Question

If a,b,c are rational then the root of equation (a+2b-3c )x²+(b+2c-3a)x+ (c+2a-3b)=0

Answer 1

$\textbf{Given:}$

$(a+2b-3c)x^2+(b+2c-3a)x+(c+2a-3b)=0$

$\textbf{To find:}$

$\text{Roots of the equation}$

$\textbf{Solution:}$

$\text{Consider,}$

$(a+2b-3c)x^2+(b+2c-3a)x+(c+2a-3b)=0$

$\text{Sum of the coefficients=}(a+2b-3c)+(b+2c-3a)+(c+2a-3b)$

$=0$

$\therefore\text{(x-1) is a factor}$

$\textbf{By synthetic division}$

$\begin{array}{r|ccc}1&a+2b-3c&b+2c-3a&c+2a-3b\\&&a+2b-3c&-c+3b-2a\\\cline{2-4}&a+2b-3c&-2a+3b-c&|0\end{array}$

$\text{Quotient=}(a+2b-3c)x+(-2a+3b-c)$

$(a+2b-3c)x+(-2a+3b-c)=0$

$\implies\,x=\dfrac{c+2a-3b}{a+2b-3c}$

$\therefore\textbf{The roots are 1 and \dfrac{c+2a-3b}{a+2b-3c} }$

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