heureka

2.
(4x+12)(x-6) = 0

\(\begin{array}{|lrcll|} \hline &(4x+12)(x-6) &=& 0 \\ 1.& 4x+12 &=& 0 \\ & 4x &=& -12 \\ & \mathbf{x}&\mathbf{=}& \mathbf{-3} \\\\ 2. & x-6 &=& 0 \\ & \mathbf{x}&\mathbf{=}& \mathbf{6} \\ \hline \end{array}\)

1.
-6x^2 + 24x = 0

\(\begin{array}{|lrcll|} \hline & -6x^2 + 24x &=& 0 \\ & x(24-6x) &=& 0 \\ 1.& \mathbf{x}&\mathbf{=}& \mathbf{0} \\\\ 2.& 24-6x &=& 0 \\ & 6x &=& 24 \\ & \mathbf{x}&\mathbf{=}& \mathbf{4} \\ \hline \end{array}\)

Solve the inequality

\(\large{\dfrac{1}{x - 1} - \dfrac{4}{x - 2} + \dfrac{4}{x - 3} - \dfrac{1}{x - 4} < \dfrac{1}{30}} \)

\(\small{ \begin{array}{|rcll|} \hline \dfrac{1}{x - 1} - \dfrac{4}{x - 2} + \dfrac{4}{x - 3} - \dfrac{1}{x - 4} &<& \dfrac{1}{30} \quad | \quad x\ne 1,\ x\ne 2,\ x\ne 3,\ x\ne 4 \\\\ \dfrac{1}{x - 1} - \dfrac{4}{x - 2} + \dfrac{4}{x - 3} - \dfrac{1}{x - 4} -\dfrac{1}{30} &<& 0 \\\\ -\dfrac{x^4-10x^3+5x^2+100x+84}{30(x - 1)(x - 2)(x - 3)(x - 4)} &<& 0 \\\\ \boxed{\small{x^4-10x^3+5x^2+100x+84 = (x - 7)(x - 6)(x + 1)(x + 2)}} \\\\ -\dfrac{(x - 7)(x - 6)(x + 1)(x + 2)}{30(x - 1)(x - 2)(x - 3)(x - 4)} &<& 0 \quad | \quad \cdot 30 \\\\ -\dfrac{(x - 7)(x - 6)(x + 1)(x + 2)}{ (x - 1)(x - 2)(x - 3)(x - 4)} &<& 0 \quad | \quad \cdot(-1)! \\\\ \dfrac{(x - 7)(x - 6)(x + 1)(x + 2)}{ (x - 1)(x - 2)(x - 3)(x - 4)} &>& 0 \quad | \quad \cdot \Big((x - 1)(x - 2)(x - 3)(x - 4)\Big)^2 \\\\ \dfrac{(x - 7)(x - 6)(x + 1)(x + 2) \Big((x - 1)(x - 2)(x - 3)(x - 4)\Big)^2 }{ (x - 1)(x - 2)(x - 3)(x - 4)} &>& 0 \\\\ \dfrac{(x - 7)(x - 6)(x + 1)(x + 2)(x - 1)^2(x - 2)^2(x - 3)^2(x - 4)^2 }{ (x - 1)(x - 2)(x - 3)(x - 4)} &>& 0 \\\\ \mathbf{(x - 7)(x - 6)(x + 1)(x + 2)(x - 1)(x - 2)(x - 3)(x - 4)} & \mathbf{>}& \mathbf{0} \\ \hline \end{array} } \)

