heureka

\sqrt{15/4-2i}

\(\begin{array}{|rcll|} \hline z &=& a+bi \qquad |z| = \sqrt{a^2+b^2} \qquad Re(z) = a \\\\ \sqrt{z} &=& \sqrt{ \frac{|z|+Re(z)}{2} }\pm i\sqrt{ \frac{|z|-Re(z)}{2} } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline z &=& \frac{15}{4}-2i \qquad a= Re(z)= \frac{15}{4} \qquad b = -2 \\\\ |z| &=& \sqrt{\left(\frac{15}{4} \right)^2+(-2)^2} \\ &=& \sqrt{\left(\frac{15^2}{4^2} \right)^2+4} \\ &=& \sqrt{\left(\frac{15^2}{4^2} \right)^2+4\cdot \frac{4^2}{4^2}} \\ &=& \sqrt{\frac{15^2+4^3}{4^2} }\\ &=& \frac{ \sqrt{15^2+4^3} } {4} \\ &=& \frac{ \sqrt{289} } {4} \\ \mathbf{|z|} &\mathbf{=}& \mathbf{\frac{17} {4}} \\\\ \sqrt{z} &=& \sqrt{ \frac{|z|+Re(z)}{2} }\pm i\sqrt{ \frac{|z|-Re(z)}{2} } \\ \sqrt{\frac{15}{4}-2i} &=& \sqrt{ \frac{\frac{17} {4}+\frac{15}{4}}{2} }\pm i\sqrt{ \frac{\frac{17} {4}-\frac{15}{4}}{2} } \\ &=& \sqrt{ \frac{\frac{17+15} {4} } {2} }\pm i \sqrt{ \frac{\frac{17-15} {4} } {2} } \\ &=& \sqrt{ \frac{\frac{32} {4} } {2} }\pm i \sqrt{ \frac{\frac{2} {4} } {2} } \\ &=& \sqrt{ \frac{ 32 } {2\cdot 4} }\pm i \sqrt{ \frac{2} {2\cdot 4} } \\ &=& \sqrt{ \frac{ 32 } {8} }\pm i \sqrt{ \frac{2} {8} } \\ &=& \sqrt{ 4 }\pm i \sqrt{ \frac{1} {4} } \\ &=& 2\pm i\cdot \frac12 \\ \hline \end{array}\)

\( \begin{array}{|rcll|} \hline \text{alle zweiten Wurzeln von } \frac{15}{4}-2i:\\ 2 + i\cdot \frac12 \\ 2 - i\cdot \frac12 \\ \hline \end{array} \)

the polynomial x^3+ax^2-bx=6 has factors of (x+1) (x+2) what is a and b

I. factors of \((x+1) \quad \Rightarrow x_1 = -1\)

\(\begin{array}{|rcll|} \hline x^3+ax^2-bx -6 &=& 0 \quad &| \quad x_1 = -1\\ ( -1)^3+a( -1)^2-b(-1) -6 &=& 0 \\ -1+a+b -6 & =& 0 \\ a+b -7 & =& 0 \quad &| \quad +7\\ a+b & =& 7 \\ \mathbf{ b} & \mathbf{=}& \mathbf{7-a} \\ \hline \end{array}\)

 II. factors of (x+2) \((x+2) \quad \Rightarrow x_2 = -2\)

\(\begin{array}{|rcll|} \hline x^3+ax^2-bx -6&=& 0 \quad &| \quad x_2 = -2\\ ( -2)^3+a( -2)^2-b(-2) -6 &=& 0 \\ -8+4a+2b -6 &=& 0 \\ 4a+2b -14 &=& 0 \quad &| \quad +14\\ 4a+2b &=& 14 \quad &| \quad : 2\\ 2a+ b &=& 7 \quad &| \quad \mathbf{ b = 7-a} \\ 2a+ 7-a &=& 7 \\ a+ 7 &=& 7 \quad &| \quad x_2 = -7\\ a &=&0 \\ \mathbf{ a} & \mathbf{=}& \mathbf{0} \\\\ b &=& 7-a \quad &| \quad \mathbf{ a} \mathbf{=} \mathbf{0}\\ b &=& 7 - 0 \\ b &=& 7\\ \mathbf{ b} & \mathbf{=}& \mathbf{7 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x^3+ax^2-bx &=& 6 \quad &| \quad \mathbf{ a} \mathbf{=} \mathbf{0} \qquad \mathbf{ b = 7 }\\ x^3+0\cdot x^2- 7\cdot x &=& 6 \\ \mathbf{x^3 - 7\cdot x }& \mathbf{=}& \mathbf{6} \\ \hline \end{array}\)

