heureka

The equation \(\dfrac{x}{x+1} + \dfrac{x}{x+2} = kx\)  has exactly two complex roots. Find all possible complex values for \(k\).

Find the ordered pair \((a,b)\) of real numbers for which \((ax + b)(x^5 + 1) - (5x + 1)\) is divisible by \(x^2 + 1\).

\(\begin{array}{|rcll|} \hline && \mathbf{(ax + b)(x^5 + 1) - (5x + 1)} \\ &=& ax^6+bx^5+ax+b-5x-1 \\ &=&\mathbf{ ax^6+bx^5+(a-5)x +(b-1) } \\\\ &=& (rx^4+sx^3+tx^2+ux+v)(x^2+1) \\ &=& rx^6+sx^5+tx^4+ux^3+vx^2+rx^4+sx^3+tx^2+ux+v \\ &=&\mathbf{ \underbrace{r}_{=a}x^6+\underbrace{s}_{=b}x^5+ \underbrace{(t+r)}_{=0}x^4+\underbrace{(u+s)}_{=0}x^3+\underbrace{(v+t)}_{=0}x^2 + \underbrace{u }_{=a-5}x+\underbrace{v}_{=b-1} } \\ \hline v &=& b-1 \\ u &=& a-5 \\ r &=& a \\ s &=& b \\\\ t+r&=& 0 \quad | \quad r = a \\ t+a &=& 0 \\ \mathbf{t} &=& \mathbf{-a} \\\\ u +s &=& 0 \quad | \quad u=a-5,\ s = b \\ a-5+b &=& 0 \\ \mathbf{a +b} &=& \mathbf{5} \qquad ( 1) \\\\ v+t &=& 0 \quad | \quad v=b-1,\ t = -a \\ b-1-a &=& 0 \\ \mathbf{b-a} &=& \mathbf{1} \qquad ( 2) \\ \hline \end{array} \)

\(\mathbf{a=\ ?} \\ \mathbf{b=\ ?}\)

\(\begin{array}{|lrcll|} \hline (1)+(2): &a +b + b-a &=& 5+1 \\ & 2b &=& 6 \\ & \mathbf{b } &=& \mathbf{3} \\ \hline (1)-(2): &a +b + -(b-a) &=& 5-1 \\ & 2a &=& 4\\ & \mathbf{a } &=& \mathbf{2} \\ \hline \end{array}\)

\((a,b) =(2,3)\)

\(\begin{array}{|rcll|} \hline && \mathbf{(ax + b)(x^5 + 1) - (5x + 1)} \\ &=& (2x + 3)(x^5 + 1) - (5x + 1) \\ &=& \mathbf{2x^6+3x^5-3x+2} \\ \hline \end{array}\)

Check:

When \(f(x) = ax^3 - 6x^2 + bx - 5\) is divided by \(x - 1\), the remainder is \(-5\).
When \(f(x)\) is divided by \(x + 2\), the remainder is \(-53\).
Find the ordered pair \((a,b)\).

In general:

If a polynomial \(f(x)\) is divided by \(x-a\) , the remainder is the constant \(f(a)\) , and \(f(x)=q(x)·(x-a)+f(a)\)

\(\begin{array}{|lrcll|} \hline (x-1),\ a= 1: & \mathbf{f(1)} &=& \mathbf{-5} \\ & a\cdot 1^3 - 6\cdot 1^2 + b\cdot 1 - 5 &=& -5 \\ & a - 6 + b - 5 &=& -5 \\ &\mathbf{ a + b } &=& \mathbf{6} \qquad (1) \\ & \mathbf{b} &=& \mathbf{6-a} \\ \hline (x+2),\ a= -2: & \mathbf{f(-2)} &=& \mathbf{-53} \\ & a\cdot(-2)^3 - 6\cdot (-2)^2 + b\cdot(-2) - 5 &=& -53 \\ & -8a - 24 -2b &=& -48 \quad | \quad : (-2) \\ & 4a +12 +b &=& 24 \\ & \mathbf{4a+b} &=& \mathbf{12} \qquad (2) \\ & 4a+b &=& 12 \quad | \quad b=6-a \\ & 4a+6-a &=& 12 \\ & 3a &=& 6 \\ &\mathbf{ a } &=& \mathbf{2} \\\\ & b &=& 6-a \\ & b &=& 6-2 \\ &\mathbf{ b } &=& \mathbf{4} \\ \hline \end{array}\)

\(\mathbf{(a,b) = (2,4)}\)

Check:

How many ordered triplets \((a,b,c) \) of rational numbers are there where\( a,b,c\) are the roots of \(x^3 + ax^2 + bx + c = 0\) ?