\(\begin{array}{|r|c|c|c|c|c|c|c|c|} \hline &(-\infty,-2) & (-2,-1) & (-1,1) & (1,2) & (2,3) & (3,4) & (4,6) & (6,7) & (7,\infty) \\ \hline x+2 & -& +& +& +& +& +& +& +& +\\ \hline x+1 & -& -& +& +& +& +& +& +& +\\ \hline x-1 & -& -& -& +& +& +& +& +& +\\ \hline x-2 & -& -& -& -& +& +& +& +& +\\ \hline x-3 & -& -& -& -& -& +& +& +& +\\ \hline x-4 & -& -& -& -& -& -& +& +& +\\ \hline x-6 & -& -& -& -& -& -& -& +& +\\ \hline x-7 & -& -& -& -& -& -& -& -& +\\ \hline \text{sign} & \\ \text{of} & \\ \text{all} & \mathbf{\large{+}} & -& \mathbf{\large{+}}& -& \mathbf{\large{+}}& -& \mathbf{\large{+}}& -& \mathbf{\large{+}} \\ & \mathbf{>0} & & \mathbf{>0}& & \mathbf{>0}& & \mathbf{>0}& & \mathbf{>0} \\ \hline \end{array}\)

\(\mathbf{(-\infty,-2) \cup (-1,1) \cup (2,3) \cup (4,6) \cup (7,\infty)}\)

Compute 

\(\mathbf{\huge{\sum \limits_{n = 1}^{9800} \dfrac{1}{\sqrt{n+\sqrt{n^2-1}} } } } \\\)

\(\begin{array}{|rcll|} \hline && \mathbf{\large{\sum \limits_{n = 1}^{9800} \dfrac{1}{\sqrt{n+\sqrt{n^2-1}} } }} \quad | \quad \sqrt{n+\sqrt{n^2-1}}\sqrt{n-\sqrt{n^2-1}}=1 \\\\ &=& \sum \limits_{n = 1}^{9800} \sqrt{n-\sqrt{n^2-1}} \\\\ &=& \sum \limits_{n = 1}^{9800} \sqrt{\dfrac{1}{2}\left( 2n-2\sqrt{n^2-1} \right) } \\\\ &=& \dfrac{1}{\sqrt{2}}\cdot \sum \limits_{n = 1}^{9800} \sqrt{ 2n-2\sqrt{n^2-1} } \quad | \quad \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2} \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \sum \limits_{n = 1}^{9800} \sqrt{ 2n-2\sqrt{n^2-1} } \quad | \quad n^2-1 = (n+1)(n-1) \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \sum \limits_{n = 1}^{9800} \sqrt{ 2n-2\sqrt{n+1}\sqrt{n-1} } \quad | \quad 2n-2\sqrt{n+1}\sqrt{n-1} = \left( \sqrt{n+1}-\sqrt{n-1} \right)^2 \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \sum \limits_{n = 1}^{9800} \sqrt{ \Big( \sqrt{n+1}-\sqrt{n-1} \Big)^2 } \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1}-\sqrt{n-1} \Big) \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big)-\sum \limits_{n = 1}^{9800} \Big( \sqrt{n-1} \Big) ~\right] \quad | \quad \sum \limits_{n = 1}^{9800} \Big( \sqrt{n-1} \Big) = 0+1+ \sum \limits_{n = 3}^{9800} \Big( \sqrt{n-1} \Big)\\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big) -\left(0+1+\sum \limits_{n = 3}^{9800} \Big( \sqrt{n-1} \Big) \right) ~\right] \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big) - \sum \limits_{n = 3}^{9800} \Big( \sqrt{n-1} \Big)-1 ~\right] \quad | \quad \sum \limits_{n = 3}^{9800} \Big( \sqrt{n-1} \Big) = \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big) \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big) - \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)-1 ~\right] \quad | \quad \sum \limits_{n = 1}^{9800} \Big( \sqrt{n+1} \Big) = \sum \limits_{n = 2}^{9801} \Big( \sqrt{n} \Big) \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 2}^{9801} \Big( \sqrt{n} \Big) - \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)-1 ~\right] \quad | \quad \sum \limits_{n = 2}^{9801} \Big( \sqrt{n} \Big) = \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big) +\sqrt{9800} +\sqrt{9801} \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big) +\sqrt{9800} +\sqrt{9801} - \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)-1 ~\right] \\\\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ \underbrace{\sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)- \sum \limits_{n = 2}^{9799} \Big( \sqrt{n} \Big)}_{=0} +\sqrt{9800} +\sqrt{9801} -1 ~\right] \\\\ &\mathbf{=}& \mathbf{\dfrac{\sqrt{2}}{2}\cdot \left[~ \sqrt{9800} +\sqrt{9801} -1 ~\right]} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\large{\sum \limits_{n = 1}^{9800} \dfrac{1}{\sqrt{n+\sqrt{n^2-1}} } }} &\mathbf{=}& \mathbf{\dfrac{\sqrt{2}}{2}\cdot \left[~ \sqrt{9800} +\sqrt{9801} -1 ~\right]} \\ &=& \dfrac{\sqrt{2}}{2}\cdot \left[~ 98.9949493661 + 99 -1 ~\right] \\ &=& \dfrac{\sqrt{2}}{2}\cdot 196.994949366 \\ &\mathbf{=}& \mathbf{139.296464556} \\ \hline \end{array} \)

