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Solving Radical Equations: Examples (page 4 of 6)
This is the same as the previous equation, except that the sign between the radicals has been reversed. And look at the graphs of the lefthand and righthand sides: So this equation does have a solution, at around x = 4. Here is the algebra: ...and here's the check: Since the solution works in the original equation, then the solution is valid, and the answer is: x = 4 Copyright © Elizabeth Stapel 20022011 All Rights Reserved You can use the Mathway widget below to practice solving a radical equation. Try the entered exercise, or type in your own exercise. Then click "Answer" to compare your answer to Mathway's. (Or skip the widget and continue with the lesson.) (Clicking on "View Steps" on the widget's answer screen will take you to the Mathway site, where you can register for a free sevenday trial of the software.)
This already has the square root by itself on one side, so I can proceed directly to squaring both sides. However, a great many students will do the following when given this type of question: <== (wrong!) Do you see how the student erroneously "distributed" the square through the parentheses? Do you see how the student squared terms, not sides? In so doing, the student has arrived at a result which, technically speaking, means that every single value of x will work, since it appears that the equation is always true everywhere. (When would zero not be equal to zero, right?) But the graph of the equations of the two sides:
...shows otherwise: And, from your experience
graphing straight
lines and radical
functions, you
should already have known that there was no way that a curvy radical
line could possibly be the same as a straight line such as y
= 3x + 2. So don't square terms; square sides! And take the time to write out the square properly: This matches the graph above. Now, checking: So the solution is x = 0. << Previous Top  1  2  3  4  5  6  Return to Index Next >>



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