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x^{3} + 1 + 2x =6x + 1/x
x^{3} + 1 + 2x = (6x^{2} + 1)/x
(x^{3} + 1 + 2x)x = 6x^{2} + 1
x^{4} - 4x^{2} - 6x^{2} -1 =0
x^{4} - 4x^{2} + x - 1 =0
x^{3} + 1 + 2x =6x + 1/x
x^{3} + 1 + 2x = (6x^{2} + 1)/x
(x^{3} + 1 + 2x)x = 6x^{2} + 1
x^{4} - 4x^{2} - 6x^{2} -1 =0
x^{4} - 4x^{2} + x - 1 =0
Degree of polynomial is highest exponent degree term i.e.,4.
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x = 1 + √2
∴ x^{4} - 4x^{3} + 4x^{2} = x^{2}(x^{2} - 4x + 4)
= x^{2}(x - 2)^{2}
= (1 + √2)^{2}(1 + √2 - 2)^{2}
x = 1 + √2
∴ x^{4} - 4x^{3} + 4x^{2} = x^{2}(x^{2} - 4x + 4)
= x^{2}(x - 2)^{2}
= (1 + √2)^{2}(1 + √2 - 2)^{2}
=(√2 + 1)^{2} (√2 - 1)^{2}
=[(√2)^{2} - (1)^{2}]^{2}
=(2 - 1)^{2} =1
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x - y = 0.9 ...(i)
and 11(x + y)^{-1}=2
⇒ 11/ (x + y) = 2
⇒ 2(x + y) =11
x - y = 0.9 ...(i)
and 11(x + y)^{-1}=2
⇒ 11/ (x + y) = 2
⇒ 2(x + y) =11
⇒ x + y = 11/2 ...(ii)
On solving Eqs.(i) and (ii),we get
x = 3.2
and y = 2.3
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Let two alternate odd integers odd integers be (2x+1) and (2x+5).
Then according to the question,
(2x + 1) (2x + 5) = 3(2x + 1) + 12
⇒ (2x + 1)(2x + 5 - 3) = 12
⇒ 2x^{2} + 3x - 5 = 0
Let two alternate odd integers odd integers be (2x+1) and (2x+5).
Then according to the question,
(2x + 1) (2x + 5) = 3(2x + 1) + 12
⇒ (2x + 1) (2x + 5 - 3) = 12
⇒ 2x^{2} + 3x - 5 = 0
On solving this quadratic equation,we get
x = 1 and x = -5/2
x = -5/2 is not a integer ∴ x = 1
Then, larger integer = 2x + 5 = 2 x 1 + 5 = 7
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