Hi Everyone im having trouble with the above equation and would appreciate any help. Thanks guys Pete
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a^2-b^2=(a-b)(a+b)
Originally Posted by petemg Hi Everyone im having trouble with the above equation and would appreciate any help. Thanks guys Pete As you have an equation in the form $\displaystyle a^2-b^2$ it makes sense to use the difference of two squares
i get that i need to use difference of squares and have completed questions with like (x+4)^2 - 3^2 no problem, its the 2 sets of brackets confusing me
$\displaystyle a = 2x+1 \text{ and } b = x-2$ Hence $\displaystyle (2x+1)^2 - (x-2)^2 = [(2x+1)-(x-2)] \times [(2x+1)+(x-2)]$ You can simplify further in the brackets
Thanks for that, would i be right with my next step, (-2x^2-x-2) (2x^2+x-2)
No. Read again what -1 wrote you... post #5
where am i going wrong? do i not multiply 2x by -x and get -2x^2
Originally Posted by petemg where am i going wrong? do i not multiply 2x by -x and get -2x^2 I think that I know your problem... read hear to understand what factor means... Divisor - Wikipedia, the free encyclopedia
Im still confused Do i multiply the left side by the right? and not 2 the two brackets to the left multiplied by each other?
Originally Posted by e^(i*pi) $\displaystyle a = 2x+1 \text{ and } b = x-2$ Hence $\displaystyle (2x+1)^2 - (x-2)^2 = [(2x+1)-(x-2)] \times [(2x+1)+(x-2)]$ You can simplify further in the brackets (2x+1)^2 - (x-2)^2 = [(2x+1)-(x-2)] *[(2x+1)+(x-2)]=(2x+1-x+2)*(2x+1+x-2)=(x+3)*(3x-1) So... (x+3) and (3x-1) are factors of (2x+1)^2 - (x-2)^2. It's like 3 and 2 are factors of 6(=2*3)
would i be left with 6x^2+3x-1?
sorry just seen you wrote the method there, how else could you get the same answer using different method?
Originally Posted by petemg sorry just seen you wrote the method there, how else could you get the same answer using different method? You can expand (2x+1)^2 - (x-2)^2 and then to find the roots using the famous formula x_1,2={-b+/-sqrt(b^2-4ac)}/2a
and how would that look? its really the 2 brackets confusing me here, i can do the (b^2-4ac)}/2a with 1 bracket
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