Three equal circles with radius r are drawn as shown,
each with its centre on the circumference of the other two circles.
A, B and C are the centres of the three circles.
Prove that an expression for the area of the shaded region is:

\(A=\frac{r^2}{2}\left(\pi -\sqrt{3}\right)\)

\(\begin{array}{|rcll|} \hline A_{\triangle ABC} &=& \frac12 \cdot r \cdot h_{\triangle ABC} \\\\ h_{\triangle ABC}^2 + (\frac{r}{2})^2 &=& r^2 \\ h_{\triangle ABC}^2 + \frac{r^2}{4} &=& r^2 \\ h_{\triangle ABC}^2 = r^2 - \frac{r^2}{4} \\ h_{\triangle ABC}^2 = \frac34 \cdot r^2 \\ h_{\triangle ABC} = \frac{\sqrt{3}}{2} \cdot r \\\\ A_{\triangle ABC} &=& \frac12 \cdot r \cdot h_{\triangle ABC} \\ A_{\triangle ABC} &=& \frac12 \cdot r \cdot \frac{\sqrt{3}}{2} \cdot r \\ \mathbf{A_{\triangle ABC} }&\mathbf{=}& \mathbf{\frac{r^2}{4} \cdot \sqrt{3} }\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline A_{Arc} &=& \pi r^2 \cdot \frac{60^{\circ}} {360^{\circ}} \\ A_{Arc} &=& \frac{\pi r^2} {6} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline A_{\text{shaded region}} &=& 3\cdot (A_{Arc}-A_{\triangle ABC})+A_{\triangle ABC} \\ A_{\text{shaded region}} &=& 3\cdot A_{Arc}-3\cdot A_{\triangle ABC}+A_{\triangle ABC} \\ A_{\text{shaded region}} &=& 3\cdot A_{Arc}-2\cdot A_{\triangle ABC} \\ A_{\text{shaded region}} &=& 3\cdot \frac{\pi r^2} {6}-2\cdot \frac{r^2}{4} \cdot \sqrt{3} \\ A_{\text{shaded region}} &=& \frac{\pi r^2} {2}- \frac{r^2}{2} \cdot \sqrt{3} \\ \mathbf{A_{\text{shaded region}} }&\mathbf{=}& \mathbf{\frac{r^2}{2} \cdot ( \pi- \sqrt{3} )} \\ \hline \end{array}\)

f(x)=(x+3)/sqrt{x-8}

What is the domain?

\(\begin{array}{|rcll|} \hline x-8 &>& 0 \quad &| \quad +8\\ \mathbf{x} &\mathbf{>}& \mathbf{8}\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \sqrt{x-8} & \ne & 0 \\ x-8 & \ne & 0^2 \quad &| \quad +8\\ \mathbf{x} & \mathbf{\ne} & \mathbf{8}\\ \hline \end{array}\)

The domain is x > 8

2^(x)*4^(3x)= 10   x=?

\(\begin{array}{|rcll|} \hline 2^{x}\cdot 4^{3x} &=& 10 \\ 2^{x}\cdot (4^{3})^x &=& 10 \quad &| \quad 4^{3} = 64\\ 2^{x}\cdot 64^x &=& 10 \\ (2\cdot 64)^{x} &=& 10 \\ 128^{x} &=& 10 \quad &| \quad \log_{10}{()}\\ \log_{10}{( 128^{x})} &=& \log_{10}{(10)} \quad &| \quad \log_{10}{(10)} = 1\\ \log_{10}{( 128^{x})} &=& 1\\ x\cdot \log_{10}{( 128 )} &=& 1\quad &| \quad : \log_{10}{( 128 )}\\ x &=& \frac{1} {\log_{10}{( 128 )}} \\ \mathbf{x} &\mathbf{=}& \mathbf{0,47456115641} \\ \hline \end{array}\)

e

Given that \((\sqrt{2}x-i\sqrt{3})\) divides P(x) find all roots of P(x)

