Basic Square Root Equations

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Algebra One $\longrightarrow$

Objectives
• Solve equations that contain one single variable root expression and no other variable terms
• Solve equations that contain one variable root expression and also have other variable terms
• Check for extraneous solutions when solving root equations
Lesson Description

One of the things we can use root properties for is solving equations that have root expressions in them. In this lesson we'll do just that - we'll practice isolating x in an equation when x is part of a root expression.

Practice Problems

Practice problems and worksheet coming soon!

As we've said in past lessons that focus on equation solving, the goal is always the same: isolate the variable. Hopefully by now we have a better understanding of what that means, and are well-practiced at using arithmetic or even things like factoring to garner an equation's solution. This lesson will get our feet wet with doing something to both sides that we really haven't yet - square both sides. This is an action we'll generally only take when we are solving equations that contain a square root.Explicitly, when we say we can "square both sides", we mean that we can raise each side to the power of $2$, and by the golden rule of equations, in conjunction with the property of exponents, the results we get on both sides will still be equal to one another, just as they were before.
You Should Know
You can always raise both sides of an equation to an integer exponent and maintain equality. This action, and a few others (reciprocating both sides, exponentiating both sides, taking the logarithm of both sides, to name some) are actions we just simply haven't seen up to this point because we haven't needed them yet. You'll see them all as you progress through your Calculus readiness courses.

Learn by Example

As we often do when learning something new, we'll walk through an example together.

Example 1Solve the following equation.$$\sqrt{x-1} = 4$$$\blacktriangleright$ If the square root of $x-1$ is $4$, then $x-1$ must have been $16$. Why? Because the square root of $16$ is $4$, so if the square root of $x-1$ is also $4$, then $x-1$ must have been $16$.Another way to say this is that we can square both sides of this equation and maintain equality:$$\sqrt{x-1} = 4$$$$\Rightarrow \big( \sqrt{x-1}\big)^2 = (4)^2$$$$x-1 = 16$$Now we have a standard issue linear equation, which we're well-practiced at solving by now.$$x = 17$$Make sure that solution truly makes the original equation true. We'll talk more below about extraneous solutions, but know now that we'll form a habit of checking our answers when we're solving these types of equations.

Let's formally define a process to solve root equations.
Define: Solving Root EquationsTo solve root equations that contain a single root, take the following steps:1. Isolate the Root2. Square both sides3. Solve the resulting equation4. Check for extraneous solutionsIn other words, once you isolate the root and square both sides, you'll be left with something familiar that you'll know how to solve. And never, ever forget to check for extraneous solutions! They can appear in any root equation. Test your solutions in the original problem to make sure they actually make the equation true.
Here's one more example to walk through together.

Example 2Solve.$$10-\sqrt{3-x} = 7$$$\blacktriangleright$ This time, the square root expression is not isolated - we cannot start this problem off by squaring both sides (see the warning note below). Instead, we should isolate the root term, and then we'll be looking at a problem like the first example, where we can proceed to square both sides.I'll move the $\sqrt{3-x}$ term to the right side, and the $7$ to the left side, so that everything remains positive. This isn't a necessity but it's nicer on the eyes.$$10 - 7 = \sqrt{3-x}$$$$3 = \sqrt{3-x}$$Now we can square both sides.$$3^2 = \left(\sqrt{3-x}\right)^2$$$$9 = 3-x$$$$x = -6$$Check the original equation to make sure this solution is valid:$$10 - \sqrt{3-(-6)} = 7$$$$10 - \sqrt{9} = 7$$$$10 - 3 = 7$$$$7=7$$
Warning!
You cannot square both sides of an equation to "undo" a square root until that square root term is the only thing on that side of the equation. If you square both sides early, you could see via FOIL multiplication that the square root term does not disappear, and you actually will not have made progress on solving the equation. e.g.$$\sqrt{x} + 1 = 4$$$$\left( \sqrt{x} + 1 \right)^2 = 4^2$$$$\left( \sqrt{x} + 1 \right) \left( \sqrt{x} + 1 \right) = 16$$$$x + 2\sqrt{x} + 1 = 16$$This is more complicated than what we started with and did not eradicate the radicand.
Once again, in summary, the key to solving equations that contain a square root is to take three fairly intuitive actions:
• Move everything but the square root expression to the other side
• Square both sides of the equation
• Isolate $x$ with familiar Algebra steps
You Should Know
This lesson focuses on solving equations that have one square root expression in them. It's possible to solve equations that have two or more separate square root equations in them, but the process is a little more involved and is covered in its own Algebra-Two lesson on solving root equations ».

