heureka

$$\\x^2=5*(5+12)\\ x^2=5*17\\ x^2=85\\ x=\sqrt{85}\\ x=9.22$$

x^3+x^2-x-5 factor   ?

1. ABC=180-2*70         =   40
2. ADC=2*ABC             =   80

$$3x^2 + 4x - 2 = 0 \qquad x?$$   

An easy way!

"p,q-formula":   $$x^2+px+q=0$$

solution:  $$x_{1,2}=-\dfrac{p}{2}\pm\sqrt{\dfrac{p^2}{4}-q}$$

check:  $$x_1+x_2=-p\qquad x_1x_2=q$$

formula:  $$3x^2 + 4x - 2 = 0$$

prepare for "p,q-formula":  $$3x^2 + 4x - 2 = 0 \quad | \quad :3$$

$$x^2+\dfrac{4}{3}x-\dfrac{2}{3} =0 \qquad p= \dfrac{4}{3} \qquad q=-\dfrac{2}{3}$$

 solution:  $$x_{1,2}=-\dfrac{1}{2}\left(\dfrac{4}{3}\right)\pm\sqrt{\dfrac{1}{4} \left(\dfrac{4}{3}\right)^2 +\dfrac{2}{3}}$$

$$x_{1,2}=-\dfrac{2}{3}\pm\sqrt{\dfrac{4}{3*3}+\dfrac{2*3}{3*3}}$$

$$x_{1,2}=-\dfrac{2}{3}\pm\dfrac{\sqrt{10} }{3}$$

$$x=x_1=\dfrac{-2+\sqrt{10}}{3} =0.38742588672$$

$$x=x_2=\dfrac{-2-\sqrt{10}}{3}=-1.72075922006$$

 check:   $$x_1+x_2= 0.38742588672-1.72075922006=-1.\bar{3}=-p$$

$$x_1x_2=0.38742588672\times(-1.72075922006)=-0.\bar{6}=q$$

Hi Melody,

the area of the two triangle together is r1*r2 = 15 * 20

$${\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{20}}}{{\mathtt{15}}}}\right)}}{{\mathtt{360}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{15}}}{{\mathtt{20}}}}\right)}}{{\mathtt{360}}}}{\mathtt{\,-\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{20}} = {\mathtt{166.041\: \!867\: \!567\: \!676\: \!442}}$$

$$\tan^{-1} ( \frac{15}{20} ) = \tan^{-1}( \frac {1}{\frac{20}{15}} ) =\cot^{-1} ( \frac{20}{15} ) =\frac{180}{2}-\tan^{-1} ( \frac{20}{15} )\\ \Rightarrow$$

$$\left({\frac{{\mathtt{\pi}}}{{\mathtt{180}}}}\right){\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{20}}}{{\mathtt{15}}}}\right)}{\mathtt{\,\times\,}}\left({\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{20}}\right){\mathtt{\,\small\textbf+\,}}{\frac{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{20}}\right)}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{20}} = {\mathtt{166.041\: \!867\: \!567\: \!676\: \!442}}$$

y = (x^2+6x-3)/(x-3)   'slant' asumtotes ?

$$y=\dfrac{x^{\textcolor[rgb]{1,0,0}{2}}+6x-3}{x^{\textcolor[rgb]{1,0,0}{1}}-3}$$

degree top = 2

degree botton = 1

'slant' asumtotes  when degree top = degree botton + 1

'slant' asumtote yes ! ( 2 = 1 + 1 )

$$\begin{array}{cccccc} (x^2&+&6x&-&3)& : (x-3)= \textcolor[rgb]{1,0,0}{x+9}+\dfrac{24}{x-3}\\ -(x^2&-&3x)\\ 0&+&9x&-&3 \\ &-&(9x&-&27)\\ &&0&+&24\\ \end{array}$$

$$\lim\limits_{ x \to \pm\infty }(\frac{24}{x-3})=0$$

'slant' asumtote  y = x + 9

3x^2 + 4x - 2 = 0     x?

$$\\3x^2+4x-2=0\\\\ 3(x^2+\frac{4}{3}x) =2 \quad | \quad :3\\\\ x^2+\frac{4}{3}x =\frac{2}{3}\\\\ \underbrace{x^2+\frac{4}{3}x+ (\frac{4}{2*3})^2}_{ =(x+\frac{4}{2*3})^2 } -(\frac{4}{2*3})^2 =\frac{2}{3}\\\\ (x+\frac{4}{2*3})^2 -(\frac{4}{2*3})^2 = \frac{2}{3}\\\\ (x+\frac{2}{3})^2 -(\frac{2}{3})^2 = \frac{2}{3}\\\\ (x+\frac{2}{3})^2 =(\frac{2}{3})^2 + \frac{2}{3}\\\\$$

$$\\(x+\frac{2}{3})^2 = \frac{2}{3}(\frac{2}{3}+1)\\\\ (x+\frac{2}{3})^2 = \frac{2}{3}*\frac{5}{3} \quad | \quad \pm\sqrt{}\\\\ x+\frac{2}{3} = \pm \sqrt{\frac{2}{3}*\frac{5}{3}}\\\\ x+\frac{2}{3} = \pm \frac{\sqrt{10}}{3}\\\\ x = -\frac{2}{3} \pm \frac{\sqrt{10}}{3}\\\\ x=x_1=\frac{-2+\sqrt{10}}{3}\\\\ x=x_2=\frac{-2-\sqrt{10}}{3}$$

$$x_1=0.38742588672$$

$$x_2=-1.72075922006$$

x^2 + 7x + 10 = (x+2)(x+5)

y=x^2+6x-3/x-3 horizontal asymptote?

$$y=\dfrac{x^{\textcolor[rgb]{1,0,0}{3}}+6x^2-3x^1-3}{x^\textcolor[rgb]{1,0,0}{1}}$$

degree top = 3

degree botton = 1

horizontal asymptote when degree top < degree botton

                                    or degree top = degree botton

no horizontal asymptote! ( 3 < 1 no )