Completing the square

By Hal 2001 on Monday, September 17, 2001 - 04:51 pm:

How do you go about completing the square for the following?

i) sin² x + 4sinx, and
ii) x⁴ + 2x² + y⁴ - 2y² + 3

Thanks for your help in advance.

By Olof Sisask on Monday, September 17, 2001 - 08:17 pm:

Hi Hal,

For (i), let

y = \sin x

, then your equation

$= y^{2} + 4 y = (y + 2)^{2} - 4$

(ii)

$= (x^{2} + 1)^{2} - 1 + (y^{2} - 1)^{2} - 1 + 3 = (x^{2} + 1)^{2} + (y^{2} - 1)^{2} + 1$

If this is equal to 0, then there are no real solutions for $x$ and $y$ , since $(x^{2} + 1)^{2} \geq 0$ and $(y^{2} - 1)^{2} \geq 0$ .

In general, if you have $x^{2} + a x + b$ , then completing the square gives $(x + a / 2)^{2} - a^{2} / 4 + b$ .
Regards,
Olof