equations with radical expressions

Solving Radical Equations: Introduction (page 1 of 6)A "radical" equation is an equation in which at least one variable expression is stuck inside a radical,usually a square root.

For instance, this is a radical equation:

...but this is not:

The "radical" in "radical equations" can be any root, whether a square root, a cube root, or some otherroot. Most of the examples in what follows use square roots as the radical, but (warning!) you should notbe surprised to see an occasional cube root or fourth root in your homework or on a test.

In general, you "solve" equations by "isolating" the variable; you isolate the variable by "undoing"whatever had been done to it.

For instance, given x + 2 = 5, you would solve byundoing the addition of the 2. That is, the additionundone by applying the opposite: subtraction:

In the same manner, given something like 3x = 12,you would solve by undoing the multiplication byapplying the opposite operation; namely, division:

When you have a variable inside a square root, youundo the root by doing the opposite: squaring. Forinstance, given , you would square both sides:

Issue 1:There are a couple of issues that frequently arise when solving radical equations. The first is that youmust square sides, not terms. Here is a classic example of why this is so:

I start with a true equation and then square both sides:

3 + 4 = 7

(3 + 4)2 = 72 49 = 49

...but if I square the terms on the left-hand side:

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3 + 4 = 7

32 + 42 "=" 72

9 + 16 "=" 49

25 "=" 49 ...............Oops!

In each case, I started with a true statement; namely, 3 + 4 = 7. When I squared both sides, I alsoended with a true statement: 49 = 49. But when I squared the terms, 32 + 42, I ended up withsomething that was not true: 25 "=" 49. The most common mistake that students make when solvingradical equations is squaring terms instead of sides. Don't make this mistake! You should alwaysremember to:

** SQUARE SIDES, NOT TERMS **

Issue 2:The other issue is that you will need to check your answers. You can always check your answers in asolved equation by plugging your answer back into the original equation and making sure that it fits.

For instance, in my first example above, you can check that I got the correctanswer by plugging 3 in for x and verifying that the equation is still true:

x + 2 = 5(3) + 2 = 5 5 = 5

You probably did some of this type of checking when you first starting solving linear equations. Buteventually you developed your skills, and you quit checking. The difficulty with radical equations is thatyou may have done every step correctly, but your answer may still be wrong. This is because the veryact of squaring the sides can create solutions that never existed before.

For instance, I could say "2 = 2", and you would know that thisis false. But look what happens when I square both sides:

(2)2 = 22 4 = 4

I started with something that was not true, squared both sides of it, and ended with something that wastrue. This is not good!

A more pertinent example would be this:

This "equation" is no more true than the "2 = 2" "equation" above, because no positive squareroot can ever equal a negative number.

But suppose I hadn't noticed that this equation has nosolution, and had proceeded to square both sides:

By squaring, I created a solution ("x = 9") that had notexisted before and is in fact not valid. But I won't discoverthis error unless I remembered to check my solution:


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So the actual answer for the equation sqrt(x) = 3 is "no solution".

There is another way to look at this "no solution" difficulty: When you are solving an equation, you canview the process as trying to find where two lines intersect on a graph.

For instance, when I was solving "x + 2 = 5" above,you could also say that I was trying to find the intersectionof y1 = x + 2 (from the left-hand side ofx + 2 = 5) and y2 = 5 (from the right-hand side):

As you can see in the graph above, the two lines intersect at x = 3, which was the solution we hadalready found. Similarly, when I was solving the equation , I was also trying to find theintersection of y1 = sqrt(x) and y2 = 4:

As the above graph displays, the solution is at x = 16.

But when I was trying to solve the equation:

...I was trying to find the intersection of y1 = sqrt(x)and y2 = 3, which do not intersect.

(Note: If you don't know how I got the curvy blue line in thegraph at right, then review how to graphradical equations.)


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So what happened when I squared both sides? I also"squared" both line equations, and got the two new linesy1 = x and y2 = 9. And, as the graph shows, these twolines actually do intersect!

This is how squaring created a solution where there hadn't been a solution before. But the after-squaring solution did not work in the before-squaring equation, because the original lines had notintersected. This illustrates why checking the solution showed that the real answer was "no solution".

Warning: Many instructors do not to show many examples (in class or in the homework) of radicalequations for which the solutions don't actually work. But then they'll put one of these on the test. Youshould expect a "no solution" radical equation on the test, so you do not want to forget to check yoursolutions!

** CHECK ALL SOLUTIONS **

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Solving Radical Equations: Examples (page 2 of 6)Solve the equation:

The two lines represented by the two sides of this equation are:

...and they graph as:

...so you can see that there should be a solution at or about x = 10. To solve this algebraically, Ineed to square each side:

x 1 = (x 7)(x 7) x 1 = x2 14x + 49

The squared expressions can be graphed as the lines y = x 1 and y = x2 14x + 49. Thesolutions of x 1 = x2 14x + 49 are the intersection points of the two lines:


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As you can see, the intersection point at x = 10, from the first graph, is still there, but now asecond, extraneous, solution has appeared at x = 5! ("Extraneous", pronounced as "ek-STRAY-nee-uss", in this context means "mathematically correct, but not relevant or useful, as far as theoriginal question is concerned".) Continuing the solution:

x 1 = x2 14x + 49

0 = x2 15x + 50

0 = (x 5)(x 10)x = 5, x = 10

So I got the result that the second graph led me to expect, but I also know, from the first graph,that "x = 5" should not be a solution. This again illustrates why you always need to check youranswers when solving radical equations: the very act of squaring has, in this case, produced anextra and incorrect "solution". Here's my check:

x = 5:

x = 10:

So the answer is x = 10.

