In the quadratic equation x2+x6=0{x}^{2}+x - 6=0, the leading coefficient, or the coefficient of x2{x}^{2}, is 11. Notice that if we compare this equation to the standard form of a quadratic equation, then a=1a=1b=1b=1, and c=bc=-b.  For a review on how to factor a quadratic polynomial with a leading coefficient of 1, you can visit this page.

Example

Factor and solve the equation: x2+x6=0{x}^{2}+x - 6=0.

Answer: This equation is already in standard form, so we will proceed with factoring the left side of the equation. To factor x2+x6=0{x}^{2}+x - 6=0, we look for two numbers whose product equals 6-6 and whose sum equals 11. Begin by looking at the possible factors of 6-6.

1(6)(6)12(3)3(2)\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}
The last pair, 3(2)3\cdot \left(-2\right) sums to 11, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.
(x2)(x+3)=0\left(x - 2\right)\left(x+3\right)=0
To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.
(x2)(x+3)=0(x2)=0(x+3)=0x=2x=3\begin{array}{lll}& \left(x - 2\right)\left(x+3\right)=0 & \hfill \\ \left(x - 2\right)\hfill=0 & & \left(x+3\right)=0 \hfill \\ x=2 & & x=-3 \hfill \end{array}
The two solutions are x=2x=2 and x=3x=-3.

If we graph the equation y=x2+x6y={x}^{2}+x - 6, we will get the parabola in the figure below (NOTE: we will learn how to graph quadratic equations in greater detail when we learn about functions). The solutions to the equation x2+x6=0{x}^{2}+x - 6=0 are the x-intercepts of y=x2+x6y={x}^{2}+x - 6. Recall that x-intercepts are where the y values are zero, therefore the points (3,0)(-3,0) and (2,0)(2,0) represent the places where the parabola crosses the x-axis.  This matches up with our solution to the equation x2+x6=0{x}^{2}+x - 6=0 because it shows that when x=2x=2 and x=3x=-3, our equation equals 00 Coordinate plane with the x-axis ranging from negative 5 to 5 and the y-axis ranging from negative 7 to 7. The function x squared plus x minus six equals zero is graphed, with the x-intercepts (-3,0) and (2,0), plotted as well. In the following video, we provide more examples of factoring to solve quadratic equations where the leading coefficient is equal to 1. https://youtu.be/bi7i_RuIGl0

Try It

[ohm_question]197328[/ohm_question]
We all know that it is rare to be given an equation to solve that has zero on one side, so let us try an example where we first have to move all the terms of the equation to the left-hand side.

Example

Solve: s24s=5s^2-4s=5

Answer: First, move all the terms to one side. The goal is to try and see if we can use the zero product principle since that is the only tool we know for solving polynomial equations.

       s24s=5       s24s5=0\begin{array}{c}\,\,\,\,\,\,\,s^2-4s=5\\\,\,\,\,\,\,\,s^2-4s-5=0\\\end{array}

We now have all the terms on the left side and zero on the other side. The polynomial s24s5s^2-4s-5 factors nicely which makes this equation a good candidate for the zero product principle.

s24s5=0(s+1)(s5)=0\begin{array}{c}s^2-4s-5=0\\\left(s+1\right)\left(s-5\right)=0\end{array}

We separate our factors into two linear equations using the principle of zero products.

(s5)=0s5=0         s=5\begin{array}{c}\left(s-5\right)=0\\s-5=0\\\,\,\,\,\,\,\,\,\,s=5\end{array} OR (s+1)=0s+1=0s=1\begin{array}{c}\left(s+1\right)=0\\s+1=0\\s=-1\end{array} Therefore, s=1 OR s=5s=-1\text{ OR }s=5.

In our next example, we solve a quadratic equation with a leading coefficient that is not equal to 1. Recall that when the leading coefficient is not 11, we factor a quadratic equation using a method called grouping, which requires four terms.  You can review how to factor by grouping here.

