heureka

If \(a + b = 2\) and \(a^3 + b^3 = 14\), then find all possible values of \(a^2 + b^2\).

\(\begin{array}{|rcll|} \hline (a+b)^3 &=& a^3+3a^2b+3ab^2+b^3 \\ (a+b)^3 &=& a^3+b^3 + 3ab(a+b) \quad & | \quad a+b = 2,\ a^3+b^3=14 \\ 2^3 &=& 14+ 3ab\times 2 \\ 8 &=& 14+ 6ab \quad & | \quad :2 \\ 4 &=& 7+ 3ab \\ 3ab &=& -3 \\ \mathbf{ab} &=& \mathbf{-1} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (a+b)^2 &=& a^2+2ab+b^2 \\ (a+b)^2 &=& a^2+b^2+2ab \quad & | \quad \mathbf{ab=-1} \\ (a+b)^2 &=& a^2+b^2+2(-1) \quad & | \quad a+b = 2 \\ 2^2 &=& a^2+b^2 -2 \\ 4 &=& a^2+b^2 -2 \\ a^2+b^2 &=& 4+2 \\ \mathbf{ a^2+b^2} &=&\mathbf{ 6} \\ \hline \end{array}\)

If one root of (a^2 + 9)x^2 + 13x + 6a = 0 is equal to the reciprocal of the other root, then find a.

\(\begin{array}{|rcll|} \hline (a^2 + 9)x^2 + 13x + 6a &=& 0 \quad & | \quad : (a^2 + 9) \\\\ x^2 + \dfrac{13}{a^2 + 9}x + \underbrace{\dfrac{6a}{a^2 + 9}}_{=r_1r_2} &=& 0 \quad & | \quad \text{vieta} \\\\ r_1r_2 &=& \dfrac{6a}{a^2 + 9} \quad |\quad r_2 = \dfrac{1}{r_1} \\ r_1\dfrac{1}{r_1} &=& \dfrac{6a}{a^2 + 9} \\ 1 &=& \dfrac{6a}{a^2 + 9} \\ a^2 + 9 &=& 6a \\ a^2-6a + 9 &=& 0 \\\\ a &=& \dfrac{6 \pm \sqrt{6^2-4(9)} } {2} \\ a &=& \dfrac{6 \pm \sqrt{0}} {2} \\ a &=& \dfrac{6}{2} \\ \mathbf{a} &=& \mathbf{3} \\ \hline \end{array} \)

How many terms are in the expansion of \(\left(a + b + c + d\right)^5\)?

The multinomial in general is: \(\left( x_1+x_2+\cdots +x_m \right)^n\)

The number of terms in a multinomial sum is \(\#_{n,m}\),

\(\#_{n,m} = \dbinom{n+m-1}{m-1}\)

Mary commutes to work.

On light traffic days it takes her only \(15\) minutes.

On moderate days it takes \(25\) minutes and

on heavy traffic days, it takes her \(60\) minutes.

Only about \(10 \%\) of the days are heavy traffic and

\(20 \%\) of the days the traffic is light.

On average, how long does it take her to get to work?

\(\begin{array}{|rcll|} \hline \mathbf{\text{average}} &=& \dfrac{15\text{min.} \times 20\% + 25\text{min.} \times (100\%-20\%-10\%) + 60\text{min.} \times 10\%} {20~\%+(100\%-20\%-10\%)+10~\%} \\\\ &=& \dfrac{15\text{min.} \times 20\% + 25\text{min.} \times 70\% + 60\text{min.} \times 10\%} {100\%} \\\\ &=& 15\text{min.} \times 20\% + 25\text{min.} \times 70\% + 60\text{min.} \times 10\% \\ &=& 3\text{min.} + 17.5\text{min.} + 6\text{min.} \\ &=& \mathbf{26.5\text{min.}} \\ \hline \end{array} \)

Two of the cars are chosen at random.

What is the probability that the two cars differ in all three categories?

I assume,

\(\begin{array}{|rcll|} \hline && { 24\times 1 \times 2 \times 3}\above 1pt \dfrac{24!}{(24-2)!} \\\\ &=& \dfrac{144}{552} \\\\ &=& 0.26086956522\ \approx 26\% \\ \hline \end{array}\)

Express \(\cos(5x)\) as a polynomial in \(\cos(x)\).

\(\begin{array}{|rcll|} \hline \mathbf{\Big(\cos(x) + i\sin(x) \Big)^5} &=& \mathbf{\cos(5x) +i\sin(5x)} \\\\ \Big(\cos(x) + i\sin(x) \Big)^5 &=& \binom50\cos^5(x) \\ &+& \binom51\cos^4(x)(i)\sin(x) \\ &+& \binom52\cos^3(x)(i^2)\sin^2(x) \quad & | \quad i^2=-1 \\ &+& \binom53\cos^2(x)(i^3)\sin^3(x) \quad & | \quad i^3=-i \\ &+& \binom54\cos(x)(i^4)\sin^4(x) \quad & | \quad i^4=1 \\ &+& \binom55(i^5)\sin^5(x) \quad & | \quad i^5=i \\\\ &=& \binom50\cos^5(x) \\ &+& \binom51\cos^4(x)(i)\sin(x) \\ &-& \binom52\cos^3(x)\sin^2(x) \\ &-& \binom53\cos^2(x)(i)\sin^3(x) \\ &+& \binom54\cos(x)\sin^4(x) \\ &+& \binom55(i)\sin^5(x) \\ \hline \end{array} \)

Compare the real parts of each side:

