 # Proving Quadratic FormulaHow does \frac{b}{2a} become b^{2}}{4a^{2}}? logosomatw 2022-01-23 Answered

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

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it bemolizisqt
Step 1
Remember how to complete the square:
$A{x}^{2}+Bx=A{\left(x+\frac{B}{2A}\right)}^{2}-\frac{{B}^{2}}{4{A}^{2}}$
So now
$a{x}^{2}+bx+c=0$ complete square
$a{\left(x+\frac{b}{2a}\right)}^{2}-\frac{{b}^{2}}{4a}=-c$
${\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}$

###### Not exactly what you’re looking for? RizerMix

Step 1
$a{x}^{2}+bx+c=0$ divide by a because $a\ne 0$
we get
${x}^{2}+\frac{b}{a}x+\frac{c}{a}=0$
${x}^{2}+2x\frac{b}{2a}+\frac{c}{a}=0$
${x}^{2}+2x\frac{b}{2a}+\frac{{b}^{2}}{4{a}^{2}}-\frac{{b}^{2}}{4{a}^{2}}+\frac{c}{a}=0$
${x}^{2}+2x\frac{b}{2a}+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}$
${x}^{2}+2x\frac{b}{2a}+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}-4ac}{4{a}^{2}}$
${x}^{2}+2x\frac{b}{2a}+\left(\frac{b}{2a}{\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}$
if in LHS we use $x=A$ and $\frac{b}{2a}=B$ then we have
${A}^{2}+2AB+{B}^{2}=\left(A+B{\right)}^{2}$
or
$\left(x+\frac{b}{2a}{\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}$
we have two values of square roote
${x}_{1}+\frac{b}{2a}=+\sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}$
and
${x}_{2}+\frac{b}{2a}=-\sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}$
or

###### Not exactly what you’re looking for? nick1337
Step 1 $a{x}^{2}+bx+c=0$ $a{x}^{2}+bx=-c$ ${x}^{2}+\frac{b}{a}x=-\frac{c}{a}$ ${x}^{2}+2x\left(\frac{b}{2a}\right)+\left(\frac{b}{2a}{\right)}^{2}-\left(\frac{b}{2a}{\right)}^{2}=-\frac{c}{a}$ $\left(x+\frac{b}{2a}{\right)}^{2}-\frac{{b}^{2}}{4{a}^{2}}=-\frac{c}{a}$ $\left(x+\frac{b}{2a}{\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}$ $x+\frac{b}{2a}=\frac{±\sqrt{{b}^{2}-4ac}}{2a}$ $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$