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Proof

Deriving the Quadratic Formula

The quadratic formula,

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

can be derived by completing the square on the general standard form a quadratic equation. Recall that completing the square is a method for solving quadratic equations. Thus, completing the square on the general form of a quadratic equation,

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

creates another way to solve all quadratic equations. By completing the square, it's possible to isolate

x .

The first step is to rewrite the equation by moving

c

and

a

to the right-hand side.

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

$LHS - c = RHS - c$

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

$LHS / a = RHS / a$

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

Write as a sum of fractions

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

Simplify quotient

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

MovePartNumRight

$\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

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

Next, a new constant term can be added to both sides of the equation, so the expression on the left becomes a perfect square trinomial. The coefficient of the

x

-term is

\frac{b}{a} .

Thus, the term needed to complete the square is

(\frac{b}{a} / 2)^{2} .

For equality to hold, this term is added on both sides.

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

$LHS + (\frac{b}{a} / 2)^{2} = RHS + (\frac{b}{a} / 2)^{2}$

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

$\frac{a}{b} / c = \frac{a}{b \cdot c}$

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

Now that the left-hand side is a perfect square trinomial, it can be written in factored form. The right-hand side can also be simplified.

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

FacPosPerfectSquare

$a^{2} + 2 a b + b^{2} = (a + b)^{2}$

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

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

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

$(a b)^{m} = a^{m} b^{m}$

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

$\frac{a}{b} = \frac{a \cdot 4 a}{b \cdot 4 a}$

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

Subtract fractions

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

Now, there is one

x

-term. To isolate

x,

it is necessary to square-root both sides of the equation. This results in both a positive and a negative term on the right-hand side.

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

$LHS = RHS$

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

$\frac{a}{b} = \frac{a}{b}$

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

$LHS - \frac{b}{2 a} = RHS - \frac{b}{2 a}$

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

MoveNegFracToNum

Put minus sign in numerator

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

Add and subtract fractions

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

Thus, it is possible to solve any quadratic equation in standard form using

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

Exercises

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