What is the minimum possible value for y:

y = x^2 + 12x + 5

\(\begin{array}{|lrcll|} \hline & y &=& x^2 + 12x + 5 \\ & &=& (x+6)^2 -36+5 \\ & &=& (x+6)^2 -31 \qquad \text{min. if } x = -6 \\ & y &=& 0 -31 \\ & y &=& -31 \\ \hline \end{array}\)

The smallest value of the expression \(y= x^2 + 12x + 5\) is -31

If x is an integer, what is the smallest value of the expression x^2 - 6x +13?

\(\begin{array}{|lrcll|} \hline & y &=& x^2 - 6x +13 \\ & &=& (x-3)^2 -9+13 \\ & &=& (x-3)^2 + 4 \qquad \text{min. if } x = 3 \\ & y &=& 0 + 4 \\ & y &=& 4 \\ \hline \end{array}\)

The smallest value of the expression \(x^2 - 6x +13\) is 4

Compute

\(\mathbf{\huge{\sum \limits_{n = 1}^{\infty} \dfrac{2n - 1}{n(n + 1)(n + 2)}}}\)

\(\begin{array}{lcll} \mathbf{ \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{3}{2 \cdot 3 \cdot 4} + \dfrac{5}{3 \cdot 4 \cdot 5} + \dfrac{7}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2n - 1}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \mathbf{ ? } } \\ \begin{array}{|lcll|} \hline s_n = \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{3}{2 \cdot 3 \cdot 4} + \dfrac{5}{3 \cdot 4 \cdot 5} + \dfrac{7}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2n - 1}{n \cdot (n+1) \cdot (n+2)} \\ \hline \end{array} \\ \end{array}\\\)

Formula:

\(\begin{array}{|lcll|} \hline \text{in general}:\ \frac{1}{n(n+d)} = \frac{1}{d}\left(\frac{1}{n}- \frac{1}{n+d} \right) \\ \hline \\ \begin{array}{lrcll} \text{we need}: & \dfrac{1}{(n+1)(n+2)} &=& \dfrac{1}{n+1}-\dfrac{1}{n+2} \\ & \dfrac{1}{n(n+1)} &=& \dfrac{1}{n}-\dfrac{1}{n+1} \\ & \dfrac{1}{n(n+2)} &=& \dfrac{1}{2} \left( \dfrac{1}{n}-\dfrac{1}{n+2} \right) \\ \end{array} \\ \hline \end{array}\)

