Find the quadratis equation having the given roots. {-3,-7}
lol loki, that's a funny face :D
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\[x^2-(-3+-7)x+(-3)(-7)\]\[=x^2+10x+21\]
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\[(x-(-3)(x-(-7))=(x+3)(x+7)\]
"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?
*but have a different vertex...
the only way I know of to pinpoint a quadratic is with 3 known points.
if we use k as a constant then: all the parabolas that fit the equation would be: k(x^2 +10x +21) right?
Yes, that's correct. I read the question as 'a' parabola...was late.
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