Refer to the example below.

a. In Part B, why did you multiply both sides of the equation by 4a?

b. In Part C, discuss why you added $b^2$ to each side to complete the square.

c. Given the expression under the radical sign, $b^2$ - 4ac, is positive, how many solutions will the quadratic formula give for a quadratic equation? Discuss.

d. If the expression under the radical sign, $b^2$ - 4ac, is 0 , how many solutions will the quadratic formula give for a quadratic equation? What if the expression is negative?

e. Another method of deriving the quadratic formula is to first divide each term by a, and then complete the square. Complete the derivation below. (In this derivation, you will use the quotient property of square roots, which says that $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$. For a square root of a fraction, this property allows you to simplify the numerator and denominator separately. For instance, $\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{\sqrt{9}}=\frac{\sqrt{5}}{3}$.)

$\begin{aligned}
a x^2+b x+c & =0 \\
a x^2+b x & =-c \quad\quad\quad \text {Subtract c from both sides.}\\
x^2+-x & =-\frac{c}{a} \quad\quad\quad \text {Divide each term by $a$.}
\end{aligned}$

$\begin{aligned}
x^2+\frac{b}{a} x+\left(\frac{b}{2 a}\right)^2 & =\left(\frac{b}{2 a}\right)^2-\frac{c}{a} \\
(\quad \quad)^2 & =\frac{b^2-4 a c}{4 a^2} \quad\quad\quad \text {Factor the left side, and write the right side as a single fraction.}
\end{aligned}$

$\begin{aligned}
& x+\frac{b}{2 a}=\pm \frac{\sqrt{b^2-4 a c}}{\sqrt{4 a^2}} \quad \quad \quad\text {Apply the quotient property of radicals.}\\
& x+\frac{b}{2 a}=\pm \frac{\sqrt{b^2-4 a c}}{} \quad \quad \quad\text {Simplify the radical in the denominator.} \\
&
\end{aligned}$

$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \quad \quad \quad\text {Solve for x.}$

Example:

Solve the general form of the quadratic equation, $a x^2+b x+c=0$, by completing the square to find the values of x in terms of a, b, and c.

A Subtract c from both sides of the equation.

$a x^2$ + bx =

B Multiply both sides of the equation by 4a to make the coefficient of $x^2$ a perfect square.

$4 a^2 x^2+\quad x=-4 a c$

C Add $b^2$ to both sides of the equation to complete the square. Then write the trinomial as the square of a binomial.

$\begin{aligned}
& 4 a^2 x^2+4 a b x+b^2=-4 a c+ \\
& (\quad \quad)^2=b^2-4 a c \\
&
\end{aligned}$

D Apply the definition of a square root and solve for $x$.

$\begin{aligned}
& =\pm \sqrt{ \quad} \\
2 a x & =-\quad \pm \sqrt{ \quad}
\end{aligned}$

x = __________

The formula $x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$ is called the quadratic formula. For any quadratic equation written in standard form, $a x^2$ + bx + c = 0, the quadratic formula gives the solutions of the equation.