anonymous
  • anonymous
math hw question
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

Luigi0210
  • Luigi0210
Difference of two perfect squares.. (x-9)(x+9)
Luigi0210
  • Luigi0210
(s-9)(s+9)
Luigi0210
  • Luigi0210
only if I get a medal.. haha jk sure :P

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
sorry lol i thought i gave you one
Luigi0210
  • Luigi0210
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: \[(x-3)^2=1\]
Luigi0210
  • Luigi0210
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 :)
anonymous
  • anonymous
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 :)
Luigi0210
  • Luigi0210
No problem, happy to help :D

Looking for something else?

Not the answer you are looking for? Search for more explanations.