Chapter P.5, Problem 68E

Solution

Derive the quadratic formula, Follow these steps to use completing the square to solve $a{x}^2+bx+c=0,a\ne 0$ .

1. Subtract c from both sides of the original equation and divide both sides of the resulting equation by a to obtain $x^2+\frac{b}{a}x=\frac{-c}{a}$
2. Add the square of one-half of the coefficient of x in (a) to both sides and simplify to obtain

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

3. Extract square roots in (b) and solve for x to obtain the quadratic formula

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

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

Given:

The equation is, $a{x}^2+bx+c=0$

Concept Used:

• Completing the Square is a method used to solve a quadratic equation by changing the form of the equation so that the left side is a perfect square trinomial.
• Transform the equation so that the constant term, c, is alone on the right side.
• If a, the leading coefficient (the coefficient of the  $x^2$ term), is not equal to 1, divide both sides by a.
• Add the square of half the coefficient of the x -term, ${\left(\frac{b}{a}\right)}^2$ to both sides of the equation.
• Factor the left side as the square of a binomial.
• Take the square root of both sides. 

Calculation:

(a)

  $\begin{array}{c}a{x}^2+bx+c=0\\ a{x}^2+bx+=-c\\ dividebothsidebya\\ x^2+\frac{bx}{a}+=-\frac{c}{a}\end{array}$

(b)

  $\begin{array}{c}$ add{(\frac{1}{2}\cdot \frac{b}{a})}^{2}bothside$$ x^2+\frac{bx}{a}+{(\frac{1}{2}\cdot \frac{b}{a})}^2=-\frac{c}{a}+{(\frac{1}{2}\cdot \frac{b}{a})}^2$$ x^2+\frac{2\cdot x\cdot b}{2a}+{(\frac{b}{2a})}^2=-\frac{c}{a}+{\frac{b}{4a^2}}^2${\left(x+\frac{b}{2a}\right)}^2=\frac{-4ac+b^2}{4a^2}\\ {\left(x+\frac{b}{2a}\right)}^2=\frac{b^2-4ac}{4a^2}\end{array}$

(c)

  $\begin{array}{c}x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\\ x=\frac{\pm \sqrt{b^2-4ac}}{2a}-\frac{b}{2a}\\ x=\frac{\pm \sqrt{b^2-4ac}-b}{2a}\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$

So the derived quadratic formula is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$