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Mathematics Question on Quadratic Equations

The value of x obtained from the equation ∣x+αβγ γx+βα αβx+γ∣=0\begin{vmatrix}x+\alpha&\beta&\gamma\\\ \gamma&x+\beta&\alpha\\\ \alpha&\beta&x+\gamma\end{vmatrix}=0 will be

A

0 and −(α+β+γ)-\left(\alpha+\beta+\gamma\right)

B

0 and α+β+γ\alpha+\beta+\gamma

C

1 and (α−β−γ)\left(\alpha-\beta-\gamma\right)

D

0andα2+β+γ20 and \alpha^{2}+\beta+\gamma^{2}

Answer

0 and −(α+β+γ)-\left(\alpha+\beta+\gamma\right)

Explanation

Solution

Given∣x+αβγ γx+βα αβx+γ∣=0Given \begin{vmatrix}x+\alpha&\beta&\gamma\\\ \gamma&x+\beta&\alpha\\\ \alpha&\beta&x+\gamma\end{vmatrix}=0 Operate C1→C1+C2+C3C_{1} \to C_{1}+C_{2}+C_{3} ∣x+α+β+γβγ x+α+β+γx+βα x+α+β+γβx+γ∣=0\begin{vmatrix}x+\alpha+\beta+\gamma&\beta&\gamma\\\ x+\alpha+\beta+\gamma&x+\beta&\alpha\\\ x+\alpha+\beta+\gamma&\beta&x+\gamma\end{vmatrix}=0 =(x+α+β+γ)∣1βγ 1x+βα 1βx+γ∣=0=\left(x+\alpha+\beta+\gamma\right) \begin{vmatrix}1&\beta&\gamma\\\ 1&x+\beta&\alpha\\\ 1&\beta&x+\gamma\end{vmatrix}=0 ⇒x+α+β+γ=0⇒ x=−(α+β+γ)\Rightarrow\quad x+\alpha+\beta+\gamma=0 \Rightarrow\, x=-\left(\alpha+\beta+\gamma\right) Again if ∣1βγ 1x+βα 1βγ∣=0⇒∣1βγ 0xα−γ 00x∣=0\begin{vmatrix}1&\beta&\gamma\\\ 1&x+\beta&\alpha\\\ 1&\beta&\gamma\end{vmatrix}=0 \Rightarrow \begin{vmatrix}1&\beta&\gamma\\\ 0&x&\alpha-\gamma\\\ 0&0&x\end{vmatrix}=0 ⇒x2=0⇒x=0\Rightarrow\quad x^{2}=0 \Rightarrow x=0 ∴\therefore\quad Solutions of the equation are x = 0, −(α+β+γ)-\left(\alpha+\beta+\gamma\right)