SOLUTION: can you help me solve the equation by finding square roots of 2x^2=12 and 147x^2=27

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Question 423835: can you help me solve the equation by finding square roots of 2x^2=12 and 147x^2=27
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Both equations are quadratic equations because of the squared terms. With quadratic equationss you start by making one side zero and then factoring (or use the Quadratic Formula).
2x%5E2=12
2x%5E2-12+=+0
2%28x%5E2-6%29+=+0
Normally one would say that x%5E2-6 will not factor further. But let's go ahead and treat it as a difference of squares and factor it with the a%5E2-b%5E2+=+%28a%2Bb%29%28a-b%29 pattern. But, you may ask, how is 6 a perfect square? Answer: It is not a square of an integer. But it is the square of something. It is the square of sqrt%286%29! So we can factor x%5E2-6 with the "b" being sqrt%286%29:
2%28x%2Bsqrt%286%29%29%28x-sqrt%286%29%29+=+0
From the Zero Product Property we know that one of these factors must be zero. The 2 is a 2, of course, and cannot be zero. But the other two factors can:
x%2Bsqrt%286%29+=+0 or x-sqrt%286%29+=+0
Solving these we get:
x+=+-sqrt%286%29 or x+=+sqrt%286%29

147x%5E2=27
147x%5E2-27=0
3%2849x%5E2-9%29=0
This time we have an obvious difference of squares:
3(7x+3)(7x-3)=0
7x+3 = 0 or 7x-3 = 0
x = -3/7 or x = 3/7