Chapter 9.6, Problem 58HP

To determine

Describe the advantages and disadvantages of each method solving quadratic equations.

Answer to Problem 58HP

The method we will choose according to the quadratic equation.

Explanation of Solution

Given:

Describe the advantages and disadvantages of each method solving quadratic equations. Which method do you prefer, and why?

Concept Used:

Factoring:

Factoring is easy if the polynomial is factorable and complicated if it is not factorable. Not all the equations are factorable.

For example $f(x)=x^2-8x+16$ factors to ${(x-4)}^2$

However, $f(x)=x^2-16x+8$ cannot be factored.

Graphing:

Graphing only gives approximate answers, but it is easy to see the number of solutions. Using square roots is easy when there is no x − term.

For example, for the quadratic $f(x)=2x^2-17x+4$ You can see the two solutions in the graph. However, it will be difficult to identify the solution x = 8.25780 and 0.24219 in the graph.

Graph of $f(x)=2x^2-17x+4$

Completing square:

Completing square can be used for any quadratic equation and exact solutions can be found, but the leading coefficient has to be 1 and $x^2$ and x term must be isolated. It is also easier if the coefficient of the x – term is even; if not, the calculations become harder when dealing with fractions.

For example $x^2-8x=9$ can be solved completing square.

  $\begin{array}{l}{x}^2-8x=9\text{Original Equation}\\ x^2-8x+16=9+16\text{Add (}\frac{8}{2}{)}^2\text{to}\text{both}\text{sides}\\ $ {(x-4)}^2=25$\sqrt{{(x-4)}^2}=\pm \sqrt{25}\text{Take}\text{the}\text{square}\text{root}\text{of}\text{each}\text{sides}\\ (x-4)=\pm 5\\ x-4+4=4\pm 5\text{Add}\text{4}\text{to}\text{both}\text{sides}\\ x=4+5\text{or}4-5\\ x=9\text{or}-1\end{array}$

Quadratic Formula:

The quadratic Formula will work for any quadratic equation and exact solutions can be found. This method can be time consuming, especially if an equation is easily factored.

For example, use the quadratic Formula to find the solutions of $f(x)=2x^2+9x-18$

  $\begin{array}{l}f(x)=2x^2+9x-18\\ 2x^2+9x-18=0[\text{make}\text{right}\text{side}\text{equal}\text{to}\text{0}]\\ a=2;b=9;c=-18\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-9\pm \sqrt{9^2-4·2·-18}}{2·2}=\frac{-9\pm \sqrt{81+144}}{4}=\frac{-9\pm \sqrt{225}}{4}=\frac{-9\pm 15}{4}\\ x=\frac{-9+15}{4}\text{and}\frac{-9-15}{4}\\ x=\frac{6}{4}\text{and}\frac{-24}{4}\\ x=\frac{3}{2}\text{and}-6\end{array}$

Thus,