Algebra
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If x = √3 + √2, then the value of x + 1 is : x
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x = √3 + √2
∴ 1 = 1 x √3 + √2 = √3 - √2 (√3 + √2)(√3 - √2)
= √3 - √2∴ x + 1 x
= √3 + √2 + √3 - √2 = 2√3Correct Option: B
x = √3 + √2
∴ 1 = 1 x √3 + √2 = √3 - √2 (√3 + √2)(√3 - √2)
= √3 - √2∴ x + 1 x
= √3 + √2 + √3 - √2 = 2√3
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If x − 1 = 5, then x2 + 1 is x x2
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x − 1 = 5 x
On squaring both sides,x2 + 1 − 2 = 25 x2 ⇒ x2 + 1 = 27 x2
Second Method :Here, x − 1 = 5 x
We know thatx2 + 1 = x − 1 2 + 2 x2 x
= 52 + 2 = 27Correct Option: C
x − 1 = 5 x
On squaring both sides,x2 + 1 − 2 = 25 x2 ⇒ x2 + 1 = 27 x2
Second Method :Here, x − 1 = 5 x
We know thatx2 + 1 = x − 1 2 + 2 x2 x
= 52 + 2 = 27
- One of the factors of the expression 4√3x2 + 5x − 2√3 is :
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4√3x² + 5x – 2√3
= 4√3 x² + 8x – 3x – 2√3
= 4x (√3x + 2) - √3( √3x + 2)
= (4x - √3) (√3 + 2) ⇒ factorsCorrect Option: D
4√3x² + 5x – 2√3
= 4√3 x² + 8x – 3x – 2√3
= 4x (√3x + 2) - √3( √3x + 2)
= (4x - √3) (√3 + 2) ⇒ factors
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If a2 – 4a – 1 = 0, then value of a2 + 1 + 3a – 3 is a2 a
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a² – 4a – 1 = 0
⇒ a² – 1 = 4a
On dividing by a, we havea - 1 = 4 a ∴ a² + 1 + 3 a - 1 a² a = a - 1 ² + 2 + 3 a - 1 a a
= 16 + 2 + 3 (4) = 30Correct Option: B
a² – 4a – 1 = 0
⇒ a² – 1 = 4a
On dividing by a, we havea - 1 = 4 a ∴ a² + 1 + 3 a - 1 a² a = a - 1 ² + 2 + 3 a - 1 a a
= 16 + 2 + 3 (4) = 30