Question

A portion of the Quadratic Formula proof is shown. Fill in the missing reason.

Answer 1

Answer:

see the procedure

Step-by-step explanation:

we have

$ax^2+bx+c=0$ ----> given

step 1

Subtract c from both sides of the equation

$ax^2+bx+c-c=-c$

$ax^2+bx=-c$

step 2

Factor the leading coefficient a

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

step 3

Complete the square and add to both sides

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

step 4

Divide both sides of the equation by a

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

step 5

Find a common denominator on the right side of the equation  and adds the fractions together on the right side of the equation

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

step 6

Rewrite the perfect square trinomial as a binomial squared on the left side of the equation

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

step 7

Take the square root of both sides of the equation

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

step 8

Subtract both sides of the equation by the term b/2a

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

step 9

Rewrite the final expression

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