Lesson Explainer: Solving Cubic Equations: Taking Cube Roots Mathematics • 8th Grade

In this explainer, we will learn how to solve cubic equations using the cube root property.

We begin by recalling that the cube root of a number π‘Ž, written οŽ’βˆšπ‘Ž, is the number whose cube is π‘Ž. In other words, ο€Ίβˆšπ‘Žο†=π‘ŽοŽ’οŠ©. We can use this to simplify or evaluate expressions. For example, we know that 2=8; hence, we can evaluate √8=2.

This is not the only use of the cube root. We can also use this idea to solve equations. For example, imagine we are told that the volume of a cube is 8 cm3. We can then say that the cube has side lengths π‘₯ cm to get the following.

The volume of this cube is then given by π‘₯, so we can construct the following equation: π‘₯=8.

We know that 2=8; however, we can also solve this equation by taking cube roots of both sides of the equation. Doing this gives √π‘₯=√8.

We know that √π‘₯=π‘₯ and √8=2; hence, π‘₯=2.

It is worth noting that we know this is the only value of π‘₯ that solves this equation, since increasing the side length will increase the volume and decreasing the side length will decrease the volume. This is true in general: increasing the value of π‘₯ will increase the value of π‘₯ and decreasing the value of π‘₯ will decrease the value of π‘₯. Hence, π‘₯=π‘ŽοŠ© will only have one solution for any real value of π‘Ž.

This is not the case with the square root, since 2=4 but (βˆ’2)=4 as well. This is the reason that we find a positive and a negative root when taking square roots of both sides of an equation but there is a unique solution when taking cube roots.

Let’s now see an example of solving some cubic equations.

Example 1: Solving a Cubic Equation by Rearranging

Solve the cubic equation π‘₯=8 in β„š.

Answer

We take the cube root of both sides of the equation to see π‘₯=√8.

We then recall that 2=8, so √8=2. Hence, π‘₯=√8=2.

In our next example, we will see how to solve a cubic equation where the variable cubed is equal to a rational number.

Example 2: Solving a Cubic Equation Involving a Fraction

Solve π‘₯=278.

Answer

We start by taking cube roots of both sides of the equation, where we note that √π‘₯=π‘₯. This gives π‘₯=ο„ž278.

We then recall that if π‘Ž and 𝑏 are perfect cubes and 𝑏≠0, then οŽ’οŽ’οŽ’ο„žπ‘Žπ‘=βˆšπ‘Žβˆšπ‘. We note that √27=3 and √8=2, so both are perfect cubes and hence π‘₯=ο„ž278=√27√8=32.

Therefore, π‘₯=32.

In our next example, we will solve a cubic equation by first rearranging.

Example 3: Solving a Cubic Equation by Rearranging

Find the value of 𝑦 given that βˆ’1000π‘¦βˆ’27=0.

Answer

Since the variable in this equation is cubed, we are going to need to take cube roots of both sides of the equation. However, in order to do this, we first need to rearrange the equation so that the cubic factor is isolated on one side of the equation.

We first add 27 to both sides of the equation to get βˆ’1000𝑦=27.

We then divide both sides of the equation through by βˆ’1000; this gives 𝑦=βˆ’271000.

We can now take cube roots of both sides of the equation to get 𝑦=ο„žβˆ’271000.

We then recall that if π‘Ž=π‘›οŠ©, 𝑏=π‘šοŠ©, and 𝑏≠0, then οŽ’οŽ’οŽ’ο„žπ‘›π‘š=βˆšπ‘›βˆšπ‘š=π‘Žπ‘. We note that οŽ’βˆšβˆ’27=βˆ’3 and √1000=10; hence, 𝑦=ο„žβˆ’271000=βˆšβˆ’27√1000=βˆ’310.

In our next example, we will solve another equation by rearranging.

Example 4: Rearranging an Equation into a Cubic Equation and Solving It

Given that π‘₯βˆˆβ„ and βˆ’π‘₯10=100π‘₯, determine the value of π‘₯.

Answer

To solve an equation of this form, we need to note that multiplying the equation through by π‘₯ will collect all of the variables into a single term. To justify multiplying the equation through by π‘₯, we note we are dividing by π‘₯ in the original equation, so π‘₯ cannot be zero. Hence, ο€»βˆ’π‘₯10×π‘₯=ο€Ό100π‘₯οˆΓ—π‘₯.

We cancel the shared factor of π‘₯ on the right-hand side of the equation, since π‘₯β‰ 0, and simplify; this gives βˆ’π‘₯10=100.

We then multiply the equation through by βˆ’10 to get π‘₯=βˆ’1000.

Finally, we take cube roots of both sides of the equation, where we note that (βˆ’10)=βˆ’1000, so οŽ’βˆšβˆ’1000=βˆ’10. This gives π‘₯=βˆšβˆ’1000=βˆ’10.

