heureka

Find all values of x such that 

\(\large{ \mathbf{\Big| ~ |x-17| - 3~ \Big| = \Big|~ |x-15| - 5 ~\Big|} }\)

\(\begin{array}{|rcll|} \hline \Big| ~ |x-17| - 3~ \Big| &=& \Big|~ |x-15| - 5 ~\Big| \quad \boxed{\text{Formula: } |a|^2 = (a)^2} \\ \left( ~ |x-17| - 3~ \right)^2 &=& \left(~ |x-15| - 5 ~\right)^2 \\ (x-17)^2-6|x-17|+9 &=& (x-15)^2-10|x-15|+25 \\ 10|x-15| &=& 6|x-17|+ (x-15)^2-(x-17)^2 + 16 \\ 10|x-15| &=& 6|x-17| +4x-48 \quad | \quad : 2 \\ \mathbf{5|x-15|} &\mathbf{=}& \mathbf{3|x-17| +2x-24} \\ \hline \end{array} \)

Here we have two different amounts.

To dissolve them, we must make a double case distinction.

Usually we did this one after the other.
For reasons of space, we start with the first case in which the content of the left amount
is greater than or equal to zero.

\(\begin{array}{|rcll|} \hline & \underline{x\ge 15:}& \\\\ & 5(x-15)\\ & =3|x-17|+2x-24 \\ \\ \underline{\text{for } x\ge 17:} && \underline{\text{for } x \lt 17: } \\\\ 5(x-15)=3(x-17)+2x-24 && 5(x-15)=-3(x-17)+2x-24 \\ 5x-75=3x-51+2x-24 && 5x-75=-3x+51+2x-24 \\ -75=-75~\text{true} && 6x=102 \\ x\ge 15 \wedge x\ge 17 && x=17 \quad (x\lt 17) ~\text{no solution} \\ \boxed{ x\ge 17 } && \\ \hline \hline & \underline{x\lt 15:}& \\\\ & -5(x-15)\\ & =3|x-17|+2x-24 \\ \\ \underline{\text{for } x\ge 17:} && \underline{\text{for } x \lt 17: } \\\\ -5(x-15)=3(x-17)+2x-24 && -5(x-15)=-3(x-17)+2x-24 \\ -5x+75=3x-51+2x-24 && -5x+75=-3x+51+2x-24 \\ -10x=-150 && -4x=-48 \\ x=15\quad (x\ge 17) ~\text{no solution} && x=12 \quad (x\lt 17) ~\text{solution}\\ && \boxed{ x=12 } \\ \hline \end{array}\)

\(\mathbf{\boxed{x=12 \lor x\ge 17}}\)

The matrices

\(\large{ \mathbf{A} = \begin{pmatrix} 1 &x \\ y & 1 \end{pmatrix} \text{ and } \mathbf{B} = \begin{pmatrix} -1 &x \\ y & -1 \end{pmatrix} }\)

are inverses.

What is \(xy\)?

\(\begin{array}{|lrcll|} \hline & AB&=&I \\ & \begin{pmatrix} 1 &x \\ y & 1 \end{pmatrix} \begin{pmatrix} -1 &x \\ y & -1 \end{pmatrix} &=& \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\\\ & \begin{pmatrix} -1+xy &x-x \\ -y+y & xy-1 \end{pmatrix} &=& \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\\\ & \begin{pmatrix} xy-1 &0 \\ 0 & xy-1 \end{pmatrix} &=& \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline xy-1 &=& 1 \\ \mathbf{xy} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}\)

Let \(\theta\) be the angle between a and b, where a and b are the vectors defined in

\(a = \begin{pmatrix} 2 \\ -6 \\ 5 \end{pmatrix} \quad \text{and} \quad b = \begin{pmatrix} -1 \\ -2 \\ 0 \end{pmatrix}\).
 

Find \(\cos^2 \theta\)

