heureka

3.

Find AC

cos-rule

\(\begin{array}{|rcll|} \hline (7\sqrt{3})^2 &=& 9^2 + AC^2 -2*9*AC\cos(150^\circ) \quad | \quad \cos(150^\circ)=-\dfrac{\sqrt{3} }{2} \\\\ 49*3 &=& 81 + AC^2 + 2*9*AC\dfrac{\sqrt{3} }{2} \\\\ 147 &=& 81 + AC^2 + 9*AC \sqrt{3} \\ 66 &=& AC^2 + 9\sqrt{3}AC \\ \mathbf{AC^2 + 9\sqrt{3}AC -66} &=& \mathbf{0} \\\\ AC &=& \dfrac{ -9\sqrt{3} \pm\sqrt{81*3-4*(-66) } } {2} \\\\ AC &=& \dfrac{ -9\sqrt{3} \pm\sqrt{81*3+264 } } {2} \\\\ AC &=& \dfrac{ -9\sqrt{3} \pm\sqrt{507} } {2} \\\\ AC &=& \dfrac{ -9\sqrt{3} \pm\sqrt{13^2*3} } {2} \\\\ AC &=& \dfrac{ -9\sqrt{3} \pm 13\sqrt{3} } {2} \\\\ AC &=& \dfrac{ -9\sqrt{3} \mathbf{+} 13\sqrt{3} } {2} \\\\ AC &=& \dfrac{ 4\sqrt{3} } {2} \\\\ \mathbf{AC} &=& \mathbf{2\sqrt{3} } \\ \hline \end{array}\)

2.

Find AC

\(\angle C = 180^\circ - (75^\circ+45^\circ) \\ \angle C = 60^\circ \)

\(\begin{array}{|rcll|} \hline \dfrac{AC}{\sin(45^\circ)} &=& \dfrac{30}{\sin(60^\circ)} \\\\ AC &=& \dfrac{30\sin(45^\circ)}{\sin(60^\circ)} \quad | \quad \sin(45^\circ)=\dfrac{\sqrt{2} }{2},\ \sin{60^\circ}=\dfrac{\sqrt{3} }{2} \\\\ AC &=& 30*\dfrac{\sqrt{2}}{2} *\dfrac{2}{\sqrt{3}} \\\\ AC &=& \dfrac{30\sqrt{2}}{\sqrt{3}} \\\\ AC &=& \dfrac{30\sqrt{2}}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\\\ AC &=& \dfrac{30\sqrt{2}\sqrt{3}}{3} \\\\ \mathbf{AC} &=& \mathbf{10\sqrt{6}} \\ \hline \end{array} \)

1.

Find BC

cos-rule

\(\begin{array}{|rcll|} \hline BC^2 &=& (7\sqrt{2})^2+6^2-2*7\sqrt{2}*6\cos(45^\circ) \quad | \quad \cos(45^\circ)=\dfrac{\sqrt{2} }{2} \\\\ BC^2 &=& 49*2+6^2-2*7\sqrt{2}*6\dfrac{\sqrt{2} }{2} \\\\ BC^2 &=& 98+36-2*7*6 \\\\ BC^2 &=& 98+36-84 \\ BC^2 &=& 50 \\ BC^2 &=& 2*25 \\ \mathbf{BC} &=& \mathbf{5\sqrt{2}} \\ \hline \end{array} \)

If

\(\dfrac{2x + 7}{2x^2 - x - 1} = \dfrac{A}{2x + 1} + \dfrac{B}{x - 1}\),

then find A and B.

\(2x^2-x-1 = (2x+1)(x-1)\)

\(\begin{array}{|lrcll|} \hline & \dfrac{2x + 7}{2x^2 - x - 1} &=& \dfrac{A}{2x + 1} + \dfrac{B}{x - 1} \\\\ & \dfrac{2x + 7}{(2x+1)(x-1)} &=& \dfrac{A}{2x + 1} + \dfrac{B}{x - 1} \quad | \quad \times (2x+1)(x-1) \\\\ & \mathbf{2x + 7} &=& \mathbf{A(x - 1) + B(2x + 1)} \quad | \quad x\ne -\dfrac{1}{2}\ \text{and}\ x\ne 1 \\ \hline x = 1: & 2*1+7 &=& A(1-1) + B ( 2*1+1) \\ & 9 &=& A*0 + 3B \\ & 9 &=& 0+3B \\ & 3B &=& 9 \quad | \quad : 3 \\ & \mathbf{B} &=& \mathbf{3} \\ \hline x = -\dfrac{1}{2}: & 2\left(-\dfrac{1}{2}\right)+7 &=& A\left(-\dfrac{1}{2}-1\right) + B \left( 2\left(-\dfrac{1}{2}\right)+1\right) \\ & -1+7 &=& A\left(-\dfrac{3}{2}\right) + B*0 \\ & 6 &=& -\dfrac{3}{2}A \\ & A &=& -\dfrac{2}{3}*6 \\ & \mathbf{A} &=& \mathbf{-4} \\ \hline \end{array}\)

Find the solutions to \(x^2 - y^2 = 100\) in positive integers.

