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plzz tell me how to complete a square 3x^2 + 6x -4 can u write it in this form? p(x-q) +r
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re-write 3x^2 + 6x -4 as 3 ( x^2 + 2x) -4 inside the parens we have x^2 + bx where b is the coefficent. we want to add (b/2)^2 to complete the square here b=2, so (b/2)^2 = 1^2= 1. But to not change the value we add 1 -1 (make sense?) 3 (x^2 +2x +1 -1) -4 if you work it out, x^2+2x+1 = (x+1)^2 so we now have 3 ( (x+1)^2 -1) -4 or multiplying by 3 3(x+1)^2 -3 -4 = 3(x+1)^2 -7
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