If | x + 1 x x x x + x x x x + 2 | = 9 8 ( 103 x + 81 ) , then 3 are the roots of the equation [2023]

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Q.
$If$ $| \begin{matrix} x + 1 & x & x \\ x & x + λ & x \\ x & x & x + λ^{2} \end{matrix} |$ $= \frac{9}{8} (103 x + 81), then λ, \frac{λ}{3} are the roots of the equation$ [2023]

1 $4 x^{2} - 24 x + 27 = 0$

2 $4 x^{2} + 24 x + 27 = 0$

3 $4 x^{2} - 24 x - 27 = 0$

4 $4 x^{2} + 24 x - 27 = 0$

Ans.
(1)

$| \begin{matrix} x + 1 & x & x \\ x & x + λ & x \\ x & x & x + λ^{2} \end{matrix} |$

$=$ $| \begin{matrix} 1 & - λ & 0 \\ 0 & λ & - λ^{2} \\ x & x & x + λ^{2} \end{matrix} |$ $[Applying R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}]$

$= (1) (x λ + λ^{3} + x λ^{2}) + λ (x λ^{2})$

$= λ^{3} + x (λ + λ^{2} + λ^{3}) = \frac{9}{8} \times 103 x + {(\frac{9}{2})}^{3} (Given)$

$\Rightarrow λ = \frac{9}{2};$ $Roots of the equation are \frac{9}{2} and \frac{3}{2}$

$Required quadratic equation is (x - \frac{9}{2}) (x - \frac{3}{2}) = 0$

$\Rightarrow (2 x - 9) (2 x - 3) = 0 \Rightarrow 4 x^{2} - 24 x + 27 = 0$

Solution

Q.
$If$ $| \begin{matrix} x + 1 & x & x \\ x & x + λ & x \\ x & x & x + λ^{2} \end{matrix} |$ $= \frac{9}{8} (103 x + 81), then λ, \frac{λ}{3} are the roots of the equation$ [2023]

1 $4 x^{2} - 24 x + 27 = 0$

2 $4 x^{2} + 24 x + 27 = 0$

3 $4 x^{2} - 24 x - 27 = 0$

4 $4 x^{2} + 24 x - 27 = 0$

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Solution

Q. If |x+1xxxx+λxxxx+λ2|=98(103x+81), then λ,λ3 are the roots of the equation [2023]

1 4x2-24x+27=0

2 4x2+24x+27=0

3 4x2-24x-27=0

4 4x2+24x-27=0

Ans. (1) |x+1xxxx+λxxxx+λ2| =|1-λ00λ-λ2xxx+λ2| [Applying R1→R1-R2, R2→R2-R3] =(1)(xλ+λ3+xλ2)+λ(xλ2) =λ3+x(λ+λ2+λ3)=98×103x+(92)3 (Given) ⇒λ=92; Roots of the equation are 92 and 32 Required quadratic equation is (x-92)(x-32)=0 ⇒(2x-9)(2x-3)=0⇒4x2-24x+27=0

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Q.
$If$ $| \begin{matrix} x + 1 & x & x \\ x & x + λ & x \\ x & x & x + λ^{2} \end{matrix} |$ $= \frac{9}{8} (103 x + 81), then λ, \frac{λ}{3} are the roots of the equation$ [2023]

1 $4 x^{2} - 24 x + 27 = 0$

2 $4 x^{2} + 24 x + 27 = 0$

3 $4 x^{2} - 24 x - 27 = 0$

4 $4 x^{2} + 24 x - 27 = 0$