
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 20022011 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:
The squared expressions can be graphed as the lines y = x – 1 and y = x^{2} – 14x + 49. The solutions of x – 1 = x^{2} – 14x + 49 are the intersection points of the two lines: As you can see, the intersection point at x = 10, from the first graph, is still there, but now a second, extraneous, solution has appeared at x = 5! ("Extraneous", pronounced as "ekSTRAYneeuss", in this context means "mathematically correct, but not relevant or useful, as far as the original question is concerned".) Continuing the solution: x – 1 = x^{2} – 14x + 49 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 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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