Question

The roots of the equations 2√5-13x-3√5=0 are:-
(A)1/√5,3√5/2 (B)-1/√5,-3√5/2 (C)-1/√5,3√5/2 (D)1/√5,-3/√5

Answer 1

Answer:

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Step-by-step explanation:

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Sol :

given that x-√5 is a factor of the cubic polynomial x3-3√5x2+13x-3√5

x-√5 ) x3-3√5x2+13x-3√5 ( x2 -2√5x + 3

x3- √5x2 ( substract )

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- 2√5x2+13x

- 2√5x2+10x ( substract )

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3x - 3√5

3x - 3√5 ( substract )

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0

∴ The quotient is x2 -2√5x + 3 = 0

Using roots of quadratic formula

a = 1, b = 2√5, c = 3

x = (-b ± √(b2 - 4ac) ) / 2a

x = (2√5 ± √((2√5)2 - 12) ) / 2

∴ the other zeros are x = √5 ± √2.

Hope it helps you!