\(\begin{array}{|rcll|} \hline \cos(5x) &=& \binom50\cos^5(x) \\ &-& \binom52\cos^3(x)\sin^2(x) \\ &+& \binom54\cos(x)\sin^4(x) \\\\ \cos(5x) &=& \binom50\cos^5(x) \quad & | \quad \binom50 = 1 \\ &-& \binom52\cos^3(x)\sin^2(x) \quad & | \quad \binom52 = 10,\ \sin^2(x)=1-\cos^2(x) \\ &+& \binom54\cos(x)\sin^2(x)\sin^2(x) \quad & | \quad \binom54 = 5,\ \sin^2(x)=1-\cos^2(x) \\\\ \cos(5x) &=& \cos^5(x) \\ &-& 10\cos^3(x)\Big(1-\cos^2(x)\Big) \\ &+& 5\cos(x)\Big(1-\cos^2(x)\Big)\Big(1-\cos^2(x)\Big) \\\\ \cos(5x) &=& \cos^5(x) \\ &-& 10\cos^3(x) + 10\cos^5(x) \\ &+& 5\cos(x)\Big(1-\cos^2(x)\Big)^2 \\\\ \cos(5x) &=& \cos^5(x) \\ &-& 10\cos^3(x) + 10\cos^5(x) \\ &+& 5\cos(x)\Big(1-2\cos^2(x)+\cos^4(x)\Big) \\\\ \cos(5x) &=& \cos^5(x) \\ &-& 10\cos^3(x) + 10\cos^5(x) \\ &+& 5\cos(x) -10\cos^3(x)+5\cos^5(x) \\\\ \mathbf{\cos(5x)} &=& \mathbf{ 16\cos^5(x) - 20\cos^3(x) + 5\cos(x) } \\ \hline \end{array}\)

Solve the equation \(9^{5-x}= \dfrac{1}{\sqrt{15^{4x}}}\).

Express your answer in terms of \(\ln3\) and \(\ln5\) .

\(\begin{array}{|rcll|} \hline 9^{5-x} &=& \dfrac{1}{\sqrt{15^{4x}}} \\ 9^{5-x} &=& \dfrac{1}{15^{\frac{4x}{2}}} \\ 9^{5-x} &=& \dfrac{1}{15^{2x}} \\ 9^59^{-x} &=& \dfrac{1}{15^{2x}} \\ 9^53^{-2x} &=& \dfrac{1}{15^{2x}} \\ 9^5 &=& \dfrac{3^{2x}}{15^{2x}} \\ 9^5 &=& \left(\dfrac{3}{15}\right)^{2x} \\ 9^5 &=& \left(\dfrac{1}{5}\right)^{2x} \\ 9^5 &=& \dfrac{1}{5^{2x}} \\ 9^55^{2x} &=& 1 \\ 3^{2\cdot 5}5^{2x} &=& 1 \\ 3^{10}5^{2x} &=& 1 \\ 5^{2x} &=& 3^{-10} \quad | \quad \ln \text{ both sides } \\ 2x\ln(5) &=& -10\ln(3) \quad | \quad : 2 \\ x\ln(5) &=& -5\ln(3) \quad | \quad : 2 \\\\ \mathbf{x} &=& \mathbf{ -\dfrac{5\ln(3)}{\ln(5)} } \\ \hline \end{array}\)

The first two positive integers \(n\) for which \(1 + 2 + \ldots + n\) is a perfect square are \(1\) and \(8\). 

What are the next two?

Definition:

\(\text{Triangular numbers: $a(n) = \dbinom{n+1}{2} =\dfrac{ n(n+1)}{2} = 1 + 2 +3+4+ \ldots + n$. }\\\\ 1, 3, 6, 10, 15, 21, 28, 36,\ldots \)

\(\text{$a(m)$-th triangular number is a square: $\\\mathbf{a(m+1) = 6*a(m)-a(m-1)+2}$, with $a(1) = 1$, $a(2) = 8$. } \\\\ 1, 8, 49, 288, 1681, 9800, 57121, 332928, \ldots \)

Example:

\(\begin{array}{|c|r|r|r|} \hline & a(m+1) & & s_n \\ m+1 & = 6*a(m)-a(m-1)+2 & n & = \dfrac{ n(n+1)}{2} \\ \hline 3 & 6*a(2)-a(1)+2 \\ & = 6*8-1+2 \\ & = 49 & 49 & 35^2 \\\\ \hline 4 & 6*a(3)-a(2)+2 \\ & 6*49-8 +2 \\ & 288 & 288 & 204^2 \\\\ \hline 5 & 6*a(4)-a(3)+2 \\ & 6*288-49 +2 \\ & 1681 & 1681 & 1189^2 \\\\ \hline \ldots \\ \hline \end{array} \)

The next two positive integers n are 49 and 288.

A number \(x\) satisfies \(x = \dfrac{1}{1 + x}\).  Determine \(x - \dfrac{1}{x}\).

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{1}{1 + x}} \quad &| \quad \times (1 + x)\\\\ x(1 + x) &=& 1 \quad &| \quad : x \\\\ 1 + x &=& \dfrac{1}{x} \quad &| \quad - \dfrac{1}{x} \\\\ 1 + x - \dfrac{1}{x} &=& 0 \quad &| \quad -1 \\\\ \mathbf{ x - \dfrac{1}{x} } &=& \mathbf{ -1 } \\ \hline \end{array}\)

What is the remainder when \(x^3 + x^6 + x^9 + x^{27}\) is divided by \(x^2 - 1\)?

\(= x^{25} + x^{23} + x^{21} + x^{19} + x^{17} + x^{15} + x^{13} + x^{11} + x^9 + 2x^7 + 2x^5 + x^4 + 2x^3 + x^2 + 3x + 1 \text{ remainder } 3x + 1 \)