# How is my book completing the square?

In my book I see:

To integrate the given function we complete the square in the denominator:

$$4x^2 - 4x + 3 = (2x-1)^2 + 2$$

How is it doing this? When I complete the square I get:

$$x^2 - x + \frac{3}{4} = 0$$ $$x^2 - x + \frac{1}{4} = \frac{-1}{2}$$ $$\left(x-\frac{1}{2}\right)^2 = \frac{-1}{2}$$ $$x - \frac{1}{2} = \sqrt{\frac{-1}{2}}$$

now I'm stuck. can someone help to show me how my book gets the $$(2x-1)^2 + 2$$?

• Why have you set the denominator equal to zero and divided through by 4? This is not a valid procedure. – Peter Foreman May 24 '19 at 15:30
• @PeterForeman I thought that's how I complete the square... – Jwan622 May 24 '19 at 15:36
• @Jwan622 Completing the square is used in this case to simplify an expression, not to solve an equation. These are different procedures. – Peter Foreman May 24 '19 at 15:37
• That's the method used for solving quadratic equations, which is something completely different. – KM101 May 24 '19 at 15:37
• This looks like a quadratic no? – Jwan622 May 24 '19 at 18:01

$$4x^2-4x+3=4x^2-4x+1+2=(2x-1)^2+2.$$
$$4x^2-4x+ 3 = 4x^2-4x + 1 + 2 = (2x-1)^2+2$$
Given expression: $$4x^2 - 4x + 3$$ $$=4\left(x^2-x+\frac{3}{4}\right)$$ $$=4\left((x^2)-2\left(\frac{1}{2}\right)(x)+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+\frac{3}{4}\right)$$ $$=4\left(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}+\frac{3}{4}\right)$$ $$=4\left(\left(x-\frac{1}{2}\right)^2+\frac{1}{2}\right)$$ $$=4\left(x-\frac{1}{2}\right)^2+2$$ $$=4\left(\frac{2x-1}{2}\right)^2+2$$ $$=(2x-1)^2+2$$
Because this expression is in the denominator of the integrand, we can further express this as $$(2x-1)^2+(\sqrt2)^2$$ And proceed with the integration, depending upon the expression in the numerator.