Extraneous Solutions

In a few advanced equation solving techniques, such as this lesson and solving rational equations », you'll need to check for extraneous solutions. Extraneous solutions are solutions that appear to be the answer to an equation as a result of the Algebra steps we take to solve it, but as crazy as it sounds, they make the equation false or undefined when we plug them back into the original equation.Extraneous solutions appear due to the nature of "squaring both sides". Similarly, if you were to multiply both sides by a variable, the potential for extraneous solutions appears as well (again, this happen when we work on solving solving rational equations » later on). Essentially, either squaring both sides or multiplying through by a variable expression creates a new possible answer in addition to what was already present in the equation. Here's a quick example:$$x + 1 = 4$$Clearly the solution to this equation is $x = 3$. However, what if we had multiplied both sides by $x$?$$x^2 + x = 4x$$Now, $x = 0$ would also make the equation true. We created an extraneous solution by multiplying by a variable. The solution $x=0$ does not make the original problem true. The same thing happens when we square both sides of an equation, but it's just not obvious that this happens.
Define: Extraneous Solutions in Root EquationsEvery instance of squaring both sides of an equation creates the potential for an extraneous solution. It does not guarantee that one would exist, but it can happen. For this reason, every time you solve any root equation, you must check your final answer(s) in the original equation to ensure that you are not reporting extraneous solutions.
FYI, we did not have extraneous solutions in Examples 1 and 2 above (these examples were chosen purposefully such that none would surface) but we should check every root equation solution every time. Notice that we still double checked that our answers to Examples 1 and 2 make the original equations true.Here's an example of a root equation that will not take us long to determine that extraneous solutions exist.

Example 3$$\sqrt{x} = -4$$$\blacktriangleright$ We know that square rooting a number will give us a positive result. Just looking at this equation, it's clear there are no solutions because there is no possible way the root of anything will give us $-4$. If we weren't paying attention to that fact and proceeded anyway, the solutions we obtain will be extraneous:$$\big( \sqrt{x} \big)^2 = (-4)^2$$$$x = 16$$But plugging $16$ into the original equation yields$$\sqrt{16} = -4$$$$4 = -4$$This equation is a falsehood.Extraneous solutions usually happen when a square root is set equal to a negative number, since this is impossible. The prevalence of extraneous solutions will increase greatly when we look at solving equations with multiple $x$ terms.
Remember!
You must check every answer to every root equation you ever solve, every time. It's never obvious when a solution is or is not extraneous, without checking by plugging it into the original equation.

Mr. Math Makes It Mean

How do teachers torture you with solving square root equations? One of the biggest traps is the need to check for extraneous solutions, as we discussed. However, there is another common way in which teachers try to throw us off - equations that have a square root variable term but also have another variable term.The major steps and intuition is the same - isolate the root expression and square both sides. However, you'll see that there is some variable expression multiplication (usually FOIL) that must happen to keep the equation true, creating a quadratic equation ». Let's see an example.

Example 4$$\sqrt{19-2x} - 2 = x - 4$$$\blacktriangleright$ We follow the same procedure of needing to isolate the root, and then squaring both sides.$$\sqrt{19-2x} = x - 2$$$$\big( \sqrt{19-2x} \big)^2 = (x-2)^2$$$$19-2x = x^2 - 4x + 4$$Then, procede to solve this quadratic expression by moving everything to the right side.$$0 = x^2 - 2x - 15$$$$0=(x-5)(x+3)$$By the zero product property, the two potential solutions are $x=5$ and $x=-3$. However, like we've discussed, any root equation is at risk for extraneous solutions. Check each one into the original equation:When $x=5$:$$\sqrt{19-2x} - 2 = x - 4$$$$\longrightarrow \sqrt{19 - 2(5)} - 2 = (5) - 4$$$$\sqrt{9} - 2 = 1$$$$1 = 1$$Since we got identity, $x=5$ is a valid solution.When $x=-3$:$$\sqrt{19-2x} - 2 = x - 4$$$$\longrightarrow \sqrt{19 - 2(-3)} - 2 = (-3) - 4$$$$\sqrt{25} - 2 = -7$$$$3 = -7$$This is a falsehood, and so $x=-3$ is not a valid solution to the equation.

Problems with one root expression and other variable expressions are all done similarly. Again, the steps are the same as we've seen throughout the lesson, it's just more work thanks to having a quadratic equation to solve.
You Should Know
Later on in Algebra Two, we'll learn how to solve equations with multiple root expressions » However, some teachers will jump right into root equations with two or more roots in Algebra One, so if your class is tackling more advanced root equations at the same time as basic ones, make sure you can get through both lessons' sets of practice problems with flawless victory.
Remember!
No matter what, if the equation you are solving has only one root expression, isolate the entire root expression and then square each side. Solving root equations is less about steps that you should memorize and more about an intuitive approach to isolating $x$.

