heureka

avatar
Имя пользователяheureka
Гол26398
Membership
Stats
Вопросов 17
ответы 5678

 #3
avatar+26398 
+2

The difference of the roots of the quadratic equation \(x^2 + bx + c = 0\) is \(|b - 2c|\).
If \(c \neq 0\), then find \(c\) in terms of \(b\).

 

\(\begin{array}{|lrcll|} \hline & x^2 + \underbrace{ b }_{-(x_1+x_2)}x + \underbrace{ c }_{x_1x_2} &=& 0 \\ \hline & b &=& -(x_1+x_2) \\ & -b &=& x_1+x_2 \\ (1) & \mathbf{x_2} &=& \mathbf{-x_1-b} \\ \hline (2) & \mathbf{c} &=& \mathbf{x_1x_2} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline | x_1-x_2| &=& |b - 2c| \quad &| \quad \mathbf{x_2=-x_1-b} \\ | x_1-(-x_1-b)| &=& |b - 2c| \\ | 2x_1+b)| &=& |b - 2c| \quad &| \quad \text{compare}\\\\ 2x_1 &=& -2c \quad &| \quad : 2 \\ \mathbf{x_1} &=& \mathbf{-c} \\\\ \mathbf{x_2} &=& \mathbf{-x_1-b} \quad (1) \quad & | \quad \mathbf{x_1=-c} \\ x_2 &=& \mathbf{-(-c)-b} \\ \mathbf{x_2} &=& \mathbf{c-b} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{c} &=& \mathbf{x_1x_2} \quad (2) \quad & | \quad \mathbf{x_1=-c},\quad \mathbf{x_2} = \mathbf{c-b} \\ c &=& (-c)(c-b)\quad &| \quad : c \qquad c \neq 0\ ! \\ 1 &=& -(c-b) \\ 1 &=& -c+b \\ \mathbf{c} &=& \mathbf{b-1} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{x_1} &=& \mathbf{-c} \quad & | \quad \mathbf{c=b-1} \\ x_1 &=& -(b-1) \\ \mathbf{x_1} &=& \mathbf{1-b} \\ \\ \mathbf{x_2} &=& \mathbf{c-b} \quad & | \quad \mathbf{c=b-1} \\ x_2 &=& b-1-b \\ \mathbf{x_2} &=& \mathbf{-1} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline | x_1-x_2| &=& |1-b-(-1)| \\ | x_1-x_2| &=& |1-b+1| \\ | x_1-x_2| &=& |2-b| \\ \hline |b-2c| &=& |b-2(b-1)| \\ |b-2c| &=& |b-2b+2)| \\ |b-2c| &=& |2-b|\ \checkmark \\ \hline \end{array}\)

 

laugh

24 окт. 2019 г.
 #4
avatar+26398 
+1

Piecewise function

If  \(z = \dfrac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} } \)  find \(\lfloor z \rfloor\).

 

The fractional part denoted by  \(\{x\}\) for real x and defined by the formula \( \{x\}=x-\lfloor x\rfloor\).

Source: https://en.wikipedia.org/wiki/Floor_and_ceiling_functions

 

\(\begin{array}{|rcll|} \hline z &=& \dfrac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} } \quad | \quad \{ \sqrt{3} \} = \sqrt{3}- \lfloor \sqrt{3}\rfloor,\ \{ \sqrt{2} \} = \sqrt{2}- \lfloor \sqrt{2}\rfloor \\\\ &=& \dfrac{ \left(\sqrt{3}- \lfloor \sqrt{3}\rfloor\right)^2 - 2 \left(\sqrt{2}- \lfloor \sqrt{2}\rfloor\right)^2 }{ \left(\sqrt{3}- \lfloor \sqrt{3}\rfloor\right) - 2 \left(\sqrt{2}- \lfloor \sqrt{2}\rfloor\right) } \quad | \quad \lfloor \sqrt{3}\rfloor = \lfloor \sqrt{2}\rfloor = 1 \\\\ &=& \dfrac{ \left(\sqrt{3}- 1\right)^2 - 2 \left(\sqrt{2}- 1\right)^2 }{ \left(\sqrt{3}- 1\right) - 2 \left(\sqrt{2}- 1\right) } \\\\ &=& \dfrac{ 3-2\sqrt{3} + 1 -2(2-2\sqrt{2}+1) } { \sqrt{3}-1-2\sqrt{2} +2 } \\\\ &=& \dfrac{ 3-2\sqrt{3} + 1 -4 +4\sqrt{2}-2 } { \sqrt{3}-2\sqrt{2} +1 } \\\\ &=& \dfrac{ -2\sqrt{3} +4\sqrt{2}-2 } { \sqrt{3}-2\sqrt{2} +1 } \\\\ &=& \dfrac{ -2\left( \sqrt{3} -2\sqrt{2}+1\right) } { \left(\sqrt{3}-2\sqrt{2} +1\right) } \\\\ &=& -2 \\ \hline \mathbf{ \lfloor z\rfloor } &=& \mathbf{ -2 } \\ \hline \end{array}\)

 

laugh

21 окт. 2019 г.