anonymous
  • anonymous
Find the quadratis equation having the given roots. {-3,-7}
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
lol loki, that's a funny face :D
anonymous
  • anonymous
mr. track, if alpha and beta are the roots of a quadratic equation, then that quadratic may be factored as\[(x-\alpha)(x-\beta)\]Expanding, you have,\[x^2-(\alpha+\beta)x+\alpha \beta\]Here, \[ \alpha = -3\]and \[\beta=-7\]so your equation is\[x^2-(-3+-7)x+(-3)(-7)\]\[=x^2+10x+21\]
anonymous
  • anonymous
That is, since your roots are -3 and -7, when you factor the quadratic that allows for this to be the case, you'd have\[(x-(-3)(x-(-7))=(x+3)(x+7)\]

Looking for something else?

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

More answers

amistre64
  • amistre64
"Find the quadratic equation having the given roots." Wouldnt that be find "a" quadratic equation given those roots? cause we can have the same root, and the same axis of symmetry, but e different vertex along the axis which would constitute a different equation with the same root. right?
amistre64
  • amistre64
*but have a different vertex...
amistre64
  • amistre64
the only way I know of to pinpoint a quadratic is with 3 known points.
amistre64
  • amistre64
if we use k as a constant then: all the parabolas that fit the equation would be: k(x^2 +10x +21) right?
anonymous
  • anonymous
Yes, that's correct. I read the question as 'a' parabola...was late.

Looking for something else?

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