



 
Hi Marquis. Whenever I see a quadratic with a positive A term, negative C term and no B term, I think "difference of squares". (Ax^{2} + Bx + C = 0) Thus, 49x^{2}  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: Since this equals zero, then one (or both) of the factors must equal zero, because only zero times something equals zero. That means either So we solve them separately: and 7x + 12 = 0 Thus the solution set (abbreviated "S.S.") is written: 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: 12x^{2}  5 = 0 The factors use the square roots of each term, so it factors to: And using the rules for working with radical expressions, these simplify Now you try your question. Cheers, Doh! 3y^{2}  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,  


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