Question

construct a polynomial whose one zero is √2 +1 and whose product of zeros is 1

​

Answer 1

$\tt \:  Let \: zeroes \: of \: polynomial \: be \: \alpha \: and \: \beta .$

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{ \alpha = \sqrt{2} + 1} \\ &\sf{ \alpha \beta = 1} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{a \: quadratic \: polynomial}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

★ Given

$: \implies \tt \: \alpha = \sqrt{2} + 1 \\ : \implies \tt \: \alpha \beta \: = 1 \: \: \: \: \: \: \: \:$

$: \implies \tt \: \beta \: = \: \dfrac{1}{ \alpha }$

$: \implies \tt \: \beta \: = \dfrac{1}{ \sqrt{2} + 1 }$

$: \implies \tt \: \beta = \dfrac{1}{ \sqrt{2} + 1 } \times \dfrac{ \sqrt{2} - 1}{ \sqrt{2} - 1}$

$: \implies \tt \: \beta = \dfrac{ \sqrt{2} - 1}{ { (\sqrt{2}) }^{2} - {1}^{2} }$

$: \implies \tt \: \beta = \dfrac{ \sqrt{2} - 1}{2 - 1}$

$: \implies \tt \: \beta = \sqrt{2} - 1$

★ Now,

$\tt \:Sum \: of \: zeros \: = \alpha + \beta = \sqrt{2} + 1 + \sqrt{2 } - 1$

$: \implies \tt \: \alpha + \beta = 2 \sqrt{2}$

★ So, we have now

$: \implies \tt \: \alpha + \beta = 2 \sqrt{2} \\ \bf \: and \\ : \implies \tt \: \: \alpha \: \beta \: = \: 1 \: \: \: \: \: \: \: \:$

★ Now, we know that quadratic polynomial is given by

$: \implies \tt \: f(x) = k \bigg( {x}^{2} - ( \alpha + \beta )x + \alpha \beta \bigg)$

★ where k is non zero real number.

$: \implies \tt \: f(x) = k \bigg( {x}^{2} - 2 \sqrt{2} \: x + 1 \bigg)$