Am I doing this right?

T

TheAmandalarson

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Sep 29, 2010
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I'm given the problem 49(x-4)^2 - 2 = 0 and it says solve.
So I started off by adding 2 to both sides.
The negative 2 cancels and I'm left with: 49(x-4)^2= 2
I then divided 49 by both sides and take the square root, so I'm left with:
(x-4)= (?2)/7

Is that my answer, or do I go further?
 
Never do that. Solve in ways that are more thorough and do not require you to keep a list. ALWAYS just reach out and grab the easy ones.

Recognize a Difference of Squares.

\(\displaystyle 49\cdot (x-4)^2 - 2 = 0\)

\(\displaystyle (7\cdot (x-4) + \sqrt{2})\cdot (7\cdot (x-4) - \sqrt{2}) = 0\)

Now what?
 
tkhunny said:
Never do that. Solve in ways that are more thorough and do not require you to keep a list. ALWAYS just reach out and grab the easy ones.

Recognize a Difference of Squares.

\(\displaystyle 49\cdot (x-4)^2 - 2 = 0\)

\(\displaystyle (7\cdot (x-4) + \sqrt{2})\cdot (7\cdot (x-4) - \sqrt{2}) = 0\)

Now what?
Click to expand...


I have no idea... I'm not use to doing it that way.
 
TheAmandalarson said:
(x-4)= (?2)/7
Is that my answer, or do I go further?
Click to expand...
Well, if you haven't been taught diff. of squares yet, what have you beem taught?
You should at least know that you're supposed to solve for x; x = something.
Can you do that?
 
TheAmandalarson said:
I have no idea...
Click to expand...

You used the magic words. Lecture time! If you REALLY have no idea, you are in the wrong class or you're simply not paying attention. Never think that again.

You've NEVER seen anything like this:

x^2 - 16 = 0

(x+4)(x-4) = 0

x+4 = 0 ==> x = -4
x-4 = 0 ==> x = 4

It's called Factoring.
 
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