heureka

I must have forgotten some rule... trying to figure out how to simplify this... 

\(x=\frac{4\left(a+b\right)±\sqrt{16\left(a+b\right)^2-48ab}}{24}\)

Simplifies to: \(x=\frac{\left(a+b\right)±\sqrt{a^2-ab+b^2}}{6}\)

\(\begin{array}{|rcll|} \hline x &=& \frac{4\left(a+b\right) \pm \sqrt{16\cdot \left(a+b\right)^2-48ab}}{24} \\ &=& \frac{4\left(a+b\right) \pm \sqrt{16\cdot \left(a+b\right)^2-16\cdot 3 \cdot ab}}{24} \\ &=& \frac{4\left(a+b\right) \pm \sqrt{ 16 \cdot[~ \left(a+b\right)^2-3\cdot ab ~]}}{24} \quad & | \quad \sqrt{a\cdot b} = \sqrt{a}\cdot \sqrt{ b} \\ &=& \frac{4\left(a+b\right) \pm \sqrt{ 16}\cdot \sqrt{ \left(a+b\right)^2-3ab } }{24} \quad & | \quad \sqrt{ 16}=4\\ &=& \frac{4\left(a+b\right) \pm 4\cdot \sqrt{ \left(a+b\right)^2-3ab } }{24} \\ &=& \frac{4 \cdot[~ \left(a+b\right) \pm \sqrt{ \left(a+b\right)^2-3ab} ~] }{24} \\ &=& \frac{4 \cdot[~ \left(a+b\right) \pm \sqrt{ \left(a+b\right)^2-3ab} ~] }{4\cdot 6} \\ &=& \frac{ \left(a+b\right) \pm \sqrt{ \left(a+b\right)^2-3ab} } { 6} \quad & | \quad (a+b)^2 = a^2 + 2ab + b^2\\ &=& \frac{ \left(a+b\right) \pm \sqrt{ a^2 + 2ab + b^2-3ab} } { 6} \\ &=& \frac{ \left(a+b\right) \pm \sqrt{ a^2 + 2ab-3ab + b^2} } { 6} \quad & | \quad 2ab-3ab = ab(2-3)= ab(-1) = -ab\\ &=& \frac{ \left(a+b\right) \pm \sqrt{ a^2 -ab + b^2} } { 6} \\ \hline \end{array}\)

Point P is 9 units from the center of a circle of radius 15.

How many different chords of the circle contain P and have integer lengths?

(15-9) * (15+9)  =   6 *  24  = 144

\(\begin{array}{|rcll|} \hline 144 &=& 2^4\cdot 3^2 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 144 &=& (2\cdot 3) \cdot ( 2^3\cdot 3 ) \\ &=& 6 \cdot 24 \\\\ 144 &=& (2^3) \cdot ( 2\cdot 3^2 ) \\ &=& 8 \cdot 18 \\\\ 144 &=& (3^2) \cdot ( 2^4 ) \\ &=& 9 \cdot 16 \\\\ 144 &=& (2^2\cdot 3) \cdot (2^2\cdot 3) \\ &=& 12 \cdot 12 \\ \hline \end{array} \)

1. chord ( length = 30 = 6+24 ) 

2. chord ( length = 26 = 8+18 )

3. chord ( length = 26 = 18+8 )

4. chord ( length = 25 = 9+16 )

5. chord ( length = 25 = 16+9 )

6. chord ( length = 24 = 12+12 )

see chord theorem:

what plus what = 7 and what times what = -8 only using two numbers

\(\begin{array}{|rclcl|} \hline x^2+p \cdot x + q &=& 0 \qquad | \qquad -p=(x_1+x_2) & & q = x_1\cdot x_2 \\ x^2 -7 x -8 &=& 0 \qquad | \qquad -p=7 & & q =-8 \\\\ x_{1,2} &=& \frac{7\pm \sqrt{49-4\cdot(-8)} }{2} \\ x_{1,2} &=& \frac{7\pm \sqrt{81} }{2} \\ x_{1,2} &=& \frac{7\pm 9 }{2} \\\\ x_1 &=& \frac{7 + 9 }{2} \\ \mathbf{x_1} &\mathbf{=}& \mathbf{8} \\\\ x_2 &=& \frac{7 - 9 }{2} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{-1} \\ \hline \end{array}\)