Example

Solve the quadratic equation: 4x2+15x+9=04{x}^{2}+15x+9=0.

Answer: We can see that this equation is already in standard for, so we will proceed with factoring. Because this equation has a leading coefficient not equal to 1, we will factor by grouping: multiply ac:4(9)=36ac:4\left(9\right)=36. Then list the factors of 3636.

1362183124966\begin{array}{l}1\cdot 36\hfill \\ 2\cdot 18\hfill \\ 3\cdot 12\hfill \\ 4\cdot 9\hfill \\ 6\cdot 6\hfill \end{array}
The only pair of factors that sums to 1515 is 3+123+12. Rewrite the equation replacing the b term, 15x15x, with two terms using 33 and 1212 as coefficients of x. Factor the first two terms, and then factor the last two terms.
4x2+3x+12x+9=0x(4x+3)+3(4x+3)=0(4x+3)(x+3)=0\begin{array}{l}4{x}^{2}+3x+12x+9=0\hfill \\ x\left(4x+3\right)+3\left(4x+3\right)=0\hfill \\ \left(4x+3\right)\left(x+3\right)=0\hfill \end{array}
Solve using the zero-product property.

(4x+3)(x+3)=0(4x+3)=0(x+3)=0x=34x=3\begin{array}{lll}&\left(4x+3\right)\left(x+3\right)=0 & \hfill \\ \left(4x+3\right)\hfill=0 & & \left(x+3\right)=0 \hfill \\ x=-\frac{3}{4} & & x=-3 \hfill \end{array}

The solutions are x=34x=-\frac{3}{4}, x=3x=-3. Coordinate plane with the x-axis ranging from negative 6 to 2 with every other tick mark labeled and the y-axis ranging from negative 6 to 2 with each tick mark numbered. The equation: four x squared plus fifteen x plus nine is graphed with its x-intercepts: (-3/4,0) and (-3,0) plotted as well. Notice that when we graph the equation  y=4x2+15x+9y=4{x}^{2}+15x+9 we can see that the x-intercepts are (3,0)(-3,0) and (34,0)(-\frac{3}{4},0). This agrees with our solutions!

  The example below shows a quadratic equation where neither side is originally equal to zero.  Remember, we cannot solve quadratic equations unless one side is equal to zero because this allows us to use the Zero Product Principle. (Note that the factoring sequence has been shortened.)

Example

Solve 5b2+4=12b5b^{2}+4=−12b for b.

Answer: The original equation has 12b−12b on the right. To make this side equal to 0, add 12b12b to both sides.

5b2+4+12b=12b+12b5b^{2}+4+12b=−12b+12b

Combine like terms.

5b2+12b+4=05b^{2}+12b+4=0

Rewrite 12b12b as 10b+2b10b+2b.

5b2+10b+2b+4=05b^{2}+10b+2b+4=0

Factor out 5b5b from the first pair and 2 from the second pair.

5b(b+2)+2(b+2)=05b\left(b+2\right)+2\left(b+2\right)=0

Factor out b+2b+2.

(5b+2)(b+2)=0\left(5b+2\right)\left(b+2\right)=0

Apply the Zero Product Property.

5b+2=0   or   b+2=05b+2=0\,\,\,\text{or}\,\,\,b+2=0

Solve each equation.

b=25   OR   b=2b=-\frac{2}{5}\,\,\,\text{OR}\,\,\,b=−2

Answer

b=25   or   b=2b=-\frac{2}{5}\,\,\,\text{or}\,\,\,b=−2

We will work through one more example that is similar to the one above, except this example has fractions, yay!

