Chapter P, Problem 53RE

Solution

The solution of the given equation using completing the square.

  $x=\frac{3}{4}\pm \frac{1}{4}\sqrt{17}$

Given information:

The equation given is:

  $2x^2-3x-1=0$ .

Formula used:

To solve an quadratic equation by using completing the square method follow the steps:

1. Rewrite the equation in form $x^2\pm bx=c$ .
2. Add/subtract on both sides of the equation to complete the square.
3. Factor the perfect square trinomial.
4. Solve the resulting equation

  ${\left(a\pm b\right)}^2=a^2+b^2\pm 2ab$

Calculation:

In order to solve the given equation:

  $2x^2-3x-1=0$ using completing the square method first rewrite it in the form $x^2\pm bx=c$ , as

  $\begin{array}{c}2x^2-3x-1=0\\ 2x^2-3x=1\\ 2\left(x^2-\frac{3}{2}x\right)=1\\ x^2-\frac{3}{2}x=\frac{1}{2}\end{array}$

Now complete the square on left side using ${\left(a\pm b\right)}^2=a^2+b^2\pm 2ab$ as:

  $\begin{array}{c}{x}^2-\frac{3}{2}x=\frac{1}{2}\\ x^2-2\times \frac{3}{4}x=\frac{1}{2}\\ x^2-2\times \frac{3}{4}x+{\left(\frac{3}{4}\right)}^2-{\left(\frac{3}{4}\right)}^2=\frac{1}{2}\\ {\left(x-\frac{3}{4}\right)}^2=\frac{1}{2}+{\left(\frac{3}{4}\right)}^2\\ {\left(x-\frac{3}{4}\right)}^2=\frac{1}{2}+\frac{9}{16}\\ {\left(x-\frac{3}{4}\right)}^2=\frac{8+9}{16}\\ {\left(x-\frac{3}{4}\right)}^2=\frac{17}{16}\end{array}$

Now solve the resulting equation

  $\begin{array}{c}{\left(x-\frac{3}{4}\right)}^2=\frac{17}{16}\\ x-\frac{3}{4}=\pm \sqrt{\frac{17}{16}}\\ x=\pm \frac{1}{4}\sqrt{17}+\frac{3}{4}\end{array}$

Thus, solution of given equation is:

  $x=\frac{3}{4}\pm \frac{1}{4}\sqrt{17}$