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If α and β are the roots of the equation x² – px + r = 0 and α/2, 2β are the roots of the equation x² – qx + r = 0. Then, the value of r is

1) 2/9 (p - q) (2q - p) 2) 2/9 (q- p) (2p - q) 3) 2/9 (q - 2p) (2q - p) 4) 2/9 (2p - q) (2q - p) Solution: (4) 2/9 (2p - q) (2q - p) α + β =... View Article

If α and β are the roots of the equation x² + ax + b = 0, then (1 / α²) + (1 / β²) =

1) [a2 - 2b] / b2 2) [b2 - 2a] / b2 3) [a2 + 2b] / b2 4) [b2 + 2a] / b2 Solution: (1) [a2 - 2b] / b2 (α + β) = α, α β = b (1 / α2) + (1 /... View Article

If α and β are the roots of the equation ax² + bx + c = 0, αβ = 3, and a, b, c are in AP, then α + β is equal to

1) - 4 2) 1 3) 4 4) - 2 Solution: (4) - 2 (α + β) = - b / a, (αβ) = 3 = c / a 2 (b / a) = 1 + (c / a) - 2 (α + β) = 1 + 3 (α + β) = - 4 / 2 = - 2... View Article

If α, β and γ are the roots of the equation x³ – 6x² + 11 x + 6 = 0, then ∑α²β + ∑αβ² is equal to

1) 80 2) 168 3) 90 4) - 84 Solution: (2) 168 Since ∑ɑ = 6 ∑αβ = 11 and ɑβɣ = - 6 ∑α2β + ∑αβ2 = 2 (α2β + β2ɣ + ɣ2α + βɣ2 + ɣα2) = 2 [αβ (α +... View Article

If α and β are the roots of the quadratic equation ax² + bx + c = 0, observe the lists given below

1) A = 5 B = 2 C = 4 D = 6 2) A = 5 B = 2 C = 1 D =4 3) A = 5 B = 4 C = 2 D = 6 4) A = 5 B = 2 C = 4 D = 1 Solution: (4) A = 5 B = 2 C = 4... View Article

The cubic equation, whose roots are thrice to each of the roots of x³ + 2x² – 4x + 1 = 0, is

1) x3 - 6x2 + 36x + 27 = 0 2) x3 + 6x2 + 36x + 27 = 0 3) x3 + 6x2 + 36x + 27 = 0 4) x3 + 6x2 - 36x + 27 = 0 Solution: (4) x3 + 6x2 - 36x + 27... View Article

If α, β and γ are the roots of the equation x³ – 8 x + 8 = 0, then ∑α² and ∑ 1/αβ are respectively

1) 0 and - 16 2) 16 and 18 3) - 16 and 0 4) 16 and 0 Solution: (4) 16 and 0 α + β + γ = 0 = ∑ ɑβ = - 8 = ɑβɣ = - 8 ∑ ɑ2 = (α + β + γ)2 - 2 ∑... View Article

If α and β are the roots of the equation ax² + bx + c = 0, then α / [ɑβ + b] + β / [aɑ + b] =

1) 2/a 2) 2/b 3) 2/c 4) - (2/a) Solution: (4) - (2 / a) It is given that α and β are the roots of the equation ax2 + bx + c = 0. aɑ2 + bɑ +... View Article

If α and β are the roots of x² – 2x cos ∅ + 1 = 0, then the equation, whose roots are αⁿ + βⁿ

1) x2 - 2 xcos n∅ - 1 = 0 2) x2 - 2 xcos n∅ + 1 = 0 3) x2 - 2 xsin n∅ + 1 = 0 4) x2 + 2 xsin n∅ - 1 = 0 Solution: (2) x2 - 2 xcos n∅ + 1 = 0... View Article

If two persons A and B solve the equation x² + ax + b = 0, while solving, A commits a mistake the coefficient of x was taken as 15 in place of -9 and finds the roots as – 7 and – 2 Then, the equation is

1) x2 + 9x + 14 = 0 2) x2 - 9x + 14 = 0 3) x2 + 9x - 14 = 0 4) x2 - 9x - 14 = 0 5) None of these Solution: (2) x2 - 9x + 14 = 0 The... View Article