\(\begin{array}{|rcll|} \hline P(x) &=& x^6-6x^5+\frac{17}{2}x^4-7x^3+\frac{21}{2}x^2+3x \\ P(x) &=& x\cdot \left( x^5-6x^4+\frac{17}{2}x^3-7x^2+\frac{21}{2}x+3 \right)\\ \hline \end{array} \)

If \((\sqrt{2}x-i\sqrt{3})\) divides P(x) then \((\sqrt{2}x+i\sqrt{3}) \) divides P(x)

\(\begin{array}{|rcll|} \hline (\sqrt{2}x-i\sqrt{3}) \cdot (\sqrt{2}x+i\sqrt{3}) &=& (\sqrt{2}x)^2 -(i\sqrt{3})^2 \\ &=& 2x^2 -(i^2\cdot 3) \quad &| \quad i^2 = -1 \\ &=& 2x^2 -[(-1)\cdot 3] \\ &=& 2x^2 +3\\ \hline \end{array}\)

{nl} \(\begin{array}{rcll} \hline \left( x^5-6x^4+\frac{17}{2}x^3-7x^2+\frac{21}{2}x+3 \right) : 2x^2 + 3 &=& \frac12x^3-3x^2+\frac72x+1\\ \hline \end{array} \)

\(\begin{array}{rcll} \Rightarrow P(x) &=& x\cdot (\sqrt{2}x-i\sqrt{3}) \cdot (\sqrt{2}x+i\sqrt{3}) \cdot \left( \frac12x^3-3x^2+\frac72x+1 \right)\\ \end{array}\)

test x=2: \(\Rightarrow \frac12x^3-3x^2+\frac72x+1 = 0 \)

\(\begin{array}{rcll} \hline \left( \frac12x^3-3x^2+\frac72x+1 \right) : x-2 &=& \frac12x^2-2x-\frac12\\ \hline \end{array}\)

\(\begin{array}{rcll} \Rightarrow P(x) &=& x\cdot (\sqrt{2}x-i\sqrt{3}) \cdot (\sqrt{2}x+i\sqrt{3}) \cdot (x-2) \cdot \left( \frac12x^2-2x-\frac12 \right)\\ P(x) &=& x\cdot (\sqrt{2}x-i\sqrt{3}) \cdot (\sqrt{2}x+i\sqrt{3}) \cdot (x-2) \cdot \frac12 \cdot \left( x^2-4x-1 \right)\\ \end{array}\)

Roots:

\(\small{ \begin{array}{|lrccccccccc|} \hline P(x)=& 0 &=& x &\cdot& (\sqrt{2}x-i\sqrt{3}) &\cdot& (\sqrt{2}x+i\sqrt{3}) &\cdot& (x-2) \cdot \frac12 \cdot &\left( x^2-4x-1 \right)&\\ \text{1. Root} & && \mathbf{x_1 = 0} \\ \text{2. Root} & & & & & \sqrt{2}x-i\sqrt{3} = 0 \\ & & & & & \sqrt{2}x=i\sqrt{3} \\ & & & & & \mathbf{x_2=i\frac{\sqrt{3} } {\sqrt{2}}} \\ \text{3. Root} & & & & & & & \sqrt{2}x+i\sqrt{3} = 0 \\ & & & & & & & \sqrt{2}x=-i\sqrt{3} \\ & & & & & & & \mathbf{x_3=-i\frac{\sqrt{3} } {\sqrt{2}}} \\ \text{4. Root} & & & & & & & & & x-2=0 \\ & & & & & & & & & \mathbf{x_4=2} \\ \text{5.6. Root} & & & & & & & & & & x^2-4x-1 = 0\\ & & & & & & & & & & x = \frac{4\pm\sqrt{4^2-4\cdot(-1)}}{2} \\ & & & & & & & & & & x = \frac{4\pm\sqrt{4\cdot 5}}{2} \\ & & & & & & & & & & x = \frac{4\pm 2 \sqrt{ 5}}{2} \\ & & & & & & & & & & x = 2\pm \sqrt{ 5} \\ & & & & & & & & & & \mathbf{x_5= 2+ \sqrt{ 5}} \\ & & & & & & & & & & \mathbf{x_6 = 2- \sqrt{ 5}} \\ \hline \end{array} } \)

1.

difference between 2 numbers is 16 and sum of their squares is 130.
what are the numbers?

