SOLUTION: give exact and approximate solutions to three decimal place y^2-14y+49=25

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Question 626231: give exact and approximate solutions to three decimal place
y^2-14y+49=25

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
y² - 14y + 49 = 25

You are able to recognize the right side, 25, as the square of an integer,
since 25 = 5², from your knowledge of basic math.

You should now be able to recognize the left side, the trinomial 
y² - 14y + 49, as the square of a binomial, from your knowledge of algebra.

That's because it has four properties:

1.  The first term is a square, (y² is the square of y)
2.  The third term +49 is also a square, (+49 is the square of 7)
3.  The middle term, ignoring the sign, is twice the product of the square roots
    of the first and third terms (-14y, ignoring the sign, 14y, is twice the
    product of the square root of y², which is y, and the square root of 49,
    which is 7, since 2·y·7 = 14y.)
4.  It factors as the square of the sum of the square roots, using the
    sign of the middle term as the sign of the second term of the binomial.

That's a lot to swallow, but it is to your advantage to be able to recognize
that the left side factors as (y - 7)². So we have

y² - 14y + 49 = 25
     (y - 7)² = 25

Now we use the principle of square roots:

      y - 7 = ±5
          y = 7 ± 5

Using the + sign, y = 7 + 5 = 12
Using the - sign, y = 7 - 5 = 2

Edwin