I assume:

\(\begin{array}{|rcll|} \hline x^3 + ax^2 + bx + c = 0 &=& (x-a)(x-b)(x-c) \\ &=& x^3\underbrace{-(a+b+c)}_{=a}x^2\underbrace{+(ab+ac+bc)}_{=b}x\underbrace{-abc}_{=c} \\ \hline \mathbf{-abc} &=& \mathbf{c} \\ -ab &=& 1 \\ \mathbf{ab} &=& \mathbf{-1} \\ \mathbf{b} &=& \mathbf{-\dfrac{1}{a}} \\\\ \mathbf{-(a+b+c)} &=&\mathbf{ a } \\ -a-b-c &=& a \\ b+c &=& -2a \quad | \quad \cdot a \\ \mathbf{ab+ac} &=& \mathbf{-2a^2} \quad | \quad ab=-1 \\ -1+ac &=& -2a^2 \\ \mathbf{ ac }&=& \mathbf{1-2a^2} \\\\ \mathbf{ab+ac+bc} &=& \mathbf{b} \quad | \quad ab+ac= -2a^2 \\ -2a^2+bc &=& b \\ -2a^2 &=& b- bc \\ -2a^2 &=& b(1-c) \quad | \quad b=-\dfrac{1}{a} \\ -2a^2 &=&-\dfrac{1}{a} (1-c) \\ 2a^2 &=& \dfrac{1}{a} (1-c) \\ 2a^3 &=& 1-c \\ \mathbf{c} &=& \mathbf{1-2a^3} \\\\ \mathbf{ ac }&=& \mathbf{1-2a^2} \quad | \quad c=1-2a^3 \\ a(1-2a^3)&=& 1-2a^2 \\ -2a^4+a &=& 1-2a^2 \\ \mathbf{-2a^4+2a^2+a-1} &=& \mathbf{0} \\ \hline \end{array}\)

\(\begin{array}{|lcll|} \hline \mathbf{-2a^4+2a^2+a-1=0},\ b=-\dfrac{1}{a},\ c=1-2a^3 \\\\ a = 1,\ b = -1,\ c = -1 \\ \text{triplet}_1 ~(a,b,c) = (1,\ -1,\ -1 ) \\\\ a =0.56519771738363939644,\ b=-1.7692923542386314152,\ c=0.6388969194713526224 \\ \text{triplet}_2 ~(a,b,c) = (0.56519771738363939644,\ -1.7692923542386314152,\ 0.6388969194713526224 ) \\ \hline \end{array} \)

check:

\(\begin{array}{|lcll|} \hline \mathbf{x^3+ x^2-x-1=0} \\ x = -1 \\ x = 1 \\ \hline \mathbf{x^3+ 0.56519771738363939644x^2-1.7692923542386314152x+0.638896919471352622=0} \\ x=-1.76929235423863142 \\ x=0.5651977173836394 \\ x=0.6388969194713526 \\ \hline \end{array}\)

If the 5th and the 8th entries of a Fibonacci Sequence are 19 and 79, respectively, then what is the first entry of this sequence?

List of Fibonacci numbers with \(a_0 = 0\) and \(a_1 = 1\):

\(\small{ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \ldots& F_{-2} & F_{-1} & F_0 & F_1 & F_2 & F_3 & F_4 & F_5 & F_6 & F_7 & F_8 &F_9 &F_{10} &F_{11} &F_{12} &F_{13} &F_{14} &F_{15}&\ldots \\ \hline \ldots& -1 & 1 & 0 & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & 144 & 233 & 377 & 610&\ldots \\ \hline \end{array} }\)

In general:

\(\begin{array}{|rcll|} \hline a_n &=& F_na_1+F_{n-1}a_0 \quad | \quad \text{with } a_0=0 \text{ and } a_1 = 1,\quad \text{yield}\quad a_n =F_n \\ \hline \end{array} \)

\(\mathbf{a_1=\ ?}\)