Divisors of 100:
1 | 2 | 4 | 5 | 10 | 20 | 25 | 50 | 100 (9 divisors)

Formula:

\(\begin{array}{|rcll|} \hline \mathbf{4xy} &=& \mathbf{ (x+y)^2-(x-y)^2 } \\\\ xy &=& \left( \dfrac{x+y}{2} \right)^2- \left( \dfrac{x-y}{2} \right)^2 \\ \hline \end{array}\)

\(\begin{array}{|l | c | c | c | } \hline x*y & \dfrac{x+y}{2} & \dfrac{x+y}{2} & \text{solutions} \\ \hline 100 = 1\times 100 & \dfrac{101}{2}\ \text{odd} & \dfrac{100}{2} \\ \hline 100 = 2\times 50 & \dfrac{52}{2}=26 & \dfrac{48}{2}=24 & 26^2-24^2 = 10^2 \\ \hline 100 = 4\times 25 & \dfrac{29}{2}\ \text{odd} & \dfrac{21}{2} \ \text{odd} \\ \hline 100 = 5\times 20 & \dfrac{25}{2}\ \text{odd} & \dfrac{15}{2}\ \text{odd} \\ \hline 100 = 10\times 10 & \dfrac{20}{2} & \dfrac{0}{2}=0\ \text{not positive integer} \\ \hline \end{array}\)

Find the sum of the infinite series
\(1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots\).

\(\text{Let $r=\dfrac{1}{1998} $}\)

\(\begin{array}{|rcll|} \hline s&=& \mathbf{1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots} \\ s&=& \mathbf{1+2r +3r^2+4r^3+\cdots + nr^{n-1} + \ldots} \\ \hline \end{array}\)

\(\begin{array}{|rcrl|} \hline s_n &=& 1+&2r +3r^2+4r^3+\cdots + nr^{n-1} \quad |\quad \cdot r \\ rs_n &=& & 1r +2r^2+3r^3+\cdots + (n-1)r^{n-1}+nr^n \\ \hline s_n-rs_n &=& 1+&r+r^2+r^3+\ldots + r^{n-1} -nr^n \\ \hline \end{array}\\ \begin{array}{|rclrcrl|} \hline s_n(1-r) &=& \underbrace{1+r+r^2+r^3+\ldots + r^{n-1}}_{=S_n(\text{ geometric progression})} -nr^n \\ s_n(1-r) &=& S_n -nr^n & S_n &=& 1+&r+r^2+r^3+\ldots + r^{n-1} \quad | \quad \cdot r\\ & & & rS_n &=& &r+r^2+r^3+\ldots + r^{n-1}+r^n \\ \hline & & & S_n-rS_n &=& 1- &r^n \\ & & & S_n(1-r) &=& 1- &r^n \\ & & & S_n &=& \dfrac{1- r^n}{1-r} \\ s_n(1-r) &=& \dfrac{1- r^n}{1-r} -nr^n \\ s_n &=& \dfrac{1- r^n}{(1-r)^2} - \dfrac{nr^n}{1-r} \\ s &=& \lim \limits_{n\to \infty} s_n \\ s &=& \lim \limits_{n\to \infty} \dfrac{1- r^n}{(1-r)^2} - \dfrac{nr^n}{1-r} \\ && \boxed{\lim \limits_{n\to \infty} r^n = \lim \limits_{n\to \infty} \left(\dfrac{1}{1998}\right)^n = 0} \\ s &=& \dfrac{1- 0}{(1-r)^2} - 0 \\ s &=& \dfrac{1}{(1-r)^2} \\ s &=& \dfrac{1}{\left(1-\dfrac{1}{1998}\right)^2} \\ s &=& \left(\dfrac{1998}{1997}\right)^2 \\ \\ s &=& \dfrac{1998^2}{1997^2} \\\\ s &=&\mathbf{ \dfrac{3992004}{3988009}} \\\\ \mathbf{s} &=& \mathbf{1.00100175300507095\ldots...} \\ \hline \end{array}\)

Find the sum of the infinite series

\(1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots\)

