જો alpha અને beta એ સમીકરણ x^2 + x + 1 = 0 | vedclass

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Question

Roots of the equation ${x^2} + x + 1 = 0$ are $\alpha  = \omega $ and $\beta  = {\omega ^2}$ where $\omega ,{\omega ^2}$

Vedclass · Accepted Answer

$y^3$

Vedclass · Answer

Roots of the equation ${x^2} + x + 1 = 0$ are $\alpha  = \omega $ and $\beta  = {\omega ^2}$ where $\omega ,{\omega ^2}$are complex cube roots of unity

$	herefore \Delta  = \left| {\begin{array}{*{20}{c}}
{y + 1}&\omega &{{\omega ^2}}\
\omega &{y + {\omega ^2}}&1\
{{\omega ^2}}&1&{y + \omega }
\end{array}} ight|$

${R_1} 	o {R_1} + {R_2} + {R_3}$

$ \Rightarrow \Delta  = \left| {\begin{array}{*{20}{c}}
1&1&1\
\omega &{y + {\omega ^2}}&1\
{{\omega ^2}}&1&{y + \omega }
\end{array}} ight|$

Expanding along ${R_1}$, we get

$\Delta  = y.{y^2} \Rightarrow D = {y^3}$

Or

If $\alpha  = {\omega ^2},\beta  = \omega $ we get same value or on expansion using $\alpha  + \beta  =  - 1,\alpha \beta  = 1$ we get value ${y^3}$.

જો $\alpha $ અને $\beta $ એ સમીકરણ $x^2 + x + 1 = 0$ ના બીજ હોય તો $y (\ne 0) \in R$ માટે $\left| {\begin{array}{*{20}{c}}
{y\, + \,1}&\alpha &\beta \\
\alpha &{y\, + \,\beta }&1\\
\beta &1&{y\, + \,\alpha }
\end{array}} \right|$ મેળવો.

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જો $a\, -\, 2b + c = 1$ હોય તો $\left| {\begin{array}{*{20}{c}}
{x + 1}&{x + 2}&{x + a} \\
{x + 2}&{x + 3}&{x + b} \\
{x + 3}&{x + 4}&{x + c}
\end{array}} \right|$ મેળવો.

સમીકરણ $\left| {\,\begin{array}{*{20}{c}}1&4&{20}\\1&{ - 2}&5\\1&{2x}&{5{x^2}}\end{array}\,} \right| = 0$ ના બીજ મેળવો.

જો $\alpha $ અને $\beta $ એ સમીકરણ $x^2 + x + 1 = 0$ ના બીજ હોય તો $y (\ne 0) \in R$ માટે $\left| {\begin{array}{*{20}{c}} {y\, + \,1}&\alpha &\beta \\ \alpha &{y\, + \,\beta }&1\\ \beta &1&{y\, + \,\alpha } \end{array}} \right|$ મેળવો.

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જો $a\, -\, 2b + c = 1$ હોય તો $\left| {\begin{array}{*{20}{c}} {x + 1}&{x + 2}&{x + a} \\ {x + 2}&{x + 3}&{x + b} \\ {x + 3}&{x + 4}&{x + c} \end{array}} \right|$ મેળવો.

સમીકરણ $\left| {\,\begin{array}{*{20}{c}}1&4&{20}\\1&{ - 2}&5\\1&{2x}&{5{x^2}}\end{array}\,} \right| = 0$ ના બીજ મેળવો.

જો $\alpha $ અને $\beta $ એ સમીકરણ $x^2 + x + 1 = 0$ ના બીજ હોય તો $y (\ne 0) \in R$ માટે $\left| {\begin{array}{*{20}{c}}
{y\, + \,1}&\alpha &\beta \\
\alpha &{y\, + \,\beta }&1\\
\beta &1&{y\, + \,\alpha }
\end{array}} \right|$ મેળવો.

જો $a\, -\, 2b + c = 1$ હોય તો $\left| {\begin{array}{*{20}{c}}
{x + 1}&{x + 2}&{x + a} \\
{x + 2}&{x + 3}&{x + b} \\
{x + 3}&{x + 4}&{x + c}
\end{array}} \right|$ મેળવો.