anonymous
  • anonymous
Please give an easy and fast way to solve this:
Mathematics
katieb
  • katieb
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

anonymous
  • anonymous
\[x ^{2}-xc-2c ^{2}=0\]
anonymous
  • anonymous
Use the quadratic formula.
anonymous
  • anonymous
\[(a+b^{2}=a ^{2} + 2ab + b ^{2}\]

Looking for something else?

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

More answers

anonymous
  • anonymous
solve what?
anonymous
  • anonymous
but isn't' a quadratic equation of the type: \[a ^{2}+2ab+b ^{2}=0\] OR \[a ^{2}-2ab+b ^{2}=0\] My equation has two -ve signs.
anonymous
  • anonymous
oh
anonymous
  • anonymous
You have to get x = 2c
anonymous
  • anonymous
(x+a)*(x-b)
anonymous
  • anonymous
Please show the steps.
anonymous
  • anonymous
x^2−xc−2*c^2=0 (x+c)(x-2c)=0
yuki
  • yuki
\[x = {c \pm \sqrt{c^2 - 4*z*(-2c^2)} \over 2}\]
yuki
  • yuki
lol
anonymous
  • anonymous
x^2-x c-2c^2=0 x=-c+-sqrt(c^2-4*1*2c^2)/2*1
anonymous
  • anonymous
OK. Thank You.
anonymous
  • anonymous
c can be treated as a constant
anonymous
  • anonymous
\[x^{2}-xc-c ^{2}=x^{2}-2xc+xc-c ^{2}=(x-2c)(x+c)=0 so we have x=2c or x=-c\]

Looking for something else?

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