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carlenstar281
math hw question
Difference of two perfect squares.. (x-9)(x+9)
only if I get a medal.. haha jk sure :P
sorry lol i thought i gave you one
Well first we have to get the c term by using (b/2)^2 \[(-\frac{ 6 }{ 2 })^2=9\] So now add that to both sides: \[(x^2-6x+9)=-8+9\] Simplify: \[(x-3)^2=1\]
Now take the square root of both sides: \[\sqrt{(x-3)^2}=\pm \sqrt{1}\] Since the square root of 1 is 1 we leave it: \[x-3= \pm1\] Now add 3: \[x-3+3=\pm1+3\] \[x=\pm1+3\] Separate and do then separately: \[x=1+3=4\] \[x-1+3=2\] And there you go :)
oh wow you make it easy thanks for showing me how you do it not i will be able to do the rest on my own thanks so much :)
No problem, happy to help :D