Monday 6 March class

(after Test 1)

Topics: (from Section 6.7 in the textbook)

• Introduction to Radical Equations

• Solving radical equations, part 1

 

Our method for solving radical equations is as follows:

• First, isolate the radical (if necessary)

• Then square both sides of the equation, using the fact that

\left(\sqrt{THING}\right)^{2} = THING

– this is just the definition of the square root!

•  Then solve the resulting equation

• And finally, check all solutions in the original equation to see if they are genuinely solutions or not (and of course, to see if we made any errors!)

 

The reason that we need to check in solving radical equations (it’s not just a good idea, it’s the law!) is because of the following phenomenon:

Suppose we want to solve \sqrt{x} = -3

We should realize at the beginning that there is no solution to this equation, because the radical sign tells us to take the principal square root, which cannot be a negative number. But suppose that (as we too often do) we just mechanically go on working this problem without realizing that. So we try squaring both sides of the equation:

\left(\sqrt{x}\right)^{2} = (-3)^{2}

 

x = 9

But this does not solve the original equation: when we go to check, we get

\sqrt{9} = -3?

 

3 \neq -3

The problem is that when we squared both sides of the equation, we lost the information that the left-hand side of the equation represented a non-negative real number, but the right-hand side was negative. After squaring, both sides became positive.

Notice that x=9 would work if the radical sign allowed us to take a negative number square root. This is the “symptom” that shows that this is what is sometimes referred to as an “extraneous solution” – that is, it is not a solution at all, but just something extra that our method generated.

 

(A similar thing causes problems when we solve rational equations. There, when we clear the denominators, we are multiplying both sides of the equation by the same thing, namely the LCM of all the denominators. This is fine as long as the thing we are multiplying by is not 0, but the problem is that we don’t know whether or not the LCM is 0 because it has a variable in it! So in that case the “symptom” of the alleged solutions that ended up having to be thrown away is that they led to a 0 denominator when we went to check.)

 

I worked out Example 1 and Exercises 14 and 16 from Section 6.7 in class, not forgetting to check before reporting the final results.

 

Homework:

• Review the examples discussed in class, including the one included in these notes.

• Do the following from the textbook: p. 547 #11, 12, 13, 15, 21, 22

• Also do (simplifying radical expressions, including rationalizing the denominator) from the Course Outline, the assigned problems from Section 6.6 (starting on p. 538)

• Do the WeBWork: some of this your should have already started and it is due by tomorrow 11 pm (but don’t wait to the last minute!) There is also a new assignment on radical equations which is not due until Sunday night: you should try to do at least the first 4 problems of that one by Wednesday’s class.

Remember that you can use the Piazza discussion board to ask questions if you get stuck on any of the WeBWorK or the other homework problems. Don’t forget to include the problem itself in your question, as that will make it easier for you to get a quick response!

 

 

 

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