\(\small{ \begin{array}{rclllllllllc} & & \mathbf{1}&\mathbf{+} & \mathbf{2\left(\dfrac{1}{1998}\right)} &\mathbf{+} & \mathbf{3\left(\dfrac{1}{1998}\right)^2} &\mathbf{+} &\mathbf{ 4\left(\dfrac{1}{1998}\right)^3} & \mathbf{+}&\mathbf{\cdots} \\\\ & & & & & & & & & & &\mathbf{\text{sum }=\dfrac{a}{1-r}} \\ &=& 1&+ & 1\left(\dfrac{1}{1998}\right)^1 &+ & 1\left(\dfrac{1}{1998}\right)^2 &+ & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{1}{1-\dfrac{1}{1998}} \\\\ & & &+ & 1\left(\dfrac{1}{1998}\right)^1 &+ & 1\left(\dfrac{1}{1998}\right)^2 &+ & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{\left(\dfrac{1}{1998}\right)^1}{1-\dfrac{1}{1998}} \\\\ & & & & &+ & 1\left(\dfrac{1}{1998}\right)^2 &+ & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{\left(\dfrac{1}{1998}\right)^2}{1-\dfrac{1}{1998}} \\\\ & & & & & & & + & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{\left(\dfrac{1}{1998}\right)^3}{1-\dfrac{1}{1998}} \\\\ & & & & & & & & & +&\ldots \\ \end{array} }\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ 1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots } \\\\ &=& \dfrac{1}{1-\dfrac{1}{1998}} + \dfrac{\left(\dfrac{1}{1998}\right)^1}{1-\dfrac{1}{1998}} + \dfrac{\left(\dfrac{1}{1998}\right)^2}{1-\dfrac{1}{1998}} + \dfrac{\left(\dfrac{1}{1998}\right)^3}{1-\dfrac{1}{1998}} +\cdots \\\\ &=& \dfrac{1}{1-\dfrac{1}{1998}} \left( 1+ \left(\dfrac{1}{1998}\right)^1+\left(\dfrac{1}{1998}\right)^2+\left(\dfrac{1}{1998}\right)^3 +\cdots \right) \\\\ &=& \dfrac{1}{\dfrac{1998-1}{1998}} \left( 1+ \left(\dfrac{1}{1998}\right)^1+\left(\dfrac{1}{1998}\right)^2+\left(\dfrac{1}{1998}\right)^3 +\cdots \right) \\\\ &=& \dfrac{1998}{1997} \left( 1+ \left(\dfrac{1}{1998}\right)^1+\left(\dfrac{1}{1998}\right)^2+\left(\dfrac{1}{1998}\right)^3 +\cdots \right) \\\\ &=& \dfrac{1998}{1997} \left( \dfrac{1}{1-\dfrac{1}{1998}} \right) \\\\ &=& \dfrac{1998}{1997} \left( \dfrac{1}{\dfrac{1998-1}{1998}} \right) \\\\ &=& \dfrac{1998}{1997}\times \dfrac{1998}{1997} \\\\ &=& \dfrac{1998^2}{1997^2} \\\\ &=&\mathbf{ \dfrac{3992004}{3988009}} \\ \hline \end{array}\)

If
\(a^2 + \dfrac{1}{a^2} \)= 1,

then find
 \(a^3 + \dfrac{1}{a^3}\).

\(\begin{array}{|rcll|} \hline \left( a^2 + \dfrac{1}{a^2} \right) \left( a+\dfrac{1}{a} \right) &=& a^3+\dfrac{a^2}{a}+\dfrac{a}{a^2} + \dfrac{1}{a^3} \\\\ \left( a^2 + \dfrac{1}{a^2} \right) \left( a+\dfrac{1}{a} \right) &=& a^3+ \dfrac{1}{a^3}+ a +\dfrac{1}{a} \\\\ \left( a^2 + \dfrac{1}{a^2} \right) \left( a+\dfrac{1}{a} \right)-\left( a+\dfrac{1}{a} \right) &=& a^3+ \dfrac{1}{a^3} \\\\ \left( a+\dfrac{1}{a} \right) \left( a^2 + \dfrac{1}{a^2} -1 \right) &=& a^3+ \dfrac{1}{a^3} \quad | \quad a^2 + \dfrac{1}{a^2} = 1 \\\\ \left( a+\dfrac{1}{a} \right) \left(1 -1 \right) &=& a^3+ \dfrac{1}{a^3} \\\\ \left( a+\dfrac{1}{a} \right) \times 0 &=& a^3+ \dfrac{1}{a^3} \\\\ 0 &=& a^3+ \dfrac{1}{a^3} \\\\ \mathbf{ a^3+ \dfrac{1}{a^3} } &=& \mathbf{0} \\ \hline \end{array}\)

If the 4014th term of a geometric sequence of non-negative numbers is 135,

and the 14th term is 375,

what is the 2014th term?

