Chapter 12.2, Problem 30WE

To determine

To Solve: The equation $a{x}^2+bx+c=0$ for x in terms of a, b and c.

Answer to Problem 30WE

The required solution set is $\left\{\frac{-b+\sqrt{-4ac+b^2}}{2a},\frac{-b-\sqrt{-4ac+b^2}}{2a}\right\}$ .

Explanation of Solution

Given information:

  $a{x}^2+bx+c=0$

Calculation:

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

(Subtract c from each side and rewrite the equation)

  $x^2+\frac{bx}{a}=\frac{-c}{a}$

(Divide each side by a so that the coefficient of $x^2$ will be 1)

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

(Add square of half the coefficient of x to both sides)

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

(Simplify)

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

(Perform square root on both sides)

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

So,

  $x=\frac{-b+\sqrt{-4ac+b^2}}{2a}$

Or

  $x=\frac{-b-\sqrt{-4ac+b^2}}{2a}$

Therefore, the solution set is $\left\{\frac{-b+\sqrt{-4ac+b^2}}{2a},\frac{-b-\sqrt{-4ac+b^2}}{2a}\right\}$ .