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Lesson Explainer: Solving Quadratic and Quadratic-Like Equations by Factoring Mathematics • 9th Grade

In this explainer, we will learn how to solve quadratic and quadratic-like equations using factoring along with questions that involve quadratics at some point in the factorization.

We begin by recalling the definition of a quadratic equation.

Definition: Quadratic Equation

A quadratic equation is any equation that can be expressed in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where π‘Ž,𝑏, and 𝑐 are constants and π‘Žβ‰ 0.

We should already be familiar with a number of methods for factoring quadratic expressions, including

  • factoring by inspection,
  • recognizing a quadratic as a perfect square,
  • recognizing a quadratic as the difference of two squares,
  • factoring by grouping,
  • factoring by completing the square.

The focus of this explainer is the application of these skills to solving quadratic equations. Suppose we have a quadratic equation in its factored form, (𝑝π‘₯+π‘ž)(π‘Ÿπ‘₯+𝑠)=0, where 𝑝 and π‘Ÿ are not equal to 0.

The key to solving such an equation is to recognize that if the product of two (or more) factors is equal to 0, then at least one of the individual factors must itself be equal to 0. So, to find all solutions to the given equation, we set each factor equal to 0, leading to two linear equations: 𝑝π‘₯+π‘ž=0π‘Ÿπ‘₯+𝑠=0.or

We then solve each equation separately: 𝑝π‘₯+π‘ž=0π‘Ÿπ‘₯+𝑠=0𝑝π‘₯=βˆ’π‘žπ‘Ÿπ‘₯=βˆ’π‘ π‘₯=βˆ’π‘žπ‘.π‘₯=βˆ’π‘ π‘Ÿ.

There are therefore two solutions, or roots, to the given quadratic equation: π‘₯=βˆ’π‘žπ‘ and π‘₯=βˆ’π‘ π‘Ÿ. In some cases, the two solutions may coincide to give one repeated root, in which case we only give this value once as the solution.

It is important to remember that not every quadratic equation is factorable, so the methods we discuss here can only be applied to those that are. In our first example, we will demonstrate the process of solving a quadratic equation given its factored form.

Example 1: Solving a Prefactored Equation

Solve the equation (2π‘₯βˆ’3)(3π‘₯+4)=0.

Answer

This quadratic equation is already in a factored form. We are given that the product of the two linear expressions (2π‘₯βˆ’3) and (3π‘₯+4) is 0. The only way the product of these two expressions can be 0 is if one of the factors individually is equal to 0. Hence, there are two possibilities: 2π‘₯βˆ’3=03π‘₯+4=0.or

We therefore have two linear equations to solve. The first equation can be solved by adding 3 to each side and then dividing by 2: 2π‘₯βˆ’3=02π‘₯=3π‘₯=32.

The second equation can be solved by subtracting 4 from each side and then dividing by 3: 3π‘₯+4=03π‘₯=βˆ’4π‘₯=βˆ’43.

The solutions to the quadratic equation (2π‘₯βˆ’3)(3π‘₯+4)=0 are π‘₯=32 and π‘₯=βˆ’43. This can be expressed as the solution set ο¬βˆ’43,32.

In the next example, we will demonstrate how to solve a quadratic equation by first recognizing it as a perfect square and hence writing the quadratic in its factored form.

Example 2: Solving a Simple Quadratic Equation by Factoring

Solve the equation π‘₯βˆ’8π‘₯+16=0 by factoring.

Answer

Upon inspection, we observe that both the first and third terms in this quadratic equation are perfect squares. 16 is equal to 4, so the equation can be expressed as π‘₯βˆ’8π‘₯+4=0.

Additionally, we observe that the middle term, βˆ’8π‘₯, is equal to the negative of twice the square root of the first term multiplied by the square root of the third term: βˆ’8π‘₯=βˆ’1Γ—2Γ—π‘₯Γ—4.

The quadratic is, therefore, a perfect square. It can be factored as ο€»βˆšπ‘₯βˆ’βˆš4=(π‘₯βˆ’4).

Hence, we have the equation (π‘₯βˆ’4)=0.

To solve a quadratic equation in its factored form, we set each factor equal to 0 and then solve the resulting linear equations. In this case, we have a repeated factor of (π‘₯βˆ’4), so we only have one equation to solve: π‘₯βˆ’4=0π‘₯=4.

This is a repeated root, but we only need to list it once as our solution.

We can check our answer by substituting π‘₯=4 back into the original quadratic equation: π‘₯βˆ’8π‘₯+16=4βˆ’(8Γ—4)+16=16βˆ’32+16=0.

The solution is π‘₯=4.

We summarize the steps involved in solving a quadratic equation by factoring in the process below.

How To: Solving a Quadratic Equation by Factoring

  • Find the factored form of the quadratic equation, using any available method.
  • Set each factor equal to 0.
  • Solve the resulting linear equations.