Let first number = n₁

Let second number = n₂

\(\begin{array}{|lrcll|} \hline (1) & n_1 - n_2 &=& 16\\ & n_2 &=& n_1-16 \\\\ (2) & n_1^2 + n_2^2 &=& 130 \quad & | \quad n_2 = n_1-16\\ & n_1^2 + (n_1-16)^2 &=& 130 \\ & n_1^2 + n_1^2 -32n_1+16^2 &=& 130 \\ & 2n_1^2 -32n_1+16^2 &=& 130 \quad & | \quad -130\\ & 2n_1^2 -32n_1+16^2-130 &=& 0\\ & 2n_1^2 -32n_1+126 &=& 0 \quad & | \quad :2\\ & n_1^2 -16n_1+63 &=& 0 \\ & n_1 &=& \frac{ 16\pm\sqrt{ (-16)^2-4\cdot 63 } } {2} \\ & n_1 &=& \frac{ 16\pm\sqrt{ 16^2-4\cdot 63 } } {2} \\ & n_1 &=& \frac{ 16\pm\sqrt{ 4 } } {2} \\ & n_1 &=& \frac{ 16\pm 2 } {2} \\ & n_1 = \frac{ 16+ 2 } {2} & \text{or} & n_1= \frac{ 16- 2 } {2} \\ & \mathbf{n_1 = 9} & \text{or} & \mathbf{n_1= 7} \\\\ & n_2 &=& n_1-16 \\ & n_2 = 9-16 & \text{or} & n_2= 7-16 \\ & \mathbf{n_2 = -7} & \text{or} & \mathbf{n_2= -9} \\ \hline \end{array}\)

check:

first number = 9

second number = -7

9 -(- 7 ) = 16

9² + (-7)² = 130

first number = 7

second number = -9

7 -(- 9 ) = 16

7² + (-9)² = 130

2.

perimeter of rectangle is 50cm and area is 144cm^2. find length and width.

Let length = a

Let width = b

\(\begin{array}{|lrcll|} \hline \text{Area: } & ab &=& 144 \quad & | \quad :a\\ & b &=& \frac{144}{a} \\\\ \text{Perimeter: } & 2a+2b &=& 50 \\ & 2\cdot(a+b) &=& 50 \quad & | \quad :2\\ & a+b &=& 25 \quad & | \quad b = \frac{144}{a}\\ & a+\frac{144}{a} &=& 25 \quad & | \quad \cdot a \\ & a^2+ 144 &=& 25a \quad & | \quad -25a \\ & a^2 -25a + 144 &=&0 \\ & a &=& \frac{ 25\pm\sqrt{ 25^2-4\cdot 144 } } {2} \\ & a &=& \frac{ 25\pm\sqrt{ 49 } } {2} \\ & a &=& \frac{ 25\pm 7 } {2} \\ & a = \frac{ 25+ 7 } {2} & \text{or} & a= \frac{ 25- 7 } {2} \\ & a = \frac{ 32 } {2} & \text{or} & a= \frac{ 18 } {2} \\ & \mathbf{a = 16} & \text{or} & \mathbf{a= 9} \\\\ & a+b &=& 25 \\ & b &=& 25-a \\ & b = 25-16 & \text{or} & b= 25-9\\ & \mathbf{b = 9} & \text{or} & \mathbf{b= 16} \\ \hline \end{array}\)

3.

man travels 196km by train and returns in a car that travels 21km/h faster.

if total journey took 11hrs, find speeds of train and car

Let time train = t₁

Let time car = t₂

\(t_1 + t_2 = 11\ h \\ t_2 = 11 - t_1\)

