Question

If one zero of the polynomial (a2 + 9)x + 13x + 6a is reciprocal of another, then find the value of a.

Answer 1

${ \bold{ \green{ \underline{ \underline{ANSWER}} : - }}}$

${ \bold{ \underline{Given \: polynomial}} : - } \\ \\ { \bold{ \orange{( {a}^{2} + 9) {x}^{2} + 13x + 6a = 0 }}} \\ \\ { \bold{ \pink{Given \: \: polynomial \: \: have \: \: raciprocal \: }}} \\ \\ { \bold{ \orange{ = > \alpha \beta = 1 \: \: \: \: (product \: \: of \: \: roots)}}} \\ \\ { \bold{ \orange{ = > \frac{c}{a} = 1 }}} \\ \\ { \bold{ \orange{ = > \frac{6a}{ { a}^{2} + 9} = 1}}} \\ \\ { \bold{ \orange{ = > {a}^{2} - 6a + 9 = 0}}} \\ \\ { \bold{ \orange{ \implies {a}^{2} - 3a - 3a + 9 = 0 }}} \\ \\ { \bold{ \orange{ \implies \: a(a - 3) - 3(a - 3) = 0}}} \\ \\ { \bold{ \orange{ \implies \: {(a - 3)}^{2} = 0}}} \\ \\ { \bold{ {\orange{ \implies { \boxed{\: a = 3}}}}}}$

${ \bold{ \boxed{\red{\huge{ \mathfrak{FOLLOW \: \: ME...}}}}}}$