Solve the equation:


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Since this equation is in the form "(square root) = (number)", I can proceed directly to squaringboth sides:

x 2 = 25

x = 27

This solution matches what I would expect from the graph of the two sides of the equation:

As you can see above, the lines:

y = 5

...intersect at x = 27, as the algebra had already shown me. Checking, I get:

So the solution is x = 27.

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I could square both sides now, but look what I would get:

So, while squaring both sides at this point would not be "wrong", it would not be the most usefulfirst step. Instead of squaring right away, I will first move the 2 over to the right-hand side, so theradical will be by itself on the left:

Now squaring both sides will work better:

Checking, I get:


Find the solution:

This problem is a bit more messy than the previous two. I cannot get the radical by itself on oneside, because there are two radicals. So how can I solve this algebraically? By squaring bothsides twice. Here's what it looks like:


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Warning: Do NOT try to do these steps in your head. Take the time to write things out completely,so you won't make mistakes!

Checking my solution, I get:

Hmm... According to this, there is no solution. I'll check the graph of the two lines:

...to see if it looks like there ought to be a solution:

No; according to the graph, it does not appear that these lines intersect (and calculus techniquescan prove this). Why did it appear that there was a solution? Look at the graphs from the secondsquaring:


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y = x2 12x + 36

y = x2 3x

So I came up with an algebraic solution because I had accidentally created one by my repeatedsquaring. But that "solution" didn't check out in the original equation, so the actual answer is thatthere is no solution.

On the other hand, look at the following...




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Solving Radical Equations: Examples (page 4 of 6)Find the solution:

This is the same as the previous equation, except that the sign between the radicals has beenreversed. And look at the graphs of the left-hand and right-hand sides:

So this equation does have a solution, at around x = 4. Here is the algebra:

...and here's the check:


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Since the solution works in the original equation, then the solution is valid, and the answer is:

x = 4

Find the solution:

This already has the square root by itself on one side, so I can proceed directly to squaring bothsides. However, a great many students will do the following when given this type of question:

This matches the graph above. Now, checking:





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Square both sides, being careful to write out the square on the right-hand side:

Then x = 8 and x = 2. Are both of these solutions valid? Graphing the lines for either side ofthe original equation:

...I get the following graph:

It appears that both solutions are valid. Here's the check:

x = 8:

x = 2:


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So the solution is x = 8 or x = 2.

The following examples are not complete. I'll leave the checking to you!

Solve:

This equation will have to be squared twice in order to solve it:

It appears that the solutions are x = 5 and x = 0. However, only one of these solutions is actuallyvalid. To find out which one, check them both.

Solve:

This equation will also have to be squared twice. Don't forget to square that 3 in front of thesquare root on the right-hand side!


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To solve this, use the Quadratic Formula. Then check your answers, because only one is actually valid.

Solve:

This one is actually simpler than the two previous examples, because the two square roots aremultiplied together, rather than added or subtracted. So this equation will need to be squaredonly once:

Then the solutions are x = 9 and x = 16. But x cannot equal 9, because this would put negativesinside both radicals in the original equation. Now you check the other solution, to see if it might work.

Solve:

Since there is a square root inside a square root, I'll have to square twice:

Using the Quadratic Formula, I get solutions of x = 401/144 and x = 3. Check these, as only one is a


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valid solution.




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Solving Radical Equations: Higher-Index Examples (page 6 of 6)

Solve the equation:

Since this is a CUBE root, ratherthan a square root, I undo theradical by cubing both sides of theequations, rather than squaring:

I must remember to check mysolution:

3 = 3 ...yes!


Solve the equation:

(Note that the "plus one" is outside the cube root.)

Since this is a cube root, I'll cubeboth sides to undo the radical. Butfirst, I want to isolate the radical:


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Remember to check the solution:

So the solution is x = 1/3.

Solve the equation:

Since this is a fourth root, I'll raise bothsides to the fourth power:

Then I'll check my answers: x = 1/2:

I'll leave the other check for you. However, the graph does indicate that both solutions are valid.

Graphing the left- and right-hand sides ofthe original equation:

...you get the picture at right:


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Zooming in, you can see that the linesseem to intersect...

...and, zooming in some more, you cansee the two solutions:

Remember that you can't have negativesinside a fourth root. That's why the greenline is broken into pieces like that: youcan only graph where x4 + 4x3 x isnon-negative, which occurs in threepieces, where the graph is at or abovethe x-axis.

Then the solution is x = 1/2, 1/3.

Since cube roots can have negative numbers inside them, you don't tend to have the difficulty with themregarding checking the answers that you did with square roots. However, you will have those difficultieswith fourth roots, sixth roots, eighth roots, etc; namely, any even-index root. Be careful!

You may or may not be required to show solutions graphically, but if you have a graphing calculator (sodrawing the graphs is just a matter of quickly punching a few buttons), you can use the graphs to checkyour work on tests. In any case, be careful with your squaring ("Square sides, not terms!"), do each stepcarefully, and don't forget to "Check your solutions!"



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equations with radical expressions

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