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 Question from marquis, a student: 3y^2 - 48 = 0

Hi Marquis.

Whenever I see a quadratic with a positive A term, negative C term and no B term, I think "difference of squares". (Ax2 + Bx + C = 0)

Thus, 49x2 - 144 = 0 is an equation involving a difference of squares. It factors into two linear terms differing only in a plus/minus for the constant:
49x2 - 144 = (7x - 12) (7x + 12)

Since this equals zero, then one (or both) of the factors must equal zero, because only zero times something equals zero. That means either
7x - 12 = 0
or
7x + 12 = 0

So we solve them separately:
7x - 12 = 0
7x = 12
x = 12/7

and

7x + 12 = 0
7x = -12
x = -12/7

Thus the solution set (abbreviated "S.S.") is written:
S.S. = { -12/7, 12/7 } or you could write it as just S.S = { ±12/7 }

I'm sure you have already done this kind of thing before. But even when the values are not "perfect" squares, it also works. Here's another example:

12x2 - 5 = 0

The factors use the square roots of each term, so it factors to:
(2x√3 - √5) ((2x√3 + √5) = 0

And using the rules for working with radical expressions, these simplify
to S.S. = { ±√(15) / 6 }

Cheers,
Stephen La Rocque.

Doh!

3y2 - 48 = 0.

Start by dividing both sides by 3, then see if it looks familiar.

(Why do I do things the hard way first?)

:) It happens to us all!

Cheers,
Stephen.

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