We will investigate how to use the difference of two squares theorem to determine a solution to the equation.
The equation at hand is as follows:
[tex](\text{ x + }\frac{1}{7})^2\text{ = 8}[/tex]We will move all the terms on the left hand side of the " = " sign as follows:
[tex](\text{ x + }\frac{1}{7})^2\text{ - 8 = 0}[/tex]The difference of two squares theorem states that:
[tex]a^2-b^2\text{ = ( a + b ) }\cdot\text{ ( a - b )}[/tex]We see from the above form that we have the following:
[tex]\begin{gathered} a\text{ = x + }\frac{1}{7} \\ \\ b\text{ = }\sqrt[]{8} \end{gathered}[/tex]Using the difference of two squares formulation we can re-write as a multiple of two factors:
[tex](\text{ x + }\frac{1}{7})^2\text{ - 8 }\equiv\text{ ( x + }\frac{1}{7}\text{ + }\sqrt[]{8}\text{ ) }\cdot\text{ ( x + }\frac{1}{7}\text{ -}\sqrt[]{8}\text{ )}[/tex]Then the factorized equation becomes:
[tex]\text{ ( x + }\frac{1}{7}\text{ + }\sqrt[]{8}\text{ ) }\cdot\text{ ( x + }\frac{1}{7}\text{ -}\sqrt[]{8}\text{ ) = 0}[/tex]The solution of the equation becomes:
[tex]\begin{gathered} x\text{ = - }\frac{1}{7}\text{ - }\sqrt[]{8} \\ \\ x\text{ = }\sqrt[]{8}-\frac{1}{7} \end{gathered}[/tex]We can condense our solution in the form:
[tex]x\text{ = - }\frac{1}{7}\pm2\sqrt[]{2}[/tex]