Question

5) find the zeroes of the polynomial √5 x 2 – 7x - 6 √5 . and verify the relationship between the zeroes and coefficients of the polynomial .​

Answer 1

Step-by-Step explanation:

Given Polynomial = √5x² – 7x – 6√5

Find zeros of polynomial,

$\sf\longrightarrow{ \sqrt{5} {x}^{2} - 7x - 6 \sqrt{5}}$

$\sf\longrightarrow{ \sqrt{5} {x}^{2} - 10x + 3x - 6 \sqrt{5}}$

$\sf\longrightarrow{ \sqrt{5} {x}(x - 2 \sqrt{5})+ 3(x - 2\sqrt{5})}$

$\sf\longrightarrow{ (\sqrt{5}{x} + 3)(x - 2 \sqrt{5}})$

Zeros =

$\sf\longrightarrow{ \sqrt{5} x + 3 = 0}$

$\sf\longrightarrow{ \sqrt{5} x = - 3}$

$\sf\longrightarrow{ x = \dfrac{ - 3}{ \sqrt{5} } }$

First zero = $\sf{\dfrac{-3}{\sqrt{5}}}$

$\sf\longrightarrow{x - 2 \sqrt{5}} = 0$

$\sf\longrightarrow{x = 2 \sqrt{5}}$

Second zero = $\sf{2\sqrt{5}}$

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Verifying the relationship between zeros and coefficients :

Let α and β be the polynomial's zeros.

In the polynomial =

• a = $\sf{\sqrt{5}}$
• b = $\sf{-7}$
• c = $\sf{-6\sqrt{5}}$

★ Sum of zeros =

• α + β

$\sf\longrightarrow{ \alpha + \beta = \dfrac{ - b}{a}}$

$\sf\longrightarrow{ \alpha + \beta = \dfrac{ - ( - 7)}{ \sqrt{5} }}$

$\sf\longrightarrow{ \alpha + \beta = \dfrac{ 7}{ \sqrt{5} } \: - - - (i)}$

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$\sf\longrightarrow{\bigg( \dfrac{ - 3}{ \sqrt{5} }\bigg)+ 2\sqrt{5}}$

$\sf\longrightarrow{\bigg( \dfrac{ - 3}{ \sqrt{5} }\bigg)+ \dfrac{10}{ \sqrt{5}}}$

$\sf\longrightarrow\dfrac{7}{ \sqrt{5}} \: - - - (ii)$

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★ Product of zeros :

$\sf\longrightarrow{ \alpha \times \beta = \dfrac{c}{a}}$

$\sf\longrightarrow{ \alpha \beta = \dfrac{ - 6 \sqrt{5} }{ \sqrt{5} }}$

$\sf\longrightarrow{ \alpha \beta = - 6} \: - - - (iii)$

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$\sf\longrightarrow{2 \sqrt{5} \times \dfrac{ - 3}{ \sqrt{5}}}$

$\sf\longrightarrow{ 2 \times ( - 3)}$

$\sf\longrightarrow{ - 6} \: - - - (iv)$

Here, i and ii are equal, and iii and iv are equally.

Hence, the relationship between zeros and coefficients is verified.

Answer 2

• $\tiny\boxed{\sf{\blue{Zeroes\:of\:the\:polynomial\:(\alpha\:and\:\beta)\:=\:\dfrac{-3}{\sqrt{5}}\:and\:2\sqrt{5}\:\:}}}$

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Explanation :

$\underline{\bf\dag{\underline{\green{Given\:Polynomial}}}:}$

$\qquad\bf\dag\:\bigg\lgroup \sf{ \sqrt{5}x^2 - 7x - 6\sqrt{5} } \bigg\rgroup$

$\underline{\bf\dag{\underline{\green{To\:Find}}}:}$

• Zeroes of the polynomial.
• Also, we have to verify the relationship between the zeroes and coefficients of the polynomial.

$\underline{\bf\dag{\underline{\green{Solution}}}:}$

★ Finding zeroes of the polynomial :-

$\qquad\leadsto\quad\tt \sqrt{5}x^2 - 7x - 6\sqrt{5} = 0$

$\quad\bf\dag\:\bigg\lgroup \sf{Using\:middle\:splitting\:method}\bigg\rgroup$

$\qquad\leadsto\quad\tt \sqrt{5}x^2 - 10x + 3x - 6\sqrt{5} = 0$

$\qquad\leadsto\quad\tt \sqrt{5}x(x - 2\sqrt{5}) + 3(x - 2\sqrt{5}) = 0$

$\qquad\leadsto\quad\tt (\sqrt{5}x + 3)(x - 2\sqrt{5}) = 0$

$\qquad\leadsto\quad\tt \sqrt{5}x + 3 = 0 \:,\:x - 2\sqrt{5} = 0$

$\qquad\leadsto\quad\tt \sqrt{5}x = -3 \:,\:x = 2\sqrt{5}$

$\qquad\leadsto\quad\bf{ x = \red{\dfrac{-3}{\sqrt{5}}} \:,\:x = \red{2\sqrt{5}}}$

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★ Verifying relation b/w the zeroes and coefficients of the polynomial :-

$\small\leadsto\quad\tt \alpha + \beta = \dfrac{-b}{a}\:,\:\alpha\beta = \dfrac{c}{a}$

★ Values that we have :-

• α and β = zeroes of the polynomial = -3/√5 and 2√5
• a = coefficient of x² = √5
• b = coefficient of x = -7
• c = constant term = -6√5

★ Putting all known values :-

$\small\leadsto\quad\tt \dfrac{-3}{\sqrt{5}} + 2\sqrt{5} = \dfrac{-(-7)}{\sqrt{5}}\:,\:\dfrac{-3}{\sqrt{5}}\:\times\:2\sqrt{5} = \dfrac{-6\sqrt{5}}{\sqrt{5}}$

$\small\leadsto\quad\tt \dfrac{-3}{\sqrt{5}} + \dfrac{10}{\sqrt{5}} = \dfrac{7}{\sqrt{5}}\:,\:\dfrac{-3}{\cancel{\sqrt{5}}}\:\times\:2\cancel{\sqrt{5}} = \dfrac{-6\cancel{\sqrt{5}}}{\cancel{\sqrt{5}}}$

$\small\leadsto\quad\tt \dfrac{-3 + 10}{\sqrt{5}} = \dfrac{7}{\sqrt{5}}\:,\:-3\:\times\:2 = -6$

$\leadsto\quad\bf{\red{ \dfrac{7}{\sqrt{5}}} = \red{\dfrac{7}{\sqrt{5}}}\:,\:\red{-6} = \red{ -6}}$

$\qquad\qquad\:\:\underline{\bf\dag{\underline{\green{Hence,\:Verified!}}}\bf\dag}$

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