Chapter 3.4, Problem 57E

To determine

To evaluate: the quadratic equation using the quadratic formula given in the question and to solve by another method and tell which method is preferable.

Answer to Problem 57E

The solutions of given quadratic equation is $x=\frac{-1+\sqrt{5}}{2}\text{or }x=\frac{-1-\sqrt{5}}{2}$ .

The method applied later is more suitable in this question.

Explanation of Solution

Given:

  $x^2=1-x$

Solving the given equation to find the solutions,

Using quadratic formula,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-1\pm \sqrt{{\left(1\right)}^2-4\left(1\right)\left(-1\right)}}{2}\\ =\frac{-1\pm \sqrt{5}}{2}\\ =\frac{-1\pm \sqrt{5}}{2}\end{array}$

So, the solutions of given quadratic equation is $x=\frac{-1+\sqrt{5}}{2}\text{or }x=\frac{-1-\sqrt{5}}{2}$ .

Now,

Solving the given quadratic equations using another method,

  $\begin{array}{l}{x}^2+x-1=0\\ x^2+x+\frac{1}{4}-\frac{1}{4}-1=0\\ {\left(x+\frac{1}{2}\right)}^2-\frac{5}{4}=0\\ \left(2x+1+\sqrt{5}\right)\left(2x+1-\sqrt{5}\right)=0\\ x=\frac{-1-\sqrt{5}}{2}\text{or }x=\frac{-1+\sqrt{5}}{2}\end{array}$

So, the solutions of given quadratic equation is $x=\frac{-1+\sqrt{5}}{2}\text{or }x=\frac{-1-\sqrt{5}}{2}$ .

The method applied later is more suitable in this question.