8-1 = 7
8*(-1) = -8

i^i

\(\begin{array}{|rcll|} \hline i^i &=& [(-1)^{\frac12}]^i \qquad | \qquad i = \sqrt{-1} = (-1)^{\frac12}\\ \mathbf{i^i} & \mathbf{=} & \mathbf{(-1)^{\frac i2}} \\ &=& e^{\frac i2 \cdot ln(-1) } \\\\ && \ln(z) = \ln(|z|) +i\cdot arg(z)\\\\ && \ln(-1+0\cdot i) = \ln(|-1+0\cdot i|) +i\cdot arg(-1+0\cdot i)\\ && \ln(-1) = \ln(|\sqrt{(-1)^2+0^2}|) +i\cdot \arctan\left(\frac{0}{-1} \right)\\ && \ln(-1) = \ln(|1|) +i\cdot \pi \qquad | \qquad \ln(|1|)=\ln(1) = 0 \\ && \ln(-1) = 0 +i\cdot \pi \\ && \ln(-1) = i\cdot \pi \\\\ i^i &=& e^{\frac i2 \cdot \ln(-1) } \\ &=& e^{\frac i2 \cdot i\cdot \pi } \\ &=& e^{\frac {i^2}2 \cdot \pi } \qquad | \qquad i^2 = -1\\ &=& e^{\frac {-1}2 \cdot \pi } \\ &=& e^{\frac {- \pi}2 } \\ \mathbf{i^i} &\mathbf{=}& \mathbf{ e^{-(\frac {\pi}2) } } \\ \mathbf{i^i} &\mathbf{=}& \mathbf{0.207879576350761908546955619834978770033877841631769608075\dots }\\ \hline \end{array} \)

Find the simplest polynomial with the given zeroes. 

2i, -2i, 3

a. x^3-3x^2+4x-12

b. x^3+3x^2+4x-12

c. x^3-3x^2-4x+12

d. x^3+3x^2-4x-12

\((x-2i)(x+2i)(x-3)\\ =(x^2-4i^2)(x-3) \qquad | \qquad i^2 = -1\\ =(x^2+4)(x-3)\\ =x^3-3x^2+4x-12\)

(a)

what will be the value of Sin2(6)-sin2(12)+sin2(18)... till 15th term , angles are in degree.

\(\small{ \begin{array}{|lcll|} \hline && \sin^2(6^{\circ})-\sin^2(12^{\circ})+\sin^2(18^{\circ})-\sin^2(24^{\circ})+\sin^2(30^{\circ})-\sin^2(36^{\circ})+\sin^2(42^{\circ}) \\ && -\sin^2(48^{\circ})+\sin^2(54^{\circ})-\sin^2(60^{\circ})+\sin^2(66^{\circ})-\sin^2(72^{\circ})+\sin^2(78^{\circ})-\sin^2(84^{\circ})+\sin^2(90^{\circ}) \\ &=& \sin^2(6^{\circ})-\sin^2(12^{\circ})+\sin^2(18^{\circ})-\sin^2(24^{\circ})+\sin^2(30^{\circ})-\sin^2(36^{\circ})+\sin^2(42^{\circ}) \\ && -\sin^2(90^{\circ}-42^{\circ})+\sin^2(90^{\circ}-36^{\circ})-\sin^2(90^{\circ}-30^{\circ}) +\sin^2(90^{\circ}-24^{\circ}) \\ && -\sin^2(90^{\circ}-18^{\circ})+\sin^2(90^{\circ}-12^{\circ})-\sin^2(90^{\circ}-6^{\circ})+\sin^2(90^{\circ}) \\ &=& \sin^2(6^{\circ})-\sin^2(12^{\circ})+\sin^2(18^{\circ})-\sin^2(24^{\circ})+\sin^2(30^{\circ})-\sin^2(36^{\circ})+\sin^2(42^{\circ}) \\ && -\cos^2(42^{\circ})+\cos^2(36^{\circ})-\cos^2(30^{\circ}) +\cos^2(24^{\circ})-\cos^2(18^{\circ})+\cos^2(12^{\circ})-\cos^2(6^{\circ})+\sin^2(90^{\circ}) \\ &=& \sin^2(6^{\circ})-\cos^2(6^{\circ})-\sin^2(12^{\circ})+\cos^2(12^{\circ})+\sin^2(18^{\circ})-\cos^2(18^{\circ})-\sin^2(24^{\circ})+\cos^2(24^{\circ}) \\ && +\sin^2(30^{\circ})-\cos^2(30^{\circ})-\sin^2(36^{\circ})+\cos^2(36^{\circ})+\sin^2(42^{\circ})-\cos^2(42^{\circ})+\sin^2(90^{\circ}) \\\\ && \boxed{~ \begin{array}{rcl} \cos(2\varphi) &=& \cos^2(\varphi)-\sin^2{\varphi}\\ -\cos(2\varphi) &=& \sin^2(\varphi)-\cos^2{\varphi}\\ \end{array} }\\ &=& -\cos(12^{\circ})+\cos(24^{\circ})-\cos(36^{\circ})+\cos(48^{\circ}) \\ && -\cos(60^{\circ})+\cos(72^{\circ})-\cos(84^{\circ})+\sin^2(90^{\circ}) \quad | \quad \sin^2(90^{\circ}) = 1^2=1\\ &=& 1 -\cos(12^{\circ})+\cos(24^{\circ})-\cos(36^{\circ})+\cos(48^{\circ}) -\cos^2(60^{\circ})+\cos(72^{\circ})-\cos(84^{\circ})\\ &=& 1 -\cos(12^{\circ})+\cos(24^{\circ})-\cos(60^{\circ}-24^{\circ})+\cos(60^{\circ}-12^{\circ}) \\ && -\cos(60^{\circ})+\cos(60^{\circ}+12^{\circ})-\cos(60^{\circ}+24^{\circ})\\ &=& 1-\cos(60^{\circ}) -\cos(12^{\circ})+ \cos(24^{\circ})-\cos(60^{\circ}-24^{\circ})-\cos(60^{\circ}+24^{\circ}) \\ && +\cos(60^{\circ}-12^{\circ})+\cos(60^{\circ}+12^{\circ})\\\\ && \boxed{~ \begin{array}{rcl} \cos(\alpha+\beta)+\cos(\alpha-\beta) &=& \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta) \\ && + \cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta) \\ &=& 2 \cos(\alpha)\cos(\beta)\\ \end{array} }\\ &=& 1-\cos(60^{\circ}) -\cos(12^{\circ})+ \cos(24^{\circ})-[\cos(60^{\circ}-24^{\circ})+\cos(60^{\circ}+24^{\circ})] \\ && +[\cos(60^{\circ}-12^{\circ})+\cos(60^{\circ}+12^{\circ})]\\ &=& 1-\cos(60^{\circ}) -\cos(12^{\circ})+ \cos(24^{\circ})-2\cdot\cos(60^{\circ})\cos(24^{\circ}) +2\cdot \cos(60^{\circ})\cos(12^{\circ})\\\\ && \boxed{~ \begin{array}{rcl} \cos(60^{\circ}) &=& \frac12 \\ \end{array} }\\ &=& 1-\frac12 -\cos(12^{\circ})+ \cos(24^{\circ})-2\cdot\frac12\cos(24^{\circ}) +2\cdot \frac12\cos(12^{\circ})\\ &=& \frac12 -\cos(12^{\circ})+ \cos(24^{\circ})-\cos(24^{\circ}) + \cos(12^{\circ})\\ &=& \frac12 -\cos(12^{\circ})+ \cos(12^{\circ})+ \cos(24^{\circ})-\cos(24^{\circ}) \\ &\mathbf{=}& \mathbf{ \dfrac12 } \\ \hline \end{array} }\)

