Question

If If one zero of polynomial (a2+1)x2+13x+6a is
reciprocal of the other, find the value of a.

Answer 1

$\begin{gathered}\Large{\purple{\underline{\textsf{\textbf{Given}}}}} \\ \end{gathered}$

One of the zero of the polynomial (a² + 1)x² + 13x + 6a is reciprocal of other .

${\underline{\pink{\boxed{\bf{\gray{Exigency To Find}}}}}}}}\:\green\bigstar \\ \end{gathered}$

The value of a .

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❍ Let's Consider one of the zero of the polynomial be \alphaα .

⠀⠀⠀⠀$\begin{gathered}\orange\bigstar\:{\underline{\pink{\boxed{\bf{\gray{Given That}}}}}}}}\:\green\bigstar \\ \end{gathered}$

⠀⠀▪︎⠀ One of the zero of the polynomial is reciprocal of other .

Therefore,

⠀⠀▪︎⠀ Second zero of polynomial is $\dfrac{1}{\alpha}$

⠀⠀⠀⠀⠀⠀⠀⠀☆ GIVEN POLYNOMIAL : (a² + 1)x² + 13x + 6a

⠀⠀Here ,

⠀⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀Cofficient of x² is ( a² + 1 )

⠀⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀Cofficient of x is 13

⠀⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀Constant term is 6a

$\begin{gathered}\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\\qquad\: \:\maltese\:\: \bf Product \:of \:zeroes\:\:: \\ \end{gathered}$

      $\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{ Product_{( Zeroes \:of\:Polynomial \:)}\:: \:\dfrac{ Constant\:Term }{Cofficient\:of\:x^2\:} }\bigg\rgroup \\\\\end{gathered}$

         $\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Product_{( Zeroes \:of\:Polynomial \:)}\:: \:\dfrac{ Constant\:Term }{Cofficient\:of\:x^2\:} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf \alpha \times \dfrac{1}{\alpha} \:: \:\dfrac{ Constant\:Term }{Cofficient\:of\:x^2\:} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf \alpha \times \dfrac{1}{\alpha} \:: \:\dfrac{ 6a }{a^2 + 1\:} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf \cancel{\alpha} \times \dfrac{1}{\cancel {\alpha}} \:: \:\dfrac{ 6a }{a^2 + 1\:} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf 1 \:: \:\dfrac{ 6a }{a^2 + 1\:} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf 1 \:: \:\dfrac{ 6a }{a^2 + 1\:} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf a^2 - 6a + 1 \\\end{gathered}$

$\begin{gathered}\qquad:\implies \pmb{\underline{\purple{\: a^2 - 6a + 1 \:\:\: }} }\bigstar \\\end{gathered}$

                             $\begin{gathered}\qquad:\implies \bf a^2 - 6a + 1 \: \qquad \longrightarrow \:\: New \: Formed \:Polynomial \:\\\end{gathered}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀Finding zeroes of new formed polynomial :

⠀⠀⠀⠀⠀⠀⠀⠀⠀☆ POLYNOMIAL : a² - 6a + 1

⠀⠀⠀⠀⠀⠀▪︎ Now , By Comparing it with ax² + bx + c :⠀

⠀⠀⠀⠀⠀⠀⠀We get ,

⠀⠀⠀⠀▪︎ ⠀a = 1

⠀⠀⠀⠀▪︎⠀b = -6

⠀⠀⠀⠀▪︎ ⠀c = 1

$\begin{gathered}\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\\qquad\: \:\maltese\:\: \bf Zeros \:of \:Polynomial \:\:: \\ \end{gathered}$

$\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{ Zeros _{(Polynomial)} = \: \dfrac { -(b) \pm \sqrt { b^2 - 4ac }}{2a} }\bigg\rgroup \\\\\end{gathered}$

             $\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { -(b)\pm \sqrt { b^2 - 4ac }}{2a} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { -(-6) + \sqrt { (-6)^2 - 4(1)(1) }}{2(1)} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { 6 + \sqrt { 36 - 4(1)(1) }}{2(1)} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { 6 + \sqrt { 36 - 4 }}{2(1)} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { 6 + \sqrt { 32 }}{2(1)} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { 6 + \sqrt { 32 }}{2} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { 6 + 4\sqrt { 2 }}{2} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { 2 (3 + 2)\sqrt { 2 }}{2} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Zeros _{(Polynomial)} = \: \dfrac { \cancel {2} (3 + 2)\sqrt { 2 }}{\cancel {2}} \\\end{gathered}$

$\begin{gathered}\qquad:\implies \sf Zeros _{(Polynomial)} = \: 3 \pm 2\sqrt { 2 } \\\end{gathered}$

$\begin{gathered}\qquad:\implies \pmb{\underline{\purple{\: Zeros _{(Polynomial)} = \: 3 \pm 2\sqrt { 2 } \:\: }} }\bigstar \\\end{gathered}$

$\huge \sf \fbox\orange{The}\fbox\pink{re}\fbox\blue{f}\fbox\red{o}\fbox \purple{r}\fbox\reder}$

$\begin{gathered}\qquad\qquad \leadsto \bf a \: = \: 3 + 2\sqrt { 2 } \:\: or\:\: 3- 2\sqrt {2} \\\end{gathered}$

$\begin{gathered}\qquad\leadsto \pmb{\underline{\purple{\: \:a \: = \: 3 + 2\sqrt { 2 } \:\: or\:\: 3- 2\sqrt {2}\: }} }\bigstar \\\end{gathered}$

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