We rearrange:
\(\begin{array}{|rcll|} \hline \dfrac{2n - 1}{n \cdot (n+1) \cdot (n+2)} &=&\left(\dfrac{2n - 1}{n}\right)\times \dfrac{1}{(n+1) \cdot (n+2)} \\\\ &=& \left(\dfrac{2n - 1}{n}\right)\times \left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\\\ &=& \left(2-\dfrac{1}{n}\right)\times \left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{2}{n+1}-\dfrac{2}{n+2}-\dfrac{1}{n(n+1)}+\dfrac{1}{n(n+2)} \\\\ &=& \dfrac{2}{n+1}-\dfrac{2}{n+2}-\left(\dfrac{1}{n}-\dfrac{1}{n+1} \right)+ \dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{1}{n}+\dfrac{1}{2}\cdot \dfrac{1}{n}+\dfrac{2}{n+1}+\dfrac{1}{n+1}-\dfrac{2}{n+2}-\dfrac{1}{2}\cdot \left(\dfrac{1}{n+2}\right) \\\\ &=& -\dfrac{1}{2n} + \dfrac{3}{n+1}-\dfrac{5}{2(n+2)} \quad | \quad \dfrac{1}{2(n+2)} = \dfrac{1}{n}\cdot \left( \dfrac12 - \dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{1}{2n} + \dfrac{3}{n+1}-5\cdot \dfrac{1}{n}\cdot \left( \dfrac12 - \dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{1}{2n} + \dfrac{3}{n+1} -\dfrac{5}{2n} + \dfrac{5}{n(n+2)} \\\\ &=& -\dfrac{3}{n} + \dfrac{3}{n+1} + \dfrac{5}{n(n+2)} \quad | \quad \dfrac{1}{n(n+2)} = \dfrac{1}{2}\cdot \left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{3}{n} + \dfrac{3}{n+1} + 5\cdot \dfrac{1}{2}\cdot \left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{3}{n} + \dfrac{3}{n+1} + \dfrac{5}{2n} - \dfrac{5}{2(n+2)} \\\\ \mathbf{\dfrac{2n - 1}{n \cdot (n+1) \cdot (n+2)} } & \mathbf{=} & \mathbf{ -\dfrac{3}{n} + \dfrac{3}{n+1} + \dfrac{5}{2n} - \dfrac{5}{2(n+2)} } \\ \hline \end{array}\)

Telescoping series:

\(\begin{array}{|rcll|} \hline s_n &=& \mathbf{-3} &\color{red}+& \color{red}\dfrac{3}{2} &+& \mathbf{\dfrac{5}{2}\cdot \dfrac{1}{1}} &\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{3} \\\\ &\color{red}-& \color{red}\dfrac{3}{2} &\color{red}+& \color{red}\dfrac{3}{3} &+&\mathbf{\dfrac{5}{2}\cdot \dfrac{1}{2}}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{4} \\\\ &\color{red}-& \color{red}\dfrac{3}{3} &\color{red}+& \color{red}\dfrac{3}{4} &\color{green}+&\color{green}\dfrac{5}{2}\cdot \dfrac{1}{3}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{5} \\\\ &\color{red}-& \color{red}\dfrac{3}{4} &\color{red}+& \color{red}\dfrac{3}{5} &\color{green}+&\color{green}\dfrac{5}{2}\cdot \dfrac{1}{4}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{6} \\\\ &\color{red}-& \color{red}\dfrac{3}{5} &\color{red}+& \color{red}\dfrac{3}{6} &\color{green}+&\color{green}\dfrac{5}{2}\cdot \dfrac{1}{5}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{7} \\\\ &\color{red}-& \color{red}\dfrac{3}{6} &\color{red}+& \color{red}\dfrac{3}{7} &\color{green}+&\color{green}\dfrac{5}{2}\cdot \dfrac{1}{6}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{8} \\\\ && \ldots \\\\ &\color{red}-& \color{red}\dfrac{3}{n-3} &\color{red}+& \color{red}\dfrac{3}{n-2} &\color{green}+&\color{green}\dfrac{5}{2(n-3)}&\color{green}-& \color{green}\dfrac{5}{2(n-1)} \\\\ &\color{red}-& \color{red}\dfrac{3}{n-2} &\color{red}+& \color{red}\dfrac{3}{n-1} &\color{green}+&\color{green}\dfrac{5}{2(n-2)}&\color{green}-& \color{green}\dfrac{5}{2n} \\\\ &\color{red}-& \color{red}\dfrac{3}{n-1} &\color{red}+& \color{red}\dfrac{3}{n} &\color{green}+&\color{green}\dfrac{5}{2(n-1)}&-& \mathbf{\dfrac{5}{2(n+1)}} \\\\ &\color{red}-& \color{red}\dfrac{3}{n} &+& \mathbf{\dfrac{3}{n+1}} &\color{green}+&\color{green}\dfrac{5}{2n}&-& \mathbf{\dfrac{5}{2(n+2)}} \\ \hline \end{array}\)