Thus far, we have only dealt with simple equations involving cubes. However, it is possible for operations to occur inside the cubic operation. For example, imagine we are told that a father is 65 years old and he is one year older than twice his son’s age cubed. We can use the cube root to determine the son’s age.

First, we need to rewrite this information into an equation. To do this, we will work backward. Let’s call the son’s age π‘₯. We are told that the father is 65 years old and he is one year older than twice his son’s age cubed. So, we need to cube twice π‘₯ and then add 1; this is then equal to 65. We have (2π‘₯)+1=65.

We can subtract 1 from both sides to get (2π‘₯)=64.

We can then take cube roots of both sides of the equation; this gives 2π‘₯=√64=4.

We then divide the equation through by 2 to get π‘₯=2.

Thus, the son’s age is 2 years.

In general, we can solve equations of the form (π‘Žπ‘₯+𝑏)+𝑐=𝑑, provided π‘Žβ‰ 0 and we can find the cube root of π‘‘βˆ’π‘. To do this, we need to rearrange the equation to find π‘₯. We can do this by using the following method.

How To: Solving a Cubic Equation

To solve a cubic equation of the form (π‘Žπ‘₯+𝑏)+𝑐=π‘‘οŠ©, where π‘Ž,𝑏,𝑐, and 𝑑 are constants and π‘Žβ‰ 0, we need to rearrange the equation for π‘₯. We can do this by following these steps:

  1. Subtract 𝑐 from both sides of the equation to get (π‘Žπ‘₯+𝑏)=π‘‘βˆ’π‘οŠ©.
  2. Take the cube root of both sides of the equation to get π‘Žπ‘₯+𝑏=βˆšπ‘‘βˆ’π‘οŽ’.
  3. Subtract 𝑏 from both sides to get π‘Žπ‘₯=βˆšπ‘‘βˆ’π‘βˆ’π‘οŽ’.
  4. Divide the equation through by π‘Ž to get π‘₯=βˆšπ‘‘βˆ’π‘βˆ’π‘π‘ŽοŽ’.

Let’s now see some examples of applying this method to solve cubic equations.

Example 5: Solving a Cubic Equation by Rearranging after Taking the Cube Root

Find the value of 𝑦 given that (2π‘¦βˆ’14)βˆ’36=28.

Answer

We first note that we have a cubic term in this equation that involves a variable. This means we will want to take cube roots of both sides of the equation to solve for 𝑦. However, we cannot do this directly to the equation since the cubed term is not isolated on one side of the equation. Therefore, we start by adding 36 to both sides of the equation to get (2π‘¦βˆ’14)=28+36=64.

We can now take cube roots of both sides of the equation to get οŽ’οŽ’ο„(2π‘¦βˆ’14)=√64.

We know that οŽ’βˆšπ‘Ž=π‘ŽοŠ©, so οŽ’ο„(2π‘¦βˆ’14)=2π‘¦βˆ’14, and that 64=4, so √64=4.

Therefore, the equation simplifies to give 2π‘¦βˆ’14=4.

We can now solve for 𝑦. We add 14 to both sides of the equation, which yields 2𝑦=4+14=18.

We then divide the equation through by 2 to get 𝑦=182=9.

In our final example, we will solve a cubic equation involving rearranging an equation where the coefficient of the variable is negative.

Example 6: Solving a Cubic Equation by Rearranging after Taking the Cube Root

Find the value of π‘₯ given that (15βˆ’3π‘₯)+2=29, where π‘₯βˆˆβ„.

Answer

We first need to rearrange the equation so that the cube term is isolated on the left-hand side of the equation. We can do this by subtracting 2 from both sides of the equation; this gives (15βˆ’3π‘₯)=29βˆ’2=27.

We can then take cube roots of both sides of the equation, where we note οŽ’ο„(15βˆ’3π‘₯)=15βˆ’3π‘₯ and √27=3. Hence, 15βˆ’3π‘₯=3.

We can now solve for π‘₯. We subtract 15 from both sides of the equation to get βˆ’3π‘₯=βˆ’12.

Then, we divide through by βˆ’3 to get π‘₯=4.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • We can solve equations by taking the cube roots of both sides of the equation. In particular, if π‘₯=π‘ŽοŠ©, then π‘₯=βˆšπ‘ŽοŽ’.
  • Unlike the square root, taking cube roots of both sides of an equation gives a unique solution.
  • To solve a cubic equation of the form (π‘Žπ‘₯+𝑏)+𝑐=π‘‘οŠ©, where π‘Ž,𝑏,𝑐, and 𝑑 are constants and π‘Žβ‰ 0, we rearrange the equation to isolate π‘₯. We get π‘₯=βˆšπ‘‘βˆ’π‘βˆ’π‘π‘Ž.

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