\(\begin{array}{|rcll|} \hline i < j < k \\ a_i &=& ar^{i-1} \\ a_j &=& ar^{j-1} \\ a_k &=& ar^{k-1} \\ \hline \dfrac{a_k}{a_i} &=& r^{(k-1)-(i-1)} \\ \mathbf{\dfrac{a_k}{a_i}} &=& \mathbf{r^{k-i}} \\\\ \dfrac{a_k}{a_j} &=& r^{(k-1)-(j-1)} \\ \mathbf{\dfrac{a_k}{a_j}} &=& \mathbf{r^{k-j}} \\ \hline r= \left(\dfrac{a_k}{a_i}\right)^{\frac{1}{k-i}} &=& \left(\dfrac{a_k}{a_j}\right)^{\frac{1}{k-j}} \\\\ a_k^{\frac{1}{k-i}} a_j^{\frac{1}{k-j}} &=& a_k^{\frac{1}{k-j}} a_i^{\frac{1}{k-i}} \\\\ a_j^{\frac{1}{k-j}} &=& a_k^{\frac{1}{k-j}-\frac{1}{k-i}} a_i^{\frac{1}{k-i}} \\\\ a_j^{\frac{1}{k-j}} &=& a_k^{\frac{k-i-k+j}{(k-j)(k-i)}} a_i^{\frac{1}{k-i}} \\\\ a_j^{\frac{1}{k-j}} &=& a_k^{\frac{j-i}{(k-j)(k-i)}} a_i^{\frac{1}{k-i}} \\\\ \mathbf{a_j} &=& \mathbf{a_k^{\frac{j-i}{k-i}} a_i^{\frac{k-j}{k-i}} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_j} &=& \mathbf{a_k^{\frac{j-i}{k-i}} a_i^{\frac{k-j}{k-i}} } \\\\ && \boxed{a_{14} =375=a_i\quad i = 14\\ a_{2014}=a_j \quad j=2014\\ a_{4014}=135=a_k\quad k=4014} \\\\ a_{2014} &=& 135^{\frac{2014-14}{4014-14}} 375^{\frac{4014-2014}{4014-14}} \\\\ a_{2014} &=& 135^{\frac{2000}{4000}} 375^{\frac{2000}{4000}} \\\\ a_{2014} &=& 135^{\frac{1}{2}} 375^{\frac{1}{2}} \\\\ a_{2014} &=& \sqrt{135*375} \\ a_{2014} &=& \sqrt{50625} \\ \mathbf{a_{2014}} &=& \mathbf{225} \\ \hline \end{array}\)

The 2014th term is 225

The quadratic polynomial p(x) leaves a remainder of 1 on division by x - 1 or x - 2.  
The product of the roots of p(x) is 1.  
What is the sum of the roots of p(x)?

The quadratic polynomial is \( p(x)=ax^2+bx+c\)

\(\begin{array}{|lrcll|} \hline & p(x) &=& q_1(x)(x-1)+1 \quad | \quad p(x)=ax^2+bx+c \\ x=1: & ax^2+bx+c &=& q_1(x)(x-1)+1 \\ & a+b+c &=& q_1(x)\cdot 0 +1 \\ &\mathbf{ a+b+c} &=& \mathbf{1} \qquad (1) \\ \hline & p(x) &=& q_2(x)(x-2)+1 \quad | \quad p(x)=ax^2+bx+c \\ x=2: & p(x)=ax^2+bx+c &=& q_2(x)(x-2)+1 \\ & 4a+2b+c &=& q_2(x)\cdot 0+1 \\ & \mathbf{4a+2b+c} &=& 1 \qquad (2) \\ \hline \end{array}\)

We have roots \(r_1\) and \(r_2\)

\(\begin{array}{|rcll|} \hline p(x)=ax^2+bx+c &=& 0 \\ ax^2+bx+c &=& 0 \quad | \quad : a \\ x^2+\underbrace{\frac{b}{a}}_{=-(r_1+r_2)}x+\underbrace{\frac{c}{a}}_{=r_1r_2} &=& 0 \\ \hline r_1r_2= 1 &=& \frac{c}{a} \\ 1 &=& \frac{c}{a} \\ \mathbf{c} &=& \mathbf{a} \\ \hline -(r_1+r_2) &=& \frac{b}{a} \\ \mathbf{r_1+r_2} &=& \mathbf{-\frac{b}{a}} \\ \hline \end{array}\)

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

\(\begin{array}{|rcll|} \hline \mathbf{r_1+r_2} &=& \mathbf{-\frac{b}{a}}\quad | \quad b=3,\ a=-1 \\ r_1+r_2 &=& -\frac{3}{-1} \\ \mathbf{r_1+r_2} &=& \mathbf{3} \\ \hline \end{array}\)

The sum of the roots of \(p(x)= -x^2+3x-1\) is 3