Question

The quadratic polynomial where α=5+2√6 and αβ=1 is​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

The quadratic polynomial whose zeroes are given by

$\sf{ \alpha = 5 + 2 \sqrt{6} \: \: \: \: and \: \: \: \: \alpha \beta = 1\: }$

EVALUATION

Here

$\sf{ \alpha = 5 + 2 \sqrt{6} \: }$

Also

$\sf{ \alpha \beta = 1 \: }$

$\implies \displaystyle \sf{ \beta = \frac{1}{ \alpha } }$

$\implies \displaystyle \sf{ \beta = \frac{1}{5 + 2 \sqrt{6} } }$

$\implies \displaystyle \sf{ \beta = \frac{5 - 2 \sqrt{6} }{(5 + 2 \sqrt{6})(5 - 2 \sqrt{6} ) } }$

$\implies \displaystyle \sf{ \beta = \frac{5 - 2 \sqrt{6} }{ {5}^{2} - {(2 \sqrt{6} )}^{2} } }$

$\implies \displaystyle \sf{ \beta = \frac{5 - 2 \sqrt{6} }{25 - 24 } }$

$\implies \displaystyle \sf{ \beta = \frac{5 - 2 \sqrt{6} }{1 } }$

$\implies \displaystyle \sf{ \beta = 5 - 2 \sqrt{6} }$

Therefore

$\sf{ \alpha + \beta \: }$

$\sf{ = 5 + 2 \sqrt{6 } + 5 - 2 \sqrt{6} \: }$

$= 10$

Hence the required Quadratic polynomial is

$\sf{ {x}^{2} - ( \alpha + \beta )x + \alpha \beta }$

$= \sf{ {x}^{2} - 10x + 1 \: }$

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Answer 2

Given by question:

α, β be the roots of the quadratic polynomial.

α = $5+2\sqrt{6}$ and αβ = 1

We have to find the quadratic polynomial.

Solution:

We have:

αβ = 1

⇒ β = $\dfrac{1}{\alpha}=\dfrac{1}{5+2\sqrt{6}}$

⇒ β = $\dfrac{1}{5+2\sqrt{6}}\times \dfrac{5-2\sqrt{6}}{5-2\sqrt{6}}$

⇒ β = $\dfrac{5-2\sqrt{6}}{5^2-(2\sqrt{6})^2}$  [ ∵ $a^{2} -b^{2} =(a+b)(a-b)$

⇒ β = $\dfrac{5-2\sqrt{6}}{25-24}$

⇒ β = $5-2\sqrt{6}$

∴ α + β = $(5+2\sqrt{6})+( 5-2\sqrt{6})$ = 10

We know that:

The quadratic polynomial is:

$x^{2} -(\alpha+\beta)x+\alpha\beta$

∴ The required quadratic polynomial is:

$x^{2} -10x+1$

Thus, the required quadratic polynomial is equal to "$x^{2} -10x+1$".