Example

Solve y25=72y+52y^2-5=-\frac{7}{2}y+\frac{5}{2}

Answer: We can solve this in one of two ways.  One way is to eliminate the fractions like you may have done when solving linear equations, and the second is to find a common denominator and factor fractions. Eliminating fractions is easier, so we will show that way. Start by multiplying the whole equation by 22 to eliminate fractions:

2(y25=72y+52)      2(y2)+2(5)=2(72y)+2(52)2y210=7y+5\begin{array}{ccc}2\left(y^2-5=-\frac{7}{2}y+\frac{5}{2}\right)\\\,\,\,\,\,\,2(y^2)+2(-5)=2\left(-\frac{7}{2}y\right)+2\left(\frac{5}{2}\right)\\2y^2-10=-7y+5\end{array}

Now we can move all the terms to one side and see if this will factor so we can use the principle of zero products.

2y210=7y+52y210+7y5=02y215+7y=02y2+7y15=0\begin{array}{c}2y^2-10=-7y+5\\2y^2-10+7y-5=0\\2y^2-15+7y=0\\2y^2+7y-15=0\end{array}

We can now check whether this polynomial will factor. Using a table we can list factors until we find two numbers with a product of 215=302\cdot-15=-30 and a sum of 7.
Factors of 215=302\cdot-15=-30 Sum of Factors
1,301,-30 29-29
1,30-1,30 2929
2,152,-15 13-13
2,15-2,15 1313
3,103,-10 7-7
3,10-3,10 77
We have found the factors that will produce the middle term we want,3,10-3,10. We need to place the factors in a way that will lead to a middle term of 7y7y: (2y3)(y+5)=0\left(2y-3\right)\left(y+5\right)=0 Now we can set each factor equal to zero and solve: (2y3)=0 OR (y+5)=02y=3 OR y=5y=32 OR y=5\begin{array}{ccc}\left(2y-3\right)=0\text{ OR }\left(y+5\right)=0\\2y=3\text{ OR }y=-5\\y=\frac{3}{2}\text{ OR }y=-5\end{array} You can always check your work to make sure your solutions are correct: Check y=32y=\frac{3}{2} (32)25=72(32)+52945=214+52 common denominator = 494204=214+104114=114\begin{array}{ccc}\left(\frac{3}{2}\right)^2-5=-\frac{7}{2}\left(\frac{3}{2}\right)+\frac{5}{2}\\\frac{9}{4}-5=-\frac{21}{4}+\frac{5}{2}\\\text{ common denominator = 4}\\\frac{9}{4}-\frac{20}{4}=-\frac{21}{4}+\frac{10}{4}\\-\frac{11}{4}=-\frac{11}{4}\end{array} y=32[/latex]isindeedasolution,nowcheck [latex]y=5y=\frac{3}{2}[/latex] is indeed a solution, now check [latex]y=-5 (5)25=72(5)+52255=352+5220=40220=20\begin{array}{ccc}\left(-5\right)^2-5=-\frac{7}{2}\left(-5\right)+\frac{5}{2}\\25-5=\frac{35}{2}+\frac{5}{2}\\20=\frac{40}{2}\\20=20\end{array} y=5y=-5 is also a solution, so we must have done something right!

Therefore, y=32 OR y=5y=\frac{3}{2}\text{ OR }y=-5.

In our last video, we show how to solve another quadratic equation that contains fractions. https://youtu.be/kDj_qdKW-ls The following video contains another example of solving a quadratic equation using factoring with grouping. https://youtu.be/04zEXaOiO4U It is useful to remember that, if you factor out a constant, the constant will never equal 00. So it can essentially be ignored when solving. See the following example.

Example

Solve for k: 2k2+90=8k-2k^2+90=-8k

Answer: We need to move all the terms to one side so we can use the zero product principle.

2k2+90=8k+8k                +8k2k2+8k+90=0\begin{array}{l}-2k^2+90=-8k\\\underline{+8k}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{+8k}\\-2k^2+8k+90=0\end{array}

You will either need to try to factor out a 2-2, or use the method where we multiply 290-2\cdot{90} and find factors that sum to 88. Each term is divisible by  22, so we can factor out 2-2.

