anonymous
  • anonymous
Can you help me see where I am stuck finding the zeros of this quadratic equation?
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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
\[3x^2 - 12x + 11 = 0\]
anonymous
  • anonymous
quadratic formula for this one use \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with \[a=3,b=-12,c=11\]
anonymous
  • anonymous
My student solutions manual says I should obtain \[(6 \pm \sqrt{3}) \div 3\]

Looking for something else?

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

More answers

anonymous
  • anonymous
indeed you should
anonymous
  • anonymous
Let's walk through this, so I can see where I am tripping over my shoelaces.
anonymous
  • anonymous
first you get \[\frac{-12\pm\sqrt{(-12)^2-4\times 3\times 11}}{6}\]
anonymous
  • anonymous
Wait a sec. I see that I made an error in the first term under the radical. I will go back to the drawing board and re-attempt this one on my own.
anonymous
  • anonymous
By the way, how are you producing the nice looking divisor line?
anonymous
  • anonymous
\frac{a}{b} looks like \[\frac{a}{b}\]
anonymous
  • anonymous
Got it. I made simple mistakes in my substitutions, then got hung up on trying to do algebra with unduly complicated figures.
anonymous
  • anonymous
THANKS @satellite73, you got me unstuck!

Looking for something else?

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