heureka

why the cubic root of the (-8) give (-2) only, where is the other complex root.

if I want to get the full roots, how I can get them.

$\begin{array}{rcll} x &=& \sqrt[3]{-8} \\ x^3 &=& -8 \end{array}\\ \boxed{~ \begin{array}{rcll} x^3 +8 &=& 0 \\ && 0 = (x-x_1)(x-x_2)(x-x_3) \qquad | \qquad x_1 = -2 \\ && 0 = (x+2)(x-x_2)(x-x_3) \end{array} ~}$

$\begin{array}{rcll} && (x+2)(x-x_2)(x-x_3) \\ &=& (x+2)[x^2-x(x_2+x_3)+x_2x_3] \\ &=& x^3-x^2(x_2+x_3)+x\cdot x_2x_3 + 2x^2-2x(x_2+x_3)+2x_2x_3 \\ &=& x^3+x^2\cdot[2-(x_2+x_3)]+x\cdot[x_2x_3-2(x_2+x_3) ] +2x_2x_3 \\ \text{compare with } x^3+8\\ &=& x^3 + 8 \\ &=& x^3+x^2\cdot\underbrace{[2-(x_2+x_3)]}_{=0}+x\cdot\underbrace{[x_2x_3-2(x_2+x_3)]}_{=0} + \underbrace{2x_2x_3}_{=8} \\ \end{array}$

$\begin{array}{lcll} (1) & 2-(x_2+x_3) &=& 0\\ & x_2+x_3 &=& 2 \\ \hline (2) & x_2x_3-2(x_2+x_3) &=& 0\\ \hline (3) & 2x_2x_3 &=& 8\\ & x_2x_3 &=& 4\\ & x_3 &=& \frac{4}{x_2} \\ \hline (1) & x_2+x_3 &=& 2 \qquad & | \qquad x_3 = \frac{4}{x_2}\\ & x_2 + \frac{4}{x_2} &=& 2 \qquad & | \qquad \cdot x_2\\ & x_2^2 + 4 &=& 2x_2\\ & x_2^2 - 2x_2 + 4 &=& 0\\ & x_2 &=& \frac{2\pm \sqrt{4-4\cdot 4} }{2} \\ & x_2 &=& 1\pm \frac{\sqrt{-3\cdot 4} }{2} \\ & x_2 &=& 1\pm \frac{\sqrt{(-1)\cdot 3\cdot 4} }{2} \\ & x_2 &=& 1\pm \frac{\sqrt{-1}\sqrt{3}\sqrt{4} }{2} \\ & x_2 &=& 1\pm \frac{i\sqrt{3}2 }{2} \\ & x_2 &=& 1\pm \sqrt{3}\cdot i \\ & \mathbf{x_2} & \mathbf{=} & \mathbf{ 1 + \sqrt{3}\cdot i } \\\\ \hline (1) & x_2+x_3 &=& 2 \qquad & | \qquad x_2 = 1 + \sqrt{3}\cdot i\\ & 1 + \sqrt{3}\cdot i+x_3 &=& 2 \\ & x_3 &=& 2-1 - \sqrt{3}\cdot i \\ & x_3 &=& 1 - \sqrt{3}\cdot i\\ & \mathbf{x_3} & \mathbf{=} & \mathbf{1 - \sqrt{3}\cdot i } \\ \end{array}$

Evaluate: 3(square root sign) b +ab if a = 27 and b = 8

$\begin{array}{rcll} && \sqrt[3]{b} + a\cdot b \\ &=& \sqrt[3]{8} + 27\cdot 8 \\ &=& \sqrt[3]{2^3} + 27\cdot 8 \qquad | \qquad 8 = 2^3\\ &=& 2^{ \frac{3}{3} }+ 27\cdot 8 \\ &=& 2^1+ 27\cdot 8 \\ &=& 2+ 27\cdot 8 \\ &=& 2+ 216 \\ &=& 218\\\\ \sqrt[3]{b} + a\cdot b &=& 218 \qquad \text{if } a = 27 \text{ and } b = 8 \end{array}$

The van traveled at 120 miles in 7 hours, how long would it take to travel 1400 miles at the same speed?

$\begin{array}{rcll} \dfrac{120\ \text{miles}} {7\ \text{hour} } \cdot x &=& 1400\ \text{miles} \\\\ x &=& 1400\ \text{miles} \cdot \dfrac {7\ \text{hour} } {120\ \text{miles}} \\\\ x &=& 1400 \cdot \dfrac {7\ \text{hour} } {120} \\\\ x &=& \dfrac{1400}{120} \cdot 7\ \text{hour} \\\\ x &=& \dfrac{140}{12} \cdot 7\ \text{hour} \\\\ x &=& \dfrac{140\cdot 7}{12} \ \text{hour} \\\\ x &=& \dfrac{980}{12} \ \text{hour} \\\\ x &=& 81\ \dfrac{2}{3} \ \text{hour} \\\\ x &=& 81\ \text{hour} + \dfrac{2}{3}\ \text{hour} \cdot \dfrac{60\ \text{minutes} } {1\ \text{hour} }\\\\ x &=& 81\ \text{hour} + \dfrac{2}{3} \cdot 60\ \text{minutes} \\\\ x &=& 81\ \text{hour} + \dfrac{120}{3}\ \text{minutes} \\\\ x &=& 81\ \text{hour} + 40\ \text{minutes} \\\\ \end{array}$

The van traveled at 120 miles in  hours. how long would it take to travel 1400 miles at the same speed?

