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3 and 4 .Determinants and Matrices
medium
જો સમીકરણ $\left| {\,\begin{array}{*{20}{c}}x&3&7\\2&x&{ - 2}\\7&8&x\end{array}\,} \right| = 0$ નું એક બીજ $ 5$ હોય , તો બાકીના બે બીજ મેળવો.
A
$-2 $ અને $7$
B
$-2$ અને $-7$
C
$2 $ અને $7$
D
$2 $ અને $-7$
Solution
(d) Given, One root =$ 5$ and equation $\left| {\,\begin{array}{*{20}{c}}x&3&7\\2&x&{ – 2}\\7&8&x\end{array}\,} \right|\, = 0$.
Expanding the given equation, we get
$ + {\log _x}z({\log _y}x{\log _z}y – {\log _z}x)$
==> ${x^3} + 16x – 6x – 42 + 112 – 49x = 0$
==> ${x^3} – 39x + 70 = 0$
Since $ 5$ is the one root of given equation, therefore ${x^3} – 5{x^2} + 5{x^2} – 25x – 14x + 70 = 0$
==> ${x^2}(x – 5) + 5x(x – 5) – 14(x – 5) = 0$
==> $(x – 5)({x^2} + 5x – 14) = 0$
==> $(x – 5)\,(x – 2)\,(x + 7) = 0$ or $x = 5,\,2$ and $ -7.$
Standard 12
Mathematics
