Kvadratsetningene. Konjugatsetningen

$=x^2+2·3·x+3^2=x^2+6x+9$

$=a^2-2·5·a+5^2=a^2-10a+25$

$=x^2-3^2=x^2-9$

$=2·\left(x^2+2·4·x+4^2\right)=2·\left(x^2+8x+16\right)=2x^2+16x+32$

$=x^2-2·\frac{1}{3}·x+{\left(\frac{1}{3}\right)}^2=x^2-\frac{2}{3}x+\frac{1}{9}$

$\begin{array}{rcl}& = & \left(x^2+2·x·2+2^2\right)+\left(x^2-2·x·3+3^2\right)+\left(x^2-2^2\right)\\ & & \\ & =& \left(x^2+4x+4\right)+\left(x^2-6x+9\right)+\left(x^2-4\right)\\ & & \\ & =& x^2+4x+4+x^2-6x+9+x^2-4\\ & & \\ & =& 3x^2-2x+9\end{array}$

$\begin{array}{rcl}& = & \left(16-a^2\right)-\left(a^2+4a+4\right)+\left(a^2-4a+4\right)\\ & & \\ & =& 16-a^2-a^2-4a-4+a^2-4a+4\\ & & \\ & =& -a^2-8a+16\end{array}$

$\begin{array}{rcl}& = & -3\left(2^2-2·2·x+x^2\right)-5\left({\left(3x\right)}^2-2·3x·4+4^2\right)+2x^2\\ & & \\ & =& -3(4-4x+x^2)-5\left(9x^2-24x+16\right)+2x^2\\ & & \\ & =& -\left(12-12x+3x^2\right)-\left(45x^2-120x+80\right)+2x^2\\ & & \\ & =& -12+12x-3x^2-45x^2+120x-80+2x^2\\ & & \\ & =& -46x^2+132x-92\end{array}$

$\begin{array}{rcl}& = & -(4^2-x^2)-\left({\left(2x\right)}^2-2·\left(2x\right)·5+5^2\right)-\left(x^2-2·x·\frac{1}{3}+{\left(\frac{1}{3}\right)}^2\right)\\ & & \\ & =& -\left(16-x^2\right)-\left(4x^2-20x+25\right)-\left(x^2-\frac{2x}{3}+\frac{1}{9}\right)\\ & & \\ & =& -16+x^2-4x^2+20x-25-x^2+\frac{2x}{3}-\frac{1}{9}\\ & & \\ & =& -\frac{16·9}{1·9}+x^2-4x^2+\frac{20x·3}{1·3}-\frac{25·9}{1·9}-x^2+\frac{2x}{3}-\frac{1}{9}\\ & & \\ & =& -\frac{144}{9}+x^2-4x^2+\frac{60x}{3}-\frac{225}{9}-x^2+\frac{2x}{3}-\frac{1}{9}\\ & & \\ & =& -4x^2+\frac{62}{3}x-\frac{370}{9}\end{array}$

$\begin{array}{rcl}& = & 4-a^2-2\left(a^2+2a+1\right)-9\left(a^2-\frac{2}{3}a+\frac{1}{9}\right)\\ & & \\ & =& 4-a^2-2a^2-4a-2-9a^2+\frac{9·2}{3}a-\frac{9·1}{9}\\ & & \\ & =& 4-a^2-2a^2-4a-2-9a^2+6a-1\\ & & \\ & =& -12a^2+2a+1\end{array}$

$\begin{array}{rcl}& = & 2\left(4a^2-1\right)-\left(a^2-4a+4\right)-4\left(a^2+a+\frac{1}{4}\right)\\ & & \\ & =& 8a^2-2-a^2+4a-4-4a^2-4a-1\\ & & \\ & =& 3a^2-7\end{array}$

$\begin{array}{rcl}& = & a^2-a+\frac{1}{4}+\frac{1}{4}\left(4a^2+4a+1\right)-\frac{1}{2}\left(4a^2-1\right)\\ & & \\ & =& a^2-a+\frac{1}{4}+a^2+a+\frac{1}{4}-2a^2+\frac{1}{2}\\ & & \\ & =& 1\end{array}$

$={\left(\sqrt{5}\right)}^2-2^2=5-4=1$

$=\left(30-1\right)\left(30+1\right)={30}^2-1^2=900-1=899$

$=\left(20-2\right)\left(20+2\right)={20}^2-2^2=400-4=396$

$=\left(20+5\right)\left(20-5\right)={20}^2-5^2=400-25=375$

$$ =\left(100+3\right)\left(100-3\right)={100}^2-3^2=10 000-9=9 991$$