The colored terms shorten out

So \(s_n\) is, we have all black terms left :
\(\begin{array}{|rcll|} \hline s_n &=& -3 + \dfrac{5}{2} + \dfrac{5}{2}\cdot \dfrac{1}{2} - \dfrac{5}{2(n+1)} + \dfrac{3}{n+1} -\dfrac{5}{2(n+2)} \\\\ \mathbf{s_n} &\mathbf{=}& \mathbf{\dfrac{3}{4} - \dfrac{5}{2(n+1)} + \dfrac{3}{n+1} -\dfrac{5}{2(n+2)} } \\ \hline \end{array} \)

\( \lim \limits_{n\to \infty} { \dfrac{5}{2(n+1)}} = 0 \quad \text{ and } \quad \lim \limits_{n\to \infty} { \dfrac{3}{n+1} } = 0 \quad \text{ and } \quad \lim \limits_{n\to \infty} { \dfrac{5}{2(n+2)} } = 0\)

\(\begin{array}{|rcll|} \hline \lim \limits_{n\to \infty} s_n &=& \dfrac{3}{4} - 0 + 0 -0 \\\\ &=& \dfrac{3}{4} \\ \hline \end{array}\)

\(\begin{array}{lcll} \mathbf{\huge{\sum \limits_{n = 1}^{\infty} \dfrac{2n - 1}{n(n + 1)(n + 2)}} = \dfrac{3}{4} } \\ \end{array}\\ \)

What is the minimum value of the expression

 \(\mathbf{\huge{2x^2+3y^2+8x-24y+62}}\) 

for real x and y?

\(\begin{array}{|rcll|} \hline \mathbf{f(x,y)} & \mathbf{=} & \mathbf{2x^2+3y^2+8x-24y+62} \\\\ f_x=\dfrac{\partial f} {\partial x} &=& 4x+8 \\ f_y=\dfrac{\partial f} {\partial y} &=& 6y-24 \\ f_{xx}= \dfrac{\partial^2 f} {\partial x^2} &=& 4 \\ f_{yy}= \dfrac{\partial^2 f} {\partial y^2} &=& 6 \\ f_{xy}=f_{yx}= \dfrac{\partial^2 f} {\partial x\partial y}= \dfrac{\partial^2 f} {\partial y\partial x} &=& 0 \\ \hline \end{array} \)

if \(f_{xx}f_{yy}-f^2_{xy}>0\)  either a maximum or a minimum.

Distinguish between these as follows:

if \(f_{xx}<0\) and \(f_{yy} <0\)  then is a maximum point.

if \(f_{xx}>0\) and \(f_{yy} >0\)  then is a minimum point.

\(\begin{array}{|rcll|} \hline && f_{xx}f_{yy}-f^2_{xy} \\ &=& 4\cdot 6 - 0^2 \\ &=& 24 \quad | \quad >0 \quad \text{either a maximum or a minimum} \\\\ && f_{xx} \\ &=& 4 \\ && f_{yy} \\ &=& 6 \quad | \quad f_{xx}>0 \text{ and } f_{yy} >0 \quad \text{it is a minimum point}\\ \hline \end{array} \)

The minimum value:

\(\begin{array}{|rcll|} \hline f_x(x,y) = 4x+8 &=& 0 \\ 4x+8 &=& 0 \\ 4x &=& -8 \\ x &=& -\dfrac{8}{4} \\ \mathbf{x_{\text{minimum}}} &\mathbf{=}& \mathbf{ -2 } \\\\ f_y(x,y) = 6y-24 &=& 0 \\ 6y-24 &=& 0 \\ 6y &=& 24 \\ y &=& \dfrac{24}{6} \\ \mathbf{y_{\text{minimum}}} &\mathbf{=}& \mathbf{ 4 } \\ \hline \end{array} \)

Circles

Give with reasons ATS and TDS in terms of x...

In a chordal quadrilateral: Opposite angles are supplementary.

\(ATS = TDS = 180^\circ - x\)