Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

A curve with the equation y= -(x+p)(x+q) cuts the x-axis at the points A(-2,0) and B(6,0). Write down the values of p and q.

I got my questions answered at in under 10 minutes. Go to 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.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

the curve cuts the axis when y=0
So sub A(-2,0) into the equation? I got stuck because there were too many unknowns :/
Write a system of two equations. For the first equation, replace x with -2 and y with 0. For the second equation replace x with 6 and y with 0. Now you have two equations with variables p and q. Solve the system.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

I got stuck again. Ended up with q+p=-4 .
p+q=-4 is correct, now substitute back in pq+36+6(p+q)=0 to get pq=-12 can u find p and q knowing p+q=-4 and pq= -12 ???
yeah I got it (: Thank you!
welcome :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question