$\begin{array}{rcll} \dfrac{120\ \text{miles}} {1\ \text{hour} } \cdot x &=& 1400\ \text{miles} \\\\ x &=& 1400\ \text{miles} \cdot \dfrac {1\ \text{hour} } {120\ \text{miles}} \\\\ x &=& 1400 \cdot \dfrac {1\ \text{hour} } {120} \\\\ x &=& \dfrac{1400}{120} \cdot 1\ \text{hour} \\\\ x &=& \dfrac{140}{12} \ \text{hour} \\\\ x &=& 11\ \dfrac{2}{3} \ \text{hour} \\\\ x &=& 11\ \text{hour} + \dfrac{2}{3}\ \text{hour} \cdot \dfrac{60\ \text{minutes} } {1\ \text{hour} }\\\\ x &=& 11\ \text{hour} + \dfrac{2}{3} \cdot 60\ \text{minutes} \\\\ x &=& 11\ \text{hour} + \dfrac{120}{3}\ \text{minutes} \\\\ x &=& 11\ \text{hour} + 40\ \text{minutes} \\\\ \end{array}$

It would take to travel 1400 miles about 11 hours and 40 minutes

It took 4 cups of flour to make 6 loaves. How many loaves can be made with 84 cups?

$\begin{array}{rcll} \dfrac{4\ \text{cups of flour}} {6\ \text{loaves} } \cdot x &=& 84\ \text{cups of flour} \\\\ x &=& 84\ \text{cups of flour} \cdot \dfrac {6\ \text{loaves} } {4\ \text{cups of flour}} \\\\ x &=& 84 \cdot \dfrac {6\ \text{loaves} } {4} \\\\ x &=& \dfrac{84}{4} \cdot 6\ \text{loaves} \\\\ x &=& 21 \cdot 6\ \text{loaves} \\\\ x &=& 126\ \text{loaves} \end{array}$

With 84 cups of flour can be made 126 loaves.

binomial expansion: What is the 4th term of (2x-3y)^10

$\begin{array}{rrcr} & (2x-3y)^{10} = \\\\ 1\text{st term:} & \dbinom{10}{0} \cdot (2x)^{10} \cdot (-3y)^0 \\ 2\text{nd term:} & +\dbinom{10}{1} \cdot (2x)^{9} \cdot (-3y)^1 \\ 3\text{rd term:} & +\dbinom{10}{2} \cdot (2x)^{8} \cdot (-3y)^2 \\ 4\text{th term:} & +\dbinom{10}{3} \cdot (2x)^{7} \cdot (-3y)^3 &=& \dfrac{10}{3} \cdot \dfrac{9}{2} \cdot \dfrac{8}{1} \cdot 2^7\cdot x^7 \cdot (-3)^3 \cdot y^3\\ &&=& \dfrac{10}{3} \cdot \dfrac{9}{2} \cdot \dfrac{8}{1} \cdot 2^7\cdot x^7 \cdot (-27) \cdot y^3\\ &&=& \dfrac{10}{3} \cdot \dfrac{9}{2} \cdot \dfrac{8}{1} \cdot 128\cdot x^7 \cdot (-27) \cdot y^3\\ &&=& 10 \cdot 3 \cdot 4 \cdot 128\cdot x^7 \cdot (-27) \cdot y^3\\ &&=& -27 \cdot 10 \cdot 3 \cdot 4 \cdot 128\cdot x^7 \cdot y^3\\ &&=& -414720\cdot x^7 \cdot y^3\\ \dots & \dots \end{array}$

wie rechnet man in einem Dreieck die Höhe aus?

1. Im rechtwinkligen Dreieck:

$h_c = \dfrac{a\cdot b}{c}$

2. Im allgemeinen Dreieck:

$h_c = \dfrac{a\cdot b}{c} \cdot \sin \gamma$

linearfaktoren der Gleichung: f(x)=5x²+130x+825 ?

$\begin{array}{rcll} 5x^2+130x+825 &=& 5\cdot (x^2 + 26x+165 ) \\ &=& 5\cdot (x + 11 ) \cdot (x +15)\\ \end{array}$

4,25% als bruch

$\begin{array}{rcll} 4,25\ \% &=& \dfrac{4,25\cdot 4}{4}\ \% \\ &=& \dfrac{17}{4}\ \% \\ &=& \dfrac{17}{400} \end{array}$

When the polynomial f(x) is divided by x3 - 2, the quotient and remainder are x2 + 4x - 1 and x, respectively. Determine the value f(3).

$\begin{array}{rcll} \dfrac{f(x)}{x^3-2} &=& x^2+4x-1+\dfrac{x}{x^3-2} \qquad | \qquad \cdot (x^3-2) \\ f(x) &=& (x^2+4x-1) \cdot (x^3-2)+ x \\ f(3) &=& (3\cdot 3+4\cdot 3-1)\cdot(3\cdot 3 \cdot 3 - 2) + 3 \\ f(3) &=& (9+12-1)\cdot (27-2) +3 \\ f(3) &=& 20\cdot 25 +3 \\ f(3) &=& 500 +3 \\ \mathbf{f(3)} &\mathbf{=}& \mathbf{503} \end{array}$