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Solving Radical Equations: Examples (page 2 of 6)
The two lines represented by the two sides of this equation are:
...and they graph as: Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved
...so you can see that there should be a solution at or about x = 10. To solve this algebraically, I need to square each side:
Before I finish solving, note that, to solve this last equation (which is the result of my having squared both sides of the original equation), it can be graphed as the lines y = x – 1 and y = x2 – 14x + 49. The solutions of x – 1 = x2 – 14x + 49 are the intersection points of the two lines:
As you can see, our original solution at x = 10 is still there, but now another, extraneous, solution has appeared at x = 5! ("Extraneous", pronouced "ek-STRAY-nee-uss", in this context means "mathematically correct, but not relevant or useful, as far as the original question is concerned".) Continuing the solution: x
– 1 = x2 – 14x + 49
So I got the result that the second graph led us to expect, but we also know, from the first graph, that "x = 5" should not be a solution. This is why you always need to check your answers when solving radical equations: the very act of squaring has, in this case, produced an extra and incorrect "solution". Here's my check: x = 5:
x = 10:
So the answer is x = 10.
Since this equation is in the form "(square root) = (number)", I can proceed directly to squaring both sides:
This solution matches what I would expect from the graph of the two sides of the equation:
As you can see above, the lines:
...intersect at x = 27, as the algebra had already shown me. Checking, I get:
So the solution is x = 27. << Previous Top | 1 | 2 | 3 | 4 | 5 | 6 | Return to Index Next >
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Copyright © 2006-2008 Elizabeth Stapel | About | Terms of Use |
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