Let - are two real and different roots of a quadratic equation and the determinant -4- 3 4- 1 -3 1 -3 is zero-1- Then the quadratic equation can be

Given : nbsp;Δ= 4 α amp;3 amp;4 β α amp;1 amp;β α^3 amp;1 amp;β^3 Applying R_1→ R_1+R_2 Δ= 4 amp;4 amp;4 α amp;1 amp;β α^3 amp;1 amp;β ...

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Standard XII

Mathematics

Evaluation of a Determinant

Let α,β are t...

Question

Let $α, β$ are two real and different roots of a quadratic equation and the determinant $Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 4 - α & 3 & 4 - β α & 1 & β α^{3} & 1 & β^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣$ is zero, $(α, β \neq 1) .$ Then the quadratic equation can be

2 x^{2} + 2 x - 3 = 0

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x^{2} - 3 x + 2 = 0

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x^{2} + x - 2 = 0

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2 x^{2} - 4 x + 1 = 0

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Solution

The correct option is C $x^{2} + x - 2 = 0$
Given : $Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 4 - α & 3 & 4 - β α & 1 & β α^{3} & 1 & β^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Applying $R_{1} \to R_{1} + R_{2}$
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} 4 & 4 & 4 α & 1 & β α^{3} & 1 & β^{3} \end{matrix} ∣ ∣ ∣ ∣$
Taking $(4)$ common from $R_{1},$ we get
$Δ = 4 ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 α & 1 & β α^{3} & 1 & β^{3} \end{matrix} ∣ ∣ ∣ ∣$
$\Rightarrow 4 (α - 1) (1 - β) (β - α) (α + β + 1) = 0 ∵ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{3} & b^{3} & c^{3} \end{matrix} ∣ ∣ ∣ ∣ = (a - b) (b - c) (c - a) (a + b + c) \Rightarrow (α - 1) (1 - β) (β - α) (α + β + 1) = 0 \Rightarrow α + β + 1 = 0 (∵ α, β \neq 1) ∴ α + β = - 1$

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Evaluation of a Determinant

Standard XII Mathematics

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Let α,β are two real and different roots of a quadratic equation and the determinant Δ=∣∣ ∣ ∣∣4−α34−βα1βα31β3∣∣ ∣ ∣∣ is zero,(α,β≠1). Then the quadratic equation can be

Let $α, β$ are two real and different roots of a quadratic equation and the determinant $Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 4 - α & 3 & 4 - β α & 1 & β α^{3} & 1 & β^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣$ is zero, $(α, β \neq 1) .$ Then the quadratic equation can be