The techniques discussed here can also be applied to higher-order equations whose factored forms involve quadratics. In our next example, we consider a cubic equation that can be factored into a linear term multiplied by a quadratic and subsequently into the product of three linear expressions.

Example 3: Solving an Equation by First Identifying the Highest Common Factor between the Terms

Find the solution set for 2π‘₯=18π‘₯.

Answer

The equation we are given is not a quadratic equation. This equation is of degree 3 and hence is a cubic equation. Subtracting 18π‘₯ from each side gives 2π‘₯βˆ’18π‘₯=0.

We observe that the two terms on the left-hand side of the equation have a common factor of 2π‘₯. The equation can therefore be factored as 2π‘₯ο€Ήπ‘₯βˆ’9=0.

We now have the product of a linear term and a quadratic expression. We observe that the quadratic expression π‘₯βˆ’9 is the difference of two squares and hence can be factored as follows: π‘₯βˆ’9=π‘₯βˆ’3=(π‘₯βˆ’3)(π‘₯+3).

The fully factored form of the equation is therefore 2π‘₯(π‘₯βˆ’3)(π‘₯+3)=0.

If the product of three factors is equal to 0, at least one of the individual factors must be equal to 0. To find all possible solutions, we set each factor equal to 0, leading to 2π‘₯=0π‘₯βˆ’3=0π‘₯+3=0.oror

Solving each of these linear equations leads to π‘₯=0π‘₯=3π‘₯=βˆ’3.oror

The solution set of the given equation is {0,3,βˆ’3}.

The techniques we have encountered so far can be applied to word and practical problems in other areas of mathematics. We now consider two such examples: a word problem and a geometric problem involving the properties of right triangles.

Example 4: Solving a Word Problem by Forming and Solving a Quadratic Equation by Factoring

Given two positive numbers whose product is 35 and whose difference is 2, find the largest number.

Answer

To approach this problem, we begin by defining the larger of the two numbers to be π‘Ž. As the difference between the two numbers is 2, it follows that an expression for the smaller of the two numbers is π‘Žβˆ’2.

The product of the two numbers is 35, so we can form an equation by multiplying these two expressions together: π‘Ž(π‘Žβˆ’2)=35.

Distributing the parentheses on the left-hand side and then subtracting 35 from each side of the equation gives π‘Žβˆ’2π‘Ž=35π‘Žβˆ’2π‘Žβˆ’35=0.

This is a quadratic equation in π‘Ž, which can be solved by factoring. To factor by inspection, we require two numbers whose sum is equal to the coefficient of π‘Ž(βˆ’2) and whose product is equal to the constant term (βˆ’35). These two numbers are βˆ’7 and 5; hence, the factored form of the quadratic equation is (π‘Žβˆ’7)(π‘Ž+5)=0.

If the product of these two linear expressions is equal to 0, then at least one of the individual expressions must be equal to 0. It follows that π‘Žβˆ’7=0π‘Ž+5=0.or

Hence, π‘Ž=7π‘Ž=βˆ’5.or

It is stated in the question that the two numbers are both positive, so the only valid solution is π‘Ž=7.

As π‘Ž represented the larger of the two numbers, we have completed the problem. However, we can check our answer by calculating the other number and confirming that the two numbers have the correct product. Our expression for the smaller number was π‘Žβˆ’2; hence, the smaller number is 7βˆ’2=5. The product of 5 and 7 is indeed 35, so this confirms that our answer is correct.

The larger of the two numbers is 7.

Example 5: Finding an Unknown in a Right Triangle by Forming and Solving a Quadratic Equation

Find the value of π‘₯ given that a right triangle has a hypotenuse of length 2π‘₯ and sides of lengths π‘₯+1 and π‘₯+3.

Answer

We have been given expressions for the lengths of each side of this right triangle in terms of an unknown variable π‘₯. To determine its value, we will need to form and solve an equation. We recall that the three side lengths in a right triangle are related by the Pythagorean theorem, which states that, in a right triangle, the sum of the squares of the two shorter sides is equal to the square of the hypotenuse.

We are told that the hypotenuse of this triangle is of length 2π‘₯, and the other two sides are of lengths π‘₯+1 and π‘₯+3. Applying the Pythagorean theorem gives (π‘₯+1)+(π‘₯+3)=(2π‘₯).

We proceed by distributing the parentheses on each side of the equation and then simplifying to give π‘₯+π‘₯+π‘₯+1+π‘₯+3π‘₯+3π‘₯+9=4π‘₯2π‘₯+8π‘₯+10=4π‘₯.

The terms can all be collected on the same side of the equation by subtracting ο€Ή2π‘₯+8π‘₯+10ο…οŠ¨ from each side, leading to 2π‘₯βˆ’8π‘₯βˆ’10=0.

