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 .l​

Answer 1

x

2

+5x+6=0

x

2

+2x+3x+6=0

x(x+2)+3(x+3)=0

(x+2)(x+3)=0

x=−2,−3

∴α=−2,β=−3

α+β=−

a

b

−2−3=−

1

5

−5=−5

αβ=

a

c

(−2)(−3)=

1

6

6=6

Answer 2

Answer:

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

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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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