find the equation of the normal to y=xsinx at the point where x=π/2

|5x-3|=-7x,  

\(x < 0\ !\)

\(\begin{array}{|rcll|} \hline |5x-3| &=& -7x \qquad & | \qquad \text{square both sides} \\ (5x-3)^2 &=& (-7x)^2 \\ 25x^2-30x +9 &=& 49x^2 \qquad & | \qquad -(25x^2-30x +9) \\ 0 &=& 49x^2 -25x^2+30x -9 \\ 0 &=& 24x^2 +30x -9 \\ 24x^2 +30x -9 &=& 0 \\\\ x_{1,2} &=& \frac{-30\pm \sqrt{30^2-4\cdot 24\cdot (-9) } }{ 2\cdot 24 } \\ x_{1,2} &=& \frac{-30\pm \sqrt{900+864 } }{ 48 } \\ x_{1,2} &=& \frac{-30\pm \sqrt{ 1764} }{ 48 } \\ x_{1,2} &=& \frac{-30\pm 42 }{ 48 } \\\\ x_1 &=& \frac{-30 + 42 } { 48 } \\ \mathbf{ x_1} &\mathbf{ =}& \mathbf{ 0.25 } \qquad \text{no solution } x > 0 \\\\ x_2 &=& \frac{-30 - 42 } { 48 } \\ \mathbf{ x_2} &\mathbf{ =}& \mathbf{ -1.5 } \\ \hline \end{array}\)

The only solution is x = -1.5

Each letter represents a non-zero digit. What is the value of t?

\(\begin{array}{|rrcll|} \hline (1): & c+o-u &=& 0\\ (2): & u+n-t &=& 0\\ (3): & t+c-s &=& 0\\ \hline (1)+(2)-(3): & ( c+o-u ) + ( u+n-t ) - ( t+c-s ) &=& 0+0 -0 \\ & \not \! c +o~ - \not \! u + \not \! u+n-t - t~ - \not \!c+s &=& 0 \\ & o +n+s -2t &=& 0 \\ & o +n+s &=& 2t \\ & 2t &=& o +n+s \quad | \quad o +n+s = 12 \\ & 2t &=& 12 \quad | \quad : 2 \\ & \mathbf{t} & \mathbf{=} & \mathbf{6} \\ \hline \end{array} \)

Hallo asinus,

ich vermute, das du einen Tippfehler hast.

Deine "blaue Funktion" (x-2010)^0.5 muss heißen (x-2016)^0.5!

Gruß Heureka