\(\begin{array}{|lrcll|} \hline n=5: & F_5a_1+F_4a_0 &=& a_5 \quad | \quad a_5 = 19 \\ n=8: & F_8a_1+F_7a_0 &=& a_8 \quad | \quad a_8 = 79 \\ \hline & \mathbf{F_5a_1+F_4a_0} &=& \mathbf{ 19 } \qquad (1) \\ & a_0 &=& \dfrac{19-F_5a_1}{F_4} \quad | \quad F_4=3,\ F_5=5 \\ & a_0 &=& \dfrac{19-5a_1}{3} \\\\ & \mathbf{F_8a_1+F_7a_0} &=& \mathbf{79} \qquad (2) \\ & F_8a_1+F_7\cdot \left( \dfrac{19-5a_1}{3} \right) &=& 79 \quad | \quad F_7=13,\ F_8=21 \\ & 21a_1+13\cdot \left( \dfrac{19-5a_1}{3} \right) &=& 79 \quad | \quad \cdot 3 \\ & 63a_1+13\cdot ( 19-5a_1 ) &=& 237 \\ & 63a_1+247 - 65a_1 &=& 237 \\ & -2a_1 &=& -10 \\ & 2a_1 &=& 10 \\ & \mathbf{ a_1 } &=& \mathbf{ 5 } \\ \hline \end{array}\)

The first entry of this sequence is 5

Let
\(f(x)=(x^2+6x+9)^{50}-4x+3\), and let \(r_1,r_2,\ldots,r_{100}\) be the roots of \(f(x)\).
Compute \((r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}\).

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{(x^2+6x+9)^{50}-4x+3} \\ &=& \left((x+3)^2\right)^{50}-4x+3 \\ \mathbf{f(x)} &=& \mathbf{\left( x+3 \right)^{100}-4x+3} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline & f(r_1) =0 &=& \left( r_1+3 \right)^{100}-4r_1+3 \\ & f(r_2) =0 &=& \left( r_2+3 \right)^{100}-4r_2+3 \\ & \ldots \\ & f(r_{100}) =0 &=& \left( r_{100}+3 \right)^{100}-4r_{100}+3 \\ & \hline \\ \text{sum} & 0 &=& (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} \\ & && -4(r_1+r_2+\ldots + r_{100}) + 3\cdot 100 \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ \hline \end{array} \)

\(\mathbf{\text{vieta:}}\)

For any polynomial equation
\(0=x^n+a_{n-1}·x^{n-1}+...+a_2·x^2+a_1·x^1+a_0 \)
with the solutions \(r_1\dots r_n\), the relatively simple formulas for \(a_0\) and \(a_{n-1}\) are:

   \(a_0=(-1)^n \prod\limits_{k=1}^{n} r_k \\ a_{n-1}= -\sum \limits_{k=1}^{n} r_k\)

\(\begin{array}{|rcll|} \hline && \left( x+3 \right)^{100} \\ \\ &=& x^{100} + \binom{100}{1}x^{99}\cdot 3 + \ldots \\ \\ &=& x^{100} + \underbrace{300}_{ \underbrace{=a_{n-1}}_{= -(r_1+r_2+\ldots + r_{100})}} x^{99} + \ldots \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ && \boxed{r_1+r_2+\ldots + r_{100} = -300} \\ (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(-300) - 300 \\ \mathbf{(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}} &=& \mathbf{-1500} \\ \hline \end{array} \)

Bestimme den Wert des Parameters a

\(\det\left( \begin{array}{rrrr}a-1 & -2 \\-\frac{2}{5} & 4 \\\end{array}\right) =\dfrac{6}{5}\)

\(\begin{array}{|rcll|} \hline \det\left( \begin{array}{rrrr}a-1 & -2 \\-\frac{2}{5} & 4 \\\end{array}\right) &=& \dfrac{6}{5} \\ (a-1)*4- \left(-\dfrac{2}{5} \right)(-2) &=& \dfrac{6}{5} \\ 4a-4 - \left(\dfrac{2}{5} \right)*2 &=& \dfrac{6}{5} \\ 4a-4 - \dfrac{4}{5} &=& \dfrac{6}{5} \quad |\quad + \dfrac{4}{5} \\ 4a-4 &=& \dfrac{6}{5} + \dfrac{4}{5} \\ 4a-4 &=& \dfrac{6+4}{5} \\ 4a-4 &=& \dfrac{10}{5} \\ 4a-4 &=& 2 \quad | \quad +4 \\ 4a &=& 6 \quad | \quad :4 \\ a &=& \dfrac{6}{4} \\ \mathbf{ a } &=& \mathbf{\dfrac{3}{2}} \\ \hline \end{array} \)

The lines  \(y = \dfrac{5}{12} x\) and  \(y = \dfrac{4}{3} x\) are drawn in the coordinate plane.