If α and β are the roots of the equations x² – (1 + n²) x+ 1 / 2 (1 + n² + n⁴) = 0, then α² + β² is equal to

1) n2 2) - n2 3) n4 4) - n4 Solution: (1) n2 (α + β) = 1 + n2; (αβ) = [1 + n2 + n4] / 2 (α2 + β2) = (α + β)2 - 2αβ = (1 + n2)2 - 2 (1 / 2)... View Article

If α and β are the roots of the equation x² – ax + b = 0 and A_n = αⁿ + βⁿ. Then A_n+1 – aA_n + bA_n-1 is equal to

1) - a 2) b 3) 0 4) a - b Solution: (3) 0 If α, β are the roots of the equation, x2 - ax + b = 0, so each of them must satisfy the equation.... View Article

Let the two numbers have arithmetic mean 9 and geometric mean 4. Then, these numbers are the roots of the quadratic equation

1) x2 - 18x - 16 = 0 2) x2 - 18x +16 = 0 3) x2 + 18x - 16 = 0 4) x2 + 18x + 16 = 0 Solution: (2) x2 - 18x + 16 = 0 Let α and β be two... View Article

If α + β = – 2 and α³ + β³ = – 56, then the quadratic equation, whose roots are α and β, is

1) x2 + 2x - 16 = 0 2) x2 + 2x - 15 = 0 3) x2 + 2x - 12 = 0 4) x2 + 2x - 8 = 0 Solution: (4) x2 + 2x - 8 = 0 (α + β)3 = α3 + β3 + 3 αβ (α +... View Article

If α, β and γ are such that α + β + γ = 2, α² + β² + γ² = 6, α³ + β³ + γ³ = 8 then α⁴ + β⁴ + γ⁴ is equal to

1) 7 2) 12 3) 18 4) 36 Solution: (3) 18 α + β + γ = 2 α2 + β2 + γ2 = 6 α3 + β3 + γ3 = 8 (α + β + γ)2 = 4 α2 + β2 + γ2 + 2 ∑αβ = 4 2 ∑αβ = 4... View Article

The number of real solutions of the equation |x|² – 3 | x | + 2 = 0 is

1) 4 2) 1 3) 3 4) 2 Solution: (1) 4 |x|2 - 3 |x| + 2 = 0 (|x| - 1) (|x| - 2) = 0 |x| = 1, 2 x = 1, -1, 2, -2 Hence, four real solutions exist.... View Article

If α, β and γ are the roots of x³ + 4x + 1 = 0, then the equation, whose roots are α² / (β + γ), β² / (α + γ), γ² / (α + β), is

1) x3 - 4x - 1 = 0 2) x3 - 4x + 1 = 0 3) x3 + 4x - 1 = 0 4) x3 + 4x + 1 = 0 Solution: (3) x3 + 4x - 1 = 0 Given α, β and γ are the roots of... View Article

If α and β are the roots of x² – 2x + 4 = 0, then the value of α⁶ + β⁶ is

1) 32 2) 64 3) 128 4) 256 Solution: (3) 128 α + β = 2 αβ = 4 α2 + β2 = (α + β)2 - 2αβ = 4 - 8 = - 4 (α2 + β2)3 = α6 + β6 + 3 α2 β2 (α2 + β2)... View Article

If x₁, x₂, x₃ and x₄ are roots of the equation x⁴ – x³ sin 2β + x² cos 2β – x cos β – sin β = 0. Then tan^-1 x₁ + tan^-1 x₂ + tan^-1 x₃ + tan^-1 x₄ =

1) β 2) π / 2 - β 3) π - β 4) - β Solution: (2) π / 2 - β ∑x1 = sin 2β ∑x1x2 = cos 2β ∑x1x2x3 = cos β x1x2x3x4 = sin β tan-1 x1 + tan-1 x2 +... View Article

If one root of equation x² + ax + 12 = 0 is 4 while the equation x² + ax + b = 0 has equal roots, then the value of b is

1) 4/49 2) 49/4 3) 7/4 4) 4/7 Solution: (2) 49 / 4 It is given that 4 is the root of the equation. x2 + ax + 12 = 0 42 + 4a + 12 = 0 4a = -... View Article

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