\(\begin{array}{|rcll|} \hline \mathbf{\cos(\theta)} &\mathbf{=}& \mathbf{\dfrac{\vec{a}\cdot \vec{b}}{|\vec{a}|\cdot |\vec{b}|} } \quad | \quad |\vec{a}| = \sqrt{2^2+(-6)^2+5^2}=\sqrt{65},\ |\vec{b}| = \sqrt{(-1)^2+(-2)^2+0^2}=\sqrt{5} \\\\ \cos(\theta) &=& \dfrac{\begin{pmatrix} 2 \\ -6 \\ 5 \end{pmatrix}\cdot \begin{pmatrix} -1 \\ -2 \\ 0 \end{pmatrix} } {\sqrt{65}\cdot \sqrt{5} } \\\\ \cos(\theta) &=& \dfrac{2\cdot (-1)+(-6)\cdot(-2)+5\cdot 0} {\sqrt{65}\cdot \sqrt{5} } \\\\ \cos(\theta) &=& \dfrac{10} {\sqrt{65}\cdot \sqrt{5} } \\\\ \cos^2(\theta) &=& \dfrac{100} {65\cdot 5 } \\\\ \cos^2(\theta) &=& \dfrac{100} {325 } \\\\ \mathbf{\cos^2(\theta)} &\mathbf{=}& \mathbf{\dfrac{4}{13}} \\ \hline \end{array}\)

Find the limit:

\(\large{ \lim \limits_{x \to 0} ~\dfrac{\sin(5x)}{\sin(8x)} } \)

Apply L'Hopital's Rule and we have:

\(\begin{array}{|rcll|} \hline && \mathbf{\lim \limits_{x \to 0} ~\dfrac{\sin(5x)}{\sin(8x)}} \\\\ &=& \lim \limits_{x \to 0} ~\dfrac{5\cos(5x)}{8\cos(8x)} \quad | \quad \lim \limits_{x \to 0} ~ \cos(5x)= 1,\ \lim \limits_{x \to 0} ~ \cos(8x)= 1 \\\\ &=& \dfrac{ 5\cdot 1}{8\cdot 1} \\\\ &\mathbf{=}& \mathbf{\dfrac{ 5 }{8 }} \\ \hline \end{array}\)

Simplify

\(\mathbf{x=\dfrac{\sum \limits_{n = 1}^{2024} \sqrt{45 + \sqrt{n}} }{\sum \limits_{n = 1}^{2024} \sqrt{45 - \sqrt{n}} } = \dfrac{\sqrt{45 + \sqrt{1}} + \sqrt{45 + \sqrt{2}} + \sqrt{45 + \sqrt{3}} + \dots + \sqrt{45 + \sqrt{2024}}}{\sqrt{45 - \sqrt{1}} + \sqrt{45 - \sqrt{2}} + \sqrt{45 - \sqrt{3}} + \dots + \sqrt{45 - \sqrt{2024}}}} \).

1.

\(\begin{array}{|rcll|} \hline \Big( \sqrt{45 + \sqrt{n}} + \sqrt{45 - \sqrt{n}} \Big)^2 &=& 45 + \sqrt{n} +2\sqrt{45 + \sqrt{n}}\sqrt{45 - \sqrt{n}} +45 -\sqrt{n} \\ &=& 90 +2\sqrt{(45 + \sqrt{n} )(45 - \sqrt{n}) } \\ &=& 90 +2\sqrt{45^2-n } \\ &=& 2\cdot \Big( 45 + \sqrt{2025-n }\Big) \\\\ \sqrt{45 + \sqrt{n}} + \sqrt{45 - \sqrt{n}} &=& \sqrt{2}\sqrt{ 45 + \sqrt{2025-n} } \\ \mathbf{\sqrt{ 45 + \sqrt{2025-n} } } &\mathbf{=} & \mathbf{\dfrac{1}{\sqrt{2}}\Big( \sqrt{45 + \sqrt{n}} + \sqrt{45 - \sqrt{n}} \Big) }~ \huge{\text{!}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x=\dfrac{\sum \limits_{n = 1}^{2024} \sqrt{45 + \sqrt{n}} }{\sum \limits_{n = 1}^{2024} \sqrt{45 - \sqrt{n}} } } &=& \mathbf{\dfrac{\sum \limits_{n = 1}^{2024} \sqrt{45 + \sqrt{2025-n}} }{\sum \limits_{n = 1}^{2024} \sqrt{45 - \sqrt{n}} } } \\\\ &=& \dfrac {\dfrac{1}{\sqrt{2}}\sum \limits_{n = 1}^{2024} \Big( \sqrt{45 + \sqrt{n}} + \sqrt{45 - \sqrt{n}} \Big)} {\sum \limits_{n = 1}^{2024} \sqrt{45 - \sqrt{n}} } \\\\ &=& \dfrac{1}{\sqrt{2}} \cdot \dfrac {\sum \limits_{n = 1}^{2024} \Big( \sqrt{45 + \sqrt{n}} \Big)+\sum \limits_{n = 1}^{2024} \Big( \sqrt{45 - \sqrt{n}} \Big)} {\sum \limits_{n = 1}^{2024} \sqrt{45 - \sqrt{n}} } \\\\ &=& \dfrac{1}{\sqrt{2}} \cdot \left( 1+ \underbrace{ \dfrac {\sum \limits_{n = 1}^{2024} \Big( \sqrt{45 + \sqrt{n}} \Big)} {\sum \limits_{n = 1}^{2024} \sqrt{45 - \sqrt{n}} } }_{\huge{=x}} \right) \\\\ x &=& \dfrac{1}{\sqrt{2}} \cdot (1+x) \\\\ x\sqrt{2} &=& 1+x \\ x\sqrt{2} -x &=& 1 \\ x(\sqrt{2}-1) &=& 1 \\ x &=& \dfrac{1}{\sqrt{2}-1}\cdot \left(\dfrac{\sqrt{2}+1}{\sqrt{2}+1}\right) \\ x &=& \dfrac{\sqrt{2}+1}{2-1} \\ \mathbf{x} & \mathbf{=} & \mathbf{ \sqrt{2}+1 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{\sqrt{45 + \sqrt{1}} + \sqrt{45 + \sqrt{2}} + \sqrt{45 + \sqrt{3}} + \dots + \sqrt{45 + \sqrt{2024}}}{\sqrt{45 - \sqrt{1}} + \sqrt{45 - \sqrt{2}} + \sqrt{45 - \sqrt{3}} + \dots + \sqrt{45 - \sqrt{2024}}} &=& \mathbf{ \sqrt{2}+1 } \\ \hline \end{array}\)