Put It To The Test

Solve each of the following equations.

Example 5$$\sqrt{x+6} - 4 = 8$$
Show solution
$\blacktriangleright$ Isolate the root by moving the $-4$ to the other side. Then proceed with our typical plan of squaring both sides and solving from there.$$\sqrt{x+6} = 12$$$$\big( \sqrt{x+6} \big)^2 = (12)^2$$$$x + 6 = 144$$$$x = 138$$As always, don't forget to check for extraneous solutions. Do this by plugging your answer into the original equation.$$\sqrt{138 + 6} - 4 = 8$$$$\sqrt{144} - 4 = 8$$$$8 = 8$$

Example 6$$2\sqrt{1-x} - 9 = 0$$
Show solution
$\blacktriangleright$ Once again isolate the root expression algebraically, then proceed with the method.$$2\sqrt{1-x} = 9$$$$\sqrt{1-x} = \frac{9}{2}$$$$\left( \sqrt{1-x} \right)^2 = \left( \frac{9}{2}\right)^2$$$$1-x = \frac{81}{4}$$$$x = 1 - \frac{81}{4}$$$$x = -\frac{77}{4}$$Even though the answer is a fraction and less fun to work with, you still need to check the solution for validity.$$2\sqrt{1-\frac{-77}{4}} - 9 = 0$$$$2\sqrt{\frac{81}{4}} - 9 = 0$$$$2 \frac{9}{2} - 9 = 0$$$$0 = 0$$

Example 7$$\sqrt{2x+11} - 2 = -5$$
Show solution
$$\blacktriangleright \,\, \sqrt{2x+11} = -3$$$$\big(\sqrt{2x+11}\big)^2 = (-3)^2$$$$2x + 11 = 9$$$$2x = -2$$$$x = -1$$Check the solution in the original problem:$$\sqrt{2(-1)+11} - 2 = -5$$$$\sqrt{9} - 2 = -5$$$$3 - 2 = -5$$$$1 = -5$$Our solution was extraneous. This equation had no solution.

Example 8$$10 - 2\sqrt{5x - 4} = -2$$
Show solution
$\blacktriangleright$ Begin once more with the goal of isolating the root expression. The cleanest way to do this is to move it to the right side and move the $-2$ to the left side:$$10 + 2 = 2\sqrt{5x-4}$$$$12 = 2\sqrt{5x-4}$$$$6 = \sqrt{5x-4}$$$$(6)^2 = \big()\sqrt{5x-4}\big)^2$$$$36 = 5x-4$$$$40 = 5x$$$$x = 8$$Check out the solution for validity:$$10 - 2\sqrt{5(8)-4} = -2$$$$10 - 2\sqrt{36} = -2$$$$10 - 2(6) = -2$$$$-2 = -2$$

Lesson Takeaways
• Understand the major steps needed to solve single root equations
• Know how to isolate the root and square both sides of the equation
• Understand how and why to check for extraneous solutions
• Be aware that equations with $x$ in more than one place are more complicated and are covered in their own lesson »
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Lesson Metrics

At the top of the lesson, you can see the lesson title, as well as the DNA of Math path to see how it fits into the big picture. You can also see the description of what will be covered in the lesson, the specific objectives that the lesson will cover, and links to the section's practice problems (if available).

Key Lesson Sections

Headlines - Every lesson is subdivided into mini sections that help you go from "no clue" to "pro, dude". If you keep getting lost skimming the lesson, start from scratch, read through, and you'll be set straight super fast.

Examples - there is no better way to learn than by doing. Some examples are instructional, while others are elective (such examples have their solutions hidden).

Perils and Pitfalls - common mistakes to avoid.

Mister Math Makes it Mean - Here's where I show you how teachers like to pour salt in your exam wounds, when applicable. Certain concepts have common ways in which teachers seek to sink your ship, and I've seen it all!

Put It To The Test - Shows you examples of the most common ways in which the concept is tested. Just knowing the concept is a great first step, but understanding the variation in how a concept can be tested will help you maximize your grades!

Lesson Takeaways - A laundry list of things you should be able to do at lesson end. Make sure you have a lock on the whole list!

Special Notes

Definitions and Theorems: These are important rules that govern how a particular topic works. Some of the more important ones will be used again in future lessons, implicitly or explicitly.

Pro-Tip: Knowing these will make your life easier.

Remember! - Remember notes need to be in your head at the peril of losing points on tests.

You Should Know - Somewhat elective information that may give you a broader understanding.

Warning! - Something you should be careful about.