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anonymous
 5 years ago
Find the quadratis equation having the given roots.
{3,7}
anonymous
 5 years ago
Find the quadratis equation having the given roots. {3,7}

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol loki, that's a funny face :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mr. 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
 5 years ago
Best ResponseYou've already chosen the best response.0That is, since your roots are 3 and 7, when you factor the quadratic that allows for this to be the case, you'd have\[(x(3)(x(7))=(x+3)(x+7)\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0"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
 5 years ago
Best ResponseYou've already chosen the best response.0*but have a different vertex...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the only way I know of to pinpoint a quadratic is with 3 known points.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if we use k as a constant then: all the parabolas that fit the equation would be: k(x^2 +10x +21) right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, that's correct. I read the question as 'a' parabola...was late.
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