- Home
- Standard 12
- Mathematics
$k $ ની કેટલી કિંમતો માટે સમીકરણ સંહતી $\left( {k + 1} \right)x + 8y = 4k\;,\;kx + \left( {k + 3} \right)y $$= 3k - 1$ ને એક પણ ઉકેલ નથી.
અંનત
$1$
$2$
$3$
Solution
${\rm{\Delta }} = \left| {\begin{array}{*{20}{c}}
{k + 1}&8\\
k&{k + 3}
\end{array}} \right|$$ = {k^2} + 4k + 3 – 8k$
$\qquad = {k^2} – 4k + 3$
$\qquad = (k – 3)(k – 1)$
${{\rm{\Delta }}_1} = \left| {\begin{array}{*{20}{c}}
{4k}&8\\
{3k – 1}&{k + 3}
\end{array}} \right|$$ = 4{k^2} + 12k – 24k + 8$
$\qquad = 4{k^2} + 12k – 24k + 8$
$\qquad = 4({k^2} – 3k + 2)$
$\qquad = 4(k – 2)(k – 1)$
${{\rm{\Delta }}_2} = \left| {\begin{array}{*{20}{c}}
{k + 1}&{4k}\\
k&{3k – 1}
\end{array}} \right|$$ = 3{k^2} + 2k – 1 – 4{k^2}$
$\qquad = – {k^2} + 2k – 1$
$\qquad = – {(k – 1)^2}$
As given no solution
${{\rm{\Delta }}_1}\;\;and\;\;{{\rm{\Delta }}_2} \ne 0$
but ${\rm{\Delta }} = 0$
$k=3$