\(\begin{array}{|lrcll|} \hline (1) & v_{\text{train}} \cdot t_1 &=& 196\ km \\ & v_{\text{train}} &=& \frac{196}{t_1} \\\\ (2) & v_{\text{car}} &=& v_{\text{train}} + 21\frac{km}{h} \\ & v_{\text{car}} \cdot t_2 &=& 196\ km \\ & (v_{\text{train}} + 21 ) \cdot t_2 &=& 196 \\ & (\frac{196}{t_1} + 21 ) \cdot t_2 &=& 196 \quad & | \quad t_2 = 11 - t_1\\ & (\frac{196}{t_1} + 21 ) \cdot (11 - t_1) &=& 196 \\ & \frac{196\cdot 11}{t_1} -196 + 21\cdot 11 -21t_1 &=& 196 \quad & | \quad - 196\\ & \frac{196\cdot 11}{t_1} -196 + 21\cdot 11-196 -21t_1 &=& 0\\ & \frac{2156}{t_1} -196 + 21\cdot 11-196 -21t_1 &=& 0\\ & \frac{2156}{t_1} -161 -21t_1 &=& 0 \quad & | \quad \cdot t_1 \\ & \frac{2156}{t_1} -161 -21t_1 &=& 0 \quad & | \quad \cdot t_1 \\ & 2156 -161t_1 -21t_1^2 &=& 0 \quad & | \quad \cdot (-1) \\ & 21t_1^2 + 161t_1 -2156 &=& 0 \\ & t_1 &=& \frac{ -161\pm\sqrt{(-161)^2-4\cdot 21 \cdot(-2156)} } {2\cdot 21} \\ & t_1 &=& \frac{ -161\pm 455 } {42} \\ & t_1 = \frac{ -161+455 } {42} & \text{or} & t_1= \frac{ -161- 455 } {42} \text{ no solution} \\ & \mathbf{t_1 = 7\ h } & & \\\\ & v_{\text{train}} &=& \frac{196}{t_1} \\ & v_{\text{train}} &=& \frac{196}{7} \\ & \mathbf{v_{\text{train}}} &\mathbf{=}& \mathbf{28\ \frac{km}{h}} \\\\ & v_{\text{car}} &=& v_{\text{train}} + 21 \\ & v_{\text{car}} &=& 28 + 21 \\ & \mathbf{v_{\text{car}}} &\mathbf{=}& \mathbf{49\ \frac{km}{h}} \\\\ \hline \end{array} \)

4.

wire 80cm in length is cut into 2 parts ad each part is bent to form a square.

if sum of areas of squares is 300cm^2, find length of the sides of the square

Let first part length = a

Let second part lenght = b

Let side of the first square = x

Let sied of the second square = y

\(\begin{array}{|lrcll|} \hline (1) & x^2 + y^2 &=& 300 \\\\ (2) & a=4x & \text{and} & b = 4y \\\\ (3) & a+b &=& 80 \\ & 4x + 4y &=& 80 \\ & 4(x+y) &=& 80 \quad &| \quad :4 \\ & x+y &=& 20 \\ & \mathbf{y} &\mathbf{=}& \mathbf{20 - x} \\ \hline (1) & x^2 + y^2 &=& 300 \\ & x^2 + (20 - x)^2 &=& 300 \\ & x^2 + 20^2-40x+x^2 &=& 300 \\ & 2x^2 -40x + 20^2 &=& 300 \\ & 2x^2 -40x + 400 &=& 300 \quad &| \quad -300 \\ & 2x^2 -40x + 400-300 &=& 0 \\ & 2x^2 -40x + 100 &=& 0 \quad &| \quad :2 \\ & x^2 -20x + 50 &=& 0 \\ & x &=& \frac{ 20\pm\sqrt{20^2-4\cdot 50} } {2} \\ & x &=& \frac{ 20\pm\sqrt{200} } {2} \\ & x &=& \frac{ 20\pm\sqrt{2\cdot 100} } {2} \\ & x &=& \frac{ 20\pm 10\cdot \sqrt{2} } {2} \\ & x &=& 10\pm 5\cdot \sqrt{2} \\ & \mathbf{x= 10+ 5\cdot \sqrt{2} } & \text{or} & \mathbf{x=10- 5\cdot \sqrt{2}} \\\\ & y &=& 20-x \\ & y = 20 - ( 10+ 5\cdot \sqrt{2} ) & \text{or} & y = 20 - ( 10- 5\cdot \sqrt{2} ) \\ & y = 20 - 10- 5\cdot \sqrt{2} & \text{or} & y = 20 - 10 + 5\cdot \sqrt{2} \\ & \mathbf{y= 10- 5\cdot \sqrt{2} } & \text{or} & \mathbf{y= 10 + 5\cdot \sqrt{2} } \\ \hline \end{array} \)