Question

Complete the square for each quadratic in Exercises 28–33. Then describe the trigonometric substitution that would be appropriate if you were solving an integral that involved that quadratic.

$2x^2-4x+1$

Short answer

The completing square of the given quadratic equation is ${\left(\sqrt{2}x-\sqrt{2}\right)}^2-1,$and the appropriate trigonometric substitution is localid="1649054906339" $\sqrt{2}x-\sqrt{2}=secu.$

Step-by-step solution

Step 1. Given Information.

The given quadratic is$2x^2-4x+1.$

Step 2. Completing the square for the given quadratic.

The complete square of the given quadratic equation is as follows:

$$\begin{gathered} 2x^2-4x+1 \\ ={\left(\sqrt{2}x\right)}^2-4x+{\left(\sqrt{2}\right)}^2+1-{\left(\sqrt{2}\right)}^2 \\ ={\left(\sqrt{2}x-\sqrt{2}\right)}^2-1 \end{gathered}$$

Step 3. Describing the trigonometric substitution. 

After completing the square for the given quadratic equation it is in the form of$x^2-a^2.$

So, for solving an integral that involved the quadratic then the trigonometric substitution can be used islocalid="1649054922648" $x=asecu.$

Here,$x=\sqrt{2}x-\sqrt{2},a=1.$

Thus, the appropriate trigonometric substitution islocalid="1649055042383" $\sqrt{2}x-\sqrt{2}=secu.$