Question

roots of the equation x 4 2x 3 13x 2 x^4-x^3-9x^2+13x-5=0

Answer 1

$</p><p>{\text{We have a bi-quadratic equation }}2{x^4} - {x^3} - 9{x^2} + 13x - 5 = 0\,{\text{which can be solved using initially Hit -Trial Method}}{\text{.}} \hfill \\</p><p>{\text{Let }}p\left( x \right) = 2{x^4} - {x^3} - 9{x^2} + 13x - 5 \hfill \\</p><p>\Rightarrow {\text{ }}p\left( 1 \right) = 2{\left( 1 \right)^4} - {1^3} - 9{\left( 1 \right)^2} + 13\left( 1 \right) - 5 = 2 - 1 - 9 + 13 - 5 = 0 \hfill \\</p><p>\Rightarrow \left( {x - 1} \right)\,\,{\text{is a factor of }}p\left( x \right) = 2{x^4} - {x^3} - 9{x^2} + 13x - 5 \hfill \\</p><p>\Rightarrow p\left( x \right) = \left( {x - 1} \right)\left( {2{x^3} + {x^2} - 8x + 5} \right) \hfill \\</p><p> {\text{Let }}q\left( x \right) = 2{x^3} + {x^2} - 8x + 5 \hfill \\</p><p>{\text{Again }}q\left( 1 \right) = 2{\left( 1 \right)^3} + {\left( 1 \right)^2} - 8\left( 1 \right) + 5 = 2 + 1 - 8 + 5 = 0 \hfill \\</p><p> \Rightarrow \left( {x - 1} \right)\,{\text{is a factor of }}q\left( x \right) = 2{x^3} + {x^2} - 8x + 5 \hfill \\</p><p>\Rightarrow q\left( x \right) = \left( {x - 1} \right)\left( {2{x^2} + 3x - 5} \right) \hfill \\</p><p> \Rightarrow q\left( x \right) = \left( {x - 1} \right)\left( {2{x^2} + 5x - 2x - 5} \right) \hfill \\</p><p> \Rightarrow q\left( x \right) = \left( {x - 1} \right)\left\{ {2x\left( {x - 1} \right) + 5\left( {x - 1} \right)} \right\} \hfill \\</p><p>\Rightarrow q\left( x \right) = \left( {x - 1} \right)\left( {2x + 5} \right)\left( {x - 1} \right) \hfill \\</p><p>\because p\left( x \right) = \left( {x - 1} \right)q\left( x \right) \hfill \\</p><p>\Rightarrow p\left( x \right) = \left( {x - 1} \right)\left( {x - 1} \right)\left( {2x + 5} \right)\left( {x - 1} \right) \hfill \\</p><p>\therefore p\left( x \right) = 0 \hfill \\</p><p>\Rightarrow \left( {x - 1} \right)\left( {x - 1} \right)\left( {2x + 5} \right)\left( {x - 1} \right) = 0 \hfill \\</p><p>\Rightarrow x = 1,1,1,\frac{{ - 5}}{2}\,{\text{are required solutions of given bi-quadratic equation}}{\text{.}} \hfill \\ </p><p>$