Question

solve 6x^4-13x^2-x+3=0 If one root is 2- root 3​

Answer 1

${ \tt{ \huge \underline{Given}}}$

${ \rm{ \large{the \: equation \: is \longrightarrow}}}$

${ \rm{ \large{ {6x}^{4} - {13x}^{2} - x + 3 }}}$

${ \rm{ \large{ the \: roots \: of \: the \: equation \: are \: 2 \: and \: 3}}}$

${ \tt{ \huge \underline{Solution}}}$

${ \rm{putting \: the \: first \: root = 2}}$

${ \rm{ { {6x}^{4} - {13x}^{2} - x + 3 }}}$

${ \rm{ { = {6 \times 2}^{4} - {13 \times 2}^{2} - 2+ 3 }}}$

${ \rm{ { = 96 - 52- 2+ 3 }}}$

${ \rm{ { = 96 - 53 }}}$

${ \rm{ { = 43 }}}$

${ \rm{putting \: the \: second \: root = 3}}$

${ \rm{ { {6x}^{4} - {13x}^{2} - x + 3 }}}$

${ \rm{ { = {6 \times 3}^{4} - {13 \times 3}^{2} - 3+ 3 }}}$

${ \rm{ { = 486 - 117 - 3+ 3 }}}$

${ \rm{ { = 486 - 117 }}}$

${ \rm{ { = 369 }}}$

${ \tt{ \large{the \: two \: different \: values\: of \: x \: = 43 \: \: and \: \: 369}}}$