Simplifying the equation by dividing the entire equation by 2 gives π‘₯βˆ’4π‘₯βˆ’5=0.

We will solve this quadratic equation by factoring, and we have a choice of methods to use. We could factor by inspection, but in this instance, we will demonstrate the method of factoring by completing the square. We wish to express the quadratic equation in the form (π‘₯+π‘Ž)+𝑏=0.

The value of π‘Ž is half the coefficient of π‘₯ in the original equation; then, we subtract the square of this value: π‘₯βˆ’4π‘₯βˆ’5=(π‘₯βˆ’2)βˆ’(βˆ’2)βˆ’5=(π‘₯βˆ’2)βˆ’4βˆ’5=(π‘₯βˆ’2)βˆ’9.

The quadratic is now in completed square form, and we observe that it is, in fact, the difference of two squares: (π‘₯βˆ’2)βˆ’9=(π‘₯βˆ’2)βˆ’3.

Hence, the factored form of this quadratic equation is ((π‘₯βˆ’2)βˆ’3)((π‘₯βˆ’2)+3)=0, which simplifies to (π‘₯βˆ’5)(π‘₯+1)=0.

To solve this equation, we first set each factor equal to 0, leading to π‘₯βˆ’5=0π‘₯+1=0.or

Solving each equation separately gives π‘₯=5π‘₯=βˆ’1.or

While these are both valid roots of the given quadratic equation, they are not both valid solutions to the problem. Recall that the expression 2π‘₯ represents the length of the hypotenuse of the right triangle, which must be strictly positive. Consequently, π‘₯ must also be strictly positive, so we disregard the solution π‘₯=βˆ’1.

The value of π‘₯ is 5.

We can check our answer to the previous problem by calculating all three side lengths and confirming that they do indeed satisfy the Pythagorean theorem. When π‘₯=5, the three side lengths are 6, 8, and 10 units, which we may recognize as a Pythagorean triple. Alternatively, we can confirm by calculation that 6+8=36+64=100=10.

Some equations we encounter may not appear to be quadratic equations at first, but by making an appropriate substitution, we see that they can be expressed as quadratic equations in another variable. We now consider one final example in which we solve a quartic (degree 4) equation by recognizing that it can, in fact, be expressed as a quadratic equation.

Example 6: Solving a Quadratic-Like Equation by Factoring

Find the solution set for π‘₯βˆ’13π‘₯+36=0οŠͺ.

Answer

The equation we have been asked to solve is of degree 4 and hence is a quartic equation. However, if we look carefully, we see that this is a quadratic-like equation. By the laws of indices, π‘₯=ο€Ήπ‘₯οŠͺ. So, this equation can be expressed as ο€Ήπ‘₯ο…βˆ’13ο€Ήπ‘₯+36=0.

We can introduce a new variable 𝑦, which we define to be equal to π‘₯. Thus, we have the quadratic equation π‘¦βˆ’13𝑦+36=0.

We now solve this equation for 𝑦; then, we subsequently solve for π‘₯.

As the coefficient of π‘¦οŠ¨ is 1, we can factor by inspection. We require two numbers whose sum is βˆ’13 and whose product is 36. These two numbers are βˆ’4 and βˆ’9; hence, the factored form of the quadratic equation is (π‘¦βˆ’4)(π‘¦βˆ’9)=0.

As the product of these two linear expressions is equal to 0, either π‘¦βˆ’4=0π‘¦βˆ’9=0.or

Hence, 𝑦=4𝑦=9.or

Finally, we reverse the substitution and solve for π‘₯. Recall that we defined 𝑦 to be equal to π‘₯, so π‘₯=4π‘₯=9.or

Each of these equations has two real roots. If π‘₯=4, then π‘₯=±√4=Β±2, and if π‘₯=9, then π‘₯=±√9=Β±3.

Hence, the solution set of the given equation is {βˆ’2,2,3,βˆ’3}.

We can check that these values are correct by substituting back into the original equation. For example, when π‘₯=2, π‘₯βˆ’13π‘₯+36=(2)βˆ’13(2)+36=16βˆ’52+36=0.οŠͺοŠͺ

Let us finish by recapping some key points.

Key Points

  • A quadratic equation is any equation that can be expressed in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where π‘Ž,𝑏, and 𝑐 are constants and π‘Žβ‰ 0.
  • To solve a quadratic equation of the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0,
    • express the quadratic equation in its factored form (if factorable),
    • set each factor equal to 0,
    • solve the resulting linear equations.
  • Some nonquadratic equations can be solved by first factoring by a highest common factor and then factoring any remaining quadratic factors.
  • Some equations are not quadratic but are β€œquadratic-like”. Such equations can be solved by making an appropriate substitution to convert to a quadratic equation in a new variable.
  • Word and practical problems can be solved by first forming a quadratic equation and then solving by factoring.

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