Look at the expression in the last row of the table. If [math]ax^2+bx+c[/math] is equivalent to [math](kx+m)^2[/math], how are [math]a[/math], [math]b[/math] and [math]c[/math] related to [math]k[/math] and [math]m[/math]?

[size=150]One way to solve the quadratic equation [math]x^2+5x+3=0[/math] is by completing the square. A partially solved equation is shown here. Study the steps.[/size][br][br][math]\displaystyle \begin {align} x^2 + 5x + 3 &= 0 &\qquad& \text{Original equation}\\\\ [br]4x^2 + 20x + 12 &= 0 &\qquad& \text{Multiply each side by 4}\\\\ [br]4x^2 + 20x &= \text-12 &\qquad& \text{Subtract 12 from each side}\\\\ [br](2x)^2 + 10(2x) &= \text-12 &\qquad& \text{Rewrite }4x^2 \text{as }(2x)^2 \text{ and }20x \text{ as }10(2x)\\\\ [br]P^2 + 10P &= \text-12 &\qquad& \text{Use } P \text{ as a placeholder for }2x\\\\[br]P^2 + 10P + \underline{\hspace{0.3in}}^2 &= \text-12 + \underline{\hspace{0.3in}}^2&\qquad& \text{ }\\\\ [br](P+\underline{\hspace{0.3in}})^2 &= \text-12 + \underline{\hspace{0.3in}}^2&\qquad& \text{}\\\\ [br]P+\underline{\hspace{0.3in}} &= \pm \sqrt {\text-12 + \underline{\hspace{0.3in}}^2}&\qquad& \text{}\\\\ [br]P &= \underline{\hspace{0.3in}} \pm \sqrt {\text-12 + \underline{\hspace{0.3in}}^2}&\qquad& \text{}\\\\ [br]P &= \underline{\hspace{0.3in}} \pm \sqrt {\underline{\hspace{0.3in}}^2 - 12}&\qquad& \text{}\\\\ [br]2x &= \underline{\hspace{0.3in}} \pm \sqrt {\underline{\hspace{0.3in}}^2 - 12}&\qquad& \text{}\\\\ [br]x &= &\qquad& \text {}\\ \end {align}[/math][br][br]Then, knowing that [math]P[/math] is a placeholder for [math]2x[/math], continue to solve for [math]x[/math] but without evaluating any part of the expression. Be prepared to explain each step.[br]

Explain how the solution is related to the quadratic formula.[br]

[list][*]First, divide each side of the equation [math]ax^2+bx+c=0[/math] by [math]a[/math] to get [math]x^2+\frac{b}{a}x+\frac{c}{a}=0[/math].[/*][size=150][*]Then, complete the square for [math]x^2+\frac{b}{a}x+\frac{c}{a}=0[/math].[/*][/size][/list]The beginning steps of this approach are shown here. Briefly explain what happens in each step.[br][br][math]\begin{align} x^2 + \frac{b}{a}x + \frac{c}{a} &= 0 &\qquad& \text{Original equation}\\ x^2 + \frac{b}{a}x &= \text-\frac{c}{a} &\qquad& \text[1]\\ x^2 + 2\left(\frac {b}{2a}\right)x + \left(\frac {b}{2a}\right)^2&=\text-\frac ca + \left(\frac {b}{2a}\right)^2 &\qquad& \text[2]\\ \left(x+\frac {b}{2a}\right)^2&= \text- \frac ca + \frac {b^2}{4a^2} &\qquad& \text[3]\\ \left(x+\frac {b}{2a}\right)^2&= \text- \frac {4ac}{4a^2} + \frac {b^2}{4a^2} &\qquad& \text[4]\\ \left(x+\frac {b}{2a}\right)^2&= \frac {b^2-4ac}{4a^2} &\qquad& \text[5]\\ x+\frac{b}{2a} &= \pm \sqrt { \frac {b^2-4ac}{4a^2}} &\qquad& \text[6]\\ x+\frac{b}{2a} &= \pm \frac {\sqrt {b^2-4ac}}{\sqrt{4a^2}} &\qquad& \text[7] \end{align}[/math]

Continue the solving process until you have the equation [math]x = \dfrac{\text-b \pm \sqrt{b^2-4ac}}{2a}[/math].[br]