Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 593: 29

$(x-\dfrac{1}{4})^2$

Given: $x^2-\dfrac{1}{2}$ We need to add $(\dfrac{-1}{4})^2$ to complete the square. Thus, $x^2-\dfrac{1}{2}x+(\dfrac{-1}{4})^2=x^2+\dfrac{1}{16}-\dfrac{1}{2}x$ Apply difference of polynomial formula. or, $x^2+(\dfrac{1}{4})^2-2(x)(\dfrac{1}{4})=(x-\dfrac{1}{4})^2$