Proving Quadratic Formula

How does

logosomatw
2022-01-23
Answered

Proving Quadratic Formula

How does

You can still ask an expert for help

bemolizisqt

Answered 2022-01-24
Author has **16** answers

Step 1

Remember how to complete the square:

$A{x}^{2}+Bx=A{(x+\frac{B}{2A})}^{2}-\frac{{B}^{2}}{4{A}^{2}}$

So now

$a{x}^{2}+bx+c=0$ complete square

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

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

$x}_{1,\text{}2}+\frac{b}{2a}=\pm \frac{\sqrt{{b}^{2}-4ac}}{2a$

$x}_{1,\text{}2}=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a$

Remember how to complete the square:

So now

RizerMix

Answered 2022-01-27
Author has **438** answers

Step 1

we get

if in LHS we use

or

we have two values of square roote

and

or

nick1337

Answered 2022-01-27
Author has **510** answers

Step 1
$a{x}^{2}+bx+c=0$
$a{x}^{2}+bx=-c$
${x}^{2}+\frac{b}{a}x=-\frac{c}{a}$
${x}^{2}+2x(\frac{b}{2a})+(\frac{b}{2a}{)}^{2}-(\frac{b}{2a}{)}^{2}=-\frac{c}{a}$
$(x+\frac{b}{2a}{)}^{2}-\frac{{b}^{2}}{4{a}^{2}}=-\frac{c}{a}$
$(x+\frac{b}{2a}{)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}$
$x+\frac{b}{2a}=\frac{\pm \sqrt{{b}^{2}-4ac}}{2a}$
$x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

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Why is sum of roots of quadratic equation power an integer coefficients of the expansion $-{f}^{\prime}\frac{x}{f\left(x\right)}$ ?

Calculate the$\sum \frac{1}{{a}^{n}}{\textstyle \phantom{\rule{2em}{0ex}}}n\in \mathbb{Z}$

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where a is the roots of a quadratic equation in the form

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