Find the slope of the line that bisects the acute angle between these lines.

\(\begin{array}{|rcll|} \hline y &=& \dfrac{5}{12} x \quad | \quad x=12 ~ \Rightarrow ~ y=5, \quad \vec{v}=\binom{12}{5}, \quad \vec{v_0}=\dfrac{1}{\sqrt{12^2+5^2} }\dbinom{12}{5}= \dbinom{\frac{12}{13}}{\frac{5}{13}}\\ y &=& \dfrac{4}{3} x \quad | \quad x=3 ~ \Rightarrow ~ y=4, \quad \vec{w}=\binom{3}{4}, \quad \vec{w_0}=\dfrac{1}{\sqrt{3^2+4^2} }\dbinom{3}{4}=\dbinom{\frac{3}{5}}{\frac{4}{5}}\\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \vec{v_0}+\vec{w_0} &=& \dbinom{\frac{12}{13}}{\frac{5}{13}} + \dbinom{\frac{3}{5}}{\frac{4}{5}} \\ \vec{v_0}+\vec{w_0} &=& \dbinom{\frac{12}{13}+\frac{3}{5}}{\frac{5}{13}+\frac{4}{5}} \\ \vec{v_0}+\vec{w_0} &=& \dbinom{\frac{99}{65}}{\frac{77}{65}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{slope of the line that bisects} &=& \dfrac{y}{x} \\\\ &=& \dfrac{\frac{77}{65}}{\frac{99}{65}} \\\\ &=& \dfrac{77}{99} \\\\ &=& \dfrac{7}{9} \\ \hline \end{array}\)

Löse die Parametergleichung ohne Fallunterscheidung nach x und y auf

\(\varphi x -\mu y=2\varphi \\ x+ \left(\dfrac{\mu}{\varphi} \right)^2y=1+\dfrac{\mu^2}{\varphi^2}\)

\(\begin{array}{|rcll|} \hline \mathbf{\varphi x -\mu y } &=& \mathbf{2\varphi} \quad | \quad :\varphi \\ x - \dfrac{\mu}{\varphi} y &=& 2 \\ \mathbf{\dfrac{\mu}{\varphi} y} &=& \mathbf{x-2} \qquad (1) \\ \hline x+ \left(\dfrac{\mu}{\varphi}\right)^2y &=& 1+\dfrac{\mu^2}{\varphi^2} \\ x+ \dfrac{\mu^2}{\varphi^2} y &=& 1+\dfrac{\mu^2}{\varphi^2} \\ x+ \dfrac{\mu }{\varphi } \left(\dfrac{\mu }{\varphi } y \right) &=& 1+\dfrac{\mu^2}{\varphi^2} \\ x+ \dfrac{\mu }{\varphi } (x-2) &=& 1+\dfrac{\mu^2}{\varphi^2} \\ x+ \dfrac{\mu }{\varphi }x -2\dfrac{\mu }{\varphi } &=& 1+\dfrac{\mu^2}{\varphi^2} \\ x+ \dfrac{\mu }{\varphi }x &=& 1+ 2\dfrac{\mu }{\varphi } + \dfrac{\mu^2}{\varphi^2} \\ x \left(1+ \dfrac{\mu }{\varphi } \right) &=& \left(1+ \dfrac{\mu }{\varphi } \right)^2 \\ \mathbf{ x } &=& \mathbf{1+ \dfrac{\mu }{\varphi }} \\ \hline \mathbf{\dfrac{\mu}{\varphi} y} &=& \mathbf{x-2} \qquad (1) \\ \dfrac{\mu}{\varphi} y &=& 1+ \dfrac{\mu }{\varphi }-2 \\ \dfrac{\mu}{\varphi} y &=& \dfrac{\mu }{\varphi }-1 \\ y &=& \dfrac{\mu }{\varphi }\dfrac{\varphi}{\mu} -\dfrac{\varphi}{\mu} \\ \mathbf{ y } &=& \mathbf{1 -\dfrac{\varphi}{\mu} } \\ \hline \end{array}\)