Find all complex numbers  such that

\(\mathbf{|z|^2-2\bar z+iz=2i}\)

Formula:

\(z=a+bi \\ \bar z = a-bi \\ |z|^2 = a^2+b^2\)

\(\begin{array}{|rcll|} \hline |z|^2-2\bar z+iz &=& 2i \\ a^2+b^2-2(a-bi )+ i(a+bi)&=& 2i \\ a^2+b^2-2a+2bi + ia+bi^2 &=& 2i \quad | \quad i^2=-1 \\ a^2+b^2-2a+2bi + ia -b &=& 2i \\ a^2+b^2-2a-b +2bi + ia &=& 2i \\ \underbrace{(a^2+b^2-2a-b)}_{=0} + \underbrace{(2b+a)}_{=2}i &=& 2i \\\\ 2b+a &=& 2 \\ 2b &=& 2-a \\ \mathbf{b} & \mathbf{=}& \mathbf{\dfrac{2-a}{2}} \\ \hline \end{array}\)

\(\mathbf{z = a + \left(\dfrac{2-a}{2}\right)i} \)

Adam forgot his 4-digit pin code. He knows it consists of three 5's and one 7.

How big of a chance is there that he types the correct code in one try?

\(\begin{array}{|r|c|} \hline \text{code}& \text{chance of correct code} \\ \hline 5557 & 25\ \% \\ \hline 5575 & 25\ \% \\ \hline 5755 & 25\ \% \\ \hline 7555 & 25\ \% \\ \hline \end{array} \)

5.
3x - x (x-4) = - 3x^2

\(\begin{array}{|lrcll|} \hline & 3x - x (x-4) &=& - 3x^2 \\ & 3x - x^2+4x &=& - 3x^2 \\ & 2x^2+7x &=& 0 \\ & x(7+2x) &=& 0\\ 1.& \mathbf{x}&\mathbf{=}& \mathbf{0} \\\\ 2.& 7+2x &=& 0 \\ & 2x &=& -7 \\ & \mathbf{x}&\mathbf{=}& -\mathbf{\dfrac{7}{2}} \\ \hline \end{array} \)

4.
(x+2)^2 + 5 = 8

\(\begin{array}{|rcll|} \hline (x+2)^2 + 5 &=& 8 \\ (x+2)^2 &=& 8-5 \\ (x+2)^2 &=& 3 \\ x+2 &=& \pm\sqrt{3} \\ x &=& -2 \pm\sqrt{3} \\\\ \mathbf{x_1}&\mathbf{=}& \mathbf{-2 +\sqrt{3}} \\ \mathbf{x_2}&\mathbf{=}& \mathbf{-2 -\sqrt{3}} \\ \hline \end{array}\)

3.
9x^2 + 6x + 1 = 0

\(\begin{array}{|rcll|} \hline 9x^2 + 6x + 1 &=& 0 \\ (3x+1)^2 &=& 0 \\ 3x+1 &=& 0 \\ 3x &=& -1 \\ \mathbf{x}&\mathbf{=}& \mathbf{-\dfrac{1}{3}} \\ \hline \end{array}\)