Proving Quadratic FormulaHow does \frac{b}{2a} become b^{2}}{4a^{2}}?

logosomatw 2022-01-23 Answered

Proving Quadratic Formula
How does $\frac{b}{2 a}$ become $b^{2} 4 a^{2}$ ?

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bemolizisqt

Answered 2022-01-24 Author has 16 answers

Step 1
Remember how to complete the square:

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

So now

a x^{2} + b x + c = 0

complete square

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

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

x_{1, 2} + \frac{b}{2 a} = \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a}

x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

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RizerMix

Answered 2022-01-27 Author has 438 answers

Step 1
$a x^{2} + b x + c = 0$ divide by a because $a \neq 0$
we get
$x^{2} + \frac{b}{a} x + \frac{c}{a} = 0$
$x^{2} + 2 x \frac{b}{2 a} + \frac{c}{a} = 0$
$x^{2} + 2 x \frac{b}{2 a} + \frac{b^{2}}{4 a^{2}} - \frac{b^{2}}{4 a^{2}} + \frac{c}{a} = 0$
$x^{2} + 2 x \frac{b}{2 a} + \frac{b^{2}}{4 a^{2}} = \frac{b^{2}}{4 a^{2}} - \frac{c}{a}$
$x^{2} + 2 x \frac{b}{2 a} + \frac{b^{2}}{4 a^{2}} = \frac{b^{2} - 4 a c}{4 a^{2}}$
$x^{2} + 2 x \frac{b}{2 a} + (\frac{b}{2 a})^{2} = \frac{b^{2} - 4 a c}{4 a^{2}}$
if in LHS we use $x = A$ and $\frac{b}{2 a} = B$ then we have
$A^{2} + 2 A B + B^{2} = (A + B)^{2}$
or
$(x + \frac{b}{2 a})^{2} = \frac{b^{2} - 4 a c}{4 a^{2}}$
we have two values of square roote
$x_{1} + \frac{b}{2 a} = + \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}}$
and
$x_{2} + \frac{b}{2 a} = - \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}}$
or
$x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

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nick1337

Answered 2022-01-27 Author has 510 answers

Step 1

a x^{2} + b x + c = 0

a x^{2} + b x = - c

x^{2} + \frac{b}{a} x = - \frac{c}{a}

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

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

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

x + \frac{b}{2 a} = \frac{\pm \sqrt{b^{2} - 4 a c}}{2 a}

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

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