Mathematics, 20.04.2020 16:04 aquinomoises518

Step A: a (x + StartFraction b Over 2 a EndFraction) squared = –c + StartFraction b squared Over 2 a EndFraction a (x + StartFraction b Over 2 a EndFraction) squared = StartFraction negative 4 a c + b squared Over 4 a EndFraction Step B: a (x + StartFraction b Over 2 a EndFraction) squared = StartFraction negative 4 a c + b squared Over 4 a EndFraction (one-half) a (x + StartFraction b Over 2 a EndFraction) squared = (StartFraction 1 Over a EndFraction)(StartFraction b squared minus 4 a c Over 4 a EndFraction) Determine the justification for the steps from the derivation of the quadratic formula. Justification of step A: Justification of step B:

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Answer from: ijohnh14

Justification of step A: common denominator

Justification of step B: multliplication property of equality

Step-by-step explanation:

i got it right.

Answer from: officialrogerfp3gf2s

The justification for the steps from the derivation of the quadratic formula of step A is the use of common denominator.

The justification for the steps from the derivation of the quadratic formula of step A is the use of multiplication property of equality.

The given equations for step A;

$a(x + \frac{b}{2a} )^2 = -c + \frac{b^2}{2a} \\\\a (x + \frac{b}{2a} )^2= \frac{-4ac + b^2}{4a}$

The given equations for step B;

$a (x + \frac{b}{2a} )^2 = \frac{-4ac + b^2}{4a} \\\\\frac{1}{2} a(x + \frac{b}{2a} )^2 = (\frac{1}{a} )(\frac{b^2- 4ac}{4a} )$

The justification for the steps from the derivation of the quadratic formula of step A is determined as follows;

The common denominator 4a was used to convert the mixed fraction into improper fraction.

The justification for the steps from the derivation of the quadratic formula of step B is determined as follows;

The use of multiplication property of equality.

Learn more about derivation of quadratic formula here:

Answer from: Quest

overpriced

step-by-step explanation:

Answer from: Quest

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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Step A: a (x + StartFraction b Over 2 a EndFraction) squared = –c + StartFraction b squared Over 2 a...