Solving equations using square roots
- Thread starter kdtmac
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- Apr 12, 2005
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1) Don't use square roots as a method. They may be the result of other, appropriate methods.
2) You simply must learn to recognize certain patterns that will simplfy your life. The "Difference of Squares" is one such pattern. They are very easy to factor.
3) This one isn't quite a difference of squares, but it can be made to be such.
3x^2 - 27 = 0
Divide by 3
x^2 - 9 = 0
That is a Difference of Squares and is easily factored.
(x+3)(x-3) = 0
Therefore, x = 3 or x = -3
I didn't see a square root in there. Why do you believe you needed one?
2) You simply must learn to recognize certain patterns that will simplfy your life. The "Difference of Squares" is one such pattern. They are very easy to factor.
3) This one isn't quite a difference of squares, but it can be made to be such.
3x^2 - 27 = 0
Divide by 3
x^2 - 9 = 0
That is a Difference of Squares and is easily factored.
(x+3)(x-3) = 0
Therefore, x = 3 or x = -3
I didn't see a square root in there. Why do you believe you needed one?
I'm guessing the teacher wanted the following:
3x^2 - 27 = 0
Factor.
3(x^2 - 9) = 0
Divide both sides by 3.
x^2 - 9 = 0
Add 9 to both sides.
x^2 = 9
Take the square root of both sides.
\(\displaystyle x=\pm 3\)
In the long run, do as TKH says. Don't use this method. Use the method he showed you. The problem with this method is we too often forget that when we take the square rood of a positive number we get two roots, one positive and the other negative.
3x^2 - 27 = 0
Factor.
3(x^2 - 9) = 0
Divide both sides by 3.
x^2 - 9 = 0
Add 9 to both sides.
x^2 = 9
Take the square root of both sides.
\(\displaystyle x=\pm 3\)
In the long run, do as TKH says. Don't use this method. Use the method he showed you. The problem with this method is we too often forget that when we take the square rood of a positive number we get two roots, one positive and the other negative.