2(k24k45)=0-2\left(k^2-4k-45\right)=0

Note how we changed the signs when we factored out a negative number. If we can factor the polynomial, we will be able to solve.   Using the shortcut for factoring we will start with the variable and place a plus and a minus sign in the binomials. We do this because 4545 is negative and the only way to get a product that is negative is if one of the factors is negative.

2(k   )(k+   )=0-2\left(k-\,\,\,\right)\left(k+\,\,\,\right)=0

We want our factors to have a product of 45-45 and a sum of 4-4:

Factors whose product is -45 Sum of the factors
145=451\cdot-45=-45 145=441-45=-44
315=453\cdot-15=-45 315=123-15=-12
59=455\cdot-9=-45 59=45-9=-4
There are more factors that will give 45-45, but we have found the ones that sum to 4-4, so we will stop. Fill in the rest of the binomials with the factors we found.

2(k9)(k+5)=0-2\left(k-9\right)\left(k+5\right)=0

Now we can set each factor equal to zero using the zero product rule.

2=0-2=0 This solution is nonsense so we discard it.

k9=0,k=9k-9=0, k=9

k+5=0,k=5k+5=0, k=-5

Answers

k=9 and k=5k=9\text{ and }k=-5

  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, which we will see later in this section. In our next example, we will solve a quadratic equation that is written as a difference of squares.  For a review of difference of squares and other special cases, you can visit this page.

Example

Solve the difference of squares equation using the zero-product property: x29=0{x}^{2}-9=0.

Answer: 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. x29=0(x3)=0(x+3)=0x=3x=3\begin{array}{lll}& {x}^{2}-9=0 & \hfill \\ \left(x - 3\right)\hfill=0 & & \left(x+3\right)=0 \hfill \\ x=3 & & x=-3 \hfill \end{array} The solutions are x=3x=3 and x=3x=-3.

Sometimes, we may be given an equation that does not look like a quadratic at first glance. In our next examples we will solve a cubic polynomial equation where the GCF of each term is x and can be factored. The result is a quadratic equation that we can solve.

Example

Solve the equation by factoring: 3x35x22x=0-3{x}^{3}-5{x}^{2}-2x=0.

Answer: This equation does not look like a quadratic, as the highest power is 33, not 22. 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-x from all of the terms and then proceed with grouping.

3x35x22x=0x(3x2+5x+2)=0\begin{array}{l}-3{x}^{3}-5{x}^{2}-2x=0\hfill \\ -x\left(3{x}^{2}+5x+2\right)=0\hfill \end{array}
Use grouping on the expression in parentheses.
x(3x2+3x+2x+2)=0x[3x(x+1)+2(x+1)]=0x(3x+2)(x+1)=0\begin{array}{l}-x\left(3{x}^{2}+3x+2x+2\right)=\hfill0\hfill \\ -x\left[3x\left(x+1\right)+2\left(x+1\right)\right]=\hfill0\hfill \\ -x\left(3x+2\right)\left(x+1\right)=\hfill0\hfill \end{array}
Now, we use the zero-product property. Notice that we have three factors.

x=0(3x+2)=0(x+1)=0x=0x=23x=1\begin{array}{ccc}-x=0&\left(3x+2\right)=0&\left(x+1\right)=0 \hfill \\ x=0 & x=-\frac{2}{3} & x=-1 \hfill \end{array}

The solutions are x=0x=0, x=23x=-\frac{2}{3}, and x=1x=-1.

  In this last video example, we solve a quadratic equation with a leading coefficient of -1 using a shortcut method of factoring and the zero product principle. https://youtu.be/nZYfgHygXis

Summary

You can find the solutions, or roots, of quadratic equations by setting one side equal to zero, factoring the polynomial, and then applying the Zero Product Property. The Principle of Zero Products states that if ab=0ab=0, then either a=0a=0 or b=0b=0, or both a and b are 00. Once the polynomial is factored, set each factor equal to zero and solve them separately. The answers will be the set of solutions for the original equation.

Contribute!

Did you have an idea for improving this content? We’d love your input.

Licenses & Attributions