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The quadratic equation, whose roots are three times the roots of 3ax² + 3bx + c = 0, is

1) ax2 + 3bx + 3c = 0 2) ax2 + 3bx + c = 0 3) 9ax2 + 9bx + c = 0 4) ax2 + bx + 3c = 0 Solution: (1) ax2 + 3bx + 3c = 0 If α, β are the roots... View Article

If α and β are the roots of x² – a (x – 1) + b = 0, then the value of 1 / [ɑ² – aɑ] + 1 / [β² – aβ] + 2 / [a + b] is

1) 4 / (a + b) 2) 1 / (a + b) 3) 0 4) - 1 Solution: (3) 0 Since a and b are the roots of the given equation, ɑ2 - a (ɑ - 1) + b = 0 β2 - a... View Article

If one root of the equation lx² + mx + n = 0 is 9/2 (where, l,m and n are positive integers and m/4n = l/m, then l + n is equal to

1) 80 2) 85 3) 90 4) 95 5) 100 Solution: (2) 85 The given equation is lx2 + mx + n = 0. m2 - 4ln = 0 It has real roots. [x - (9 / 2)]2 = 0... View Article

If 2 – i is a root of the equation ax² + 12x² + b = 0(where,a and b are real), then the value of ab is

1) 45 2) 15 3) - 15 4) - 45 5) 25 Solution: (1) 45 Imaginary roots exist in conjugate pairs. The other root is 2 + i. 2 + i + 2 - i = - 12... View Article

If the roots of the equation 1 / (x + p) + 1 / (x + q) = 1 / r (x ≠ – p, x ≠ – q, r ≠ 0) are equal in magnitude but opposite in sign, then p + q =

1) r 2) 2r 3) r 4) 1/r 5) 2/r Solution: (2) 2r 1 / (x + p) + 1 / (x + q) = 1 / r ⇒ r (2x + p + q) = [x2 + (p + q) x + pq] ⇒ x2 + (p + q -... View Article

If α and β are the roots of the quadratic equation ƛ (x² – x) + x + 5 = 0 and ƛ₁, ƛ₂ are two values ƛ obtained from (ɑ / β) + (β / ɑ) = 4 / 5, then ƛ₁ / ƛ₁² + ƛ₂ / ƛ₂² are two values of ƛ

1) 4192 2) 4144 3) 4096 4) 4048 Solution: (4) 4048 (ɑ + β) = [ƛ - 1] / ƛ; ɑβ = j / ƛ (ɑ / β) + (β / ɑ) = 4 / 5 [ɑ2 + β2] / ɑβ = 4 / 5 [(ɑβ)2... View Article

If the sum of the roots of the equation ax² + 2x + 3a = 0 is equal to their product, then value of a is

1) -2/3 2) -3 3) 4 4) -1/2 Solution: (1) -2 / 3 Let α and β be the roots of the equation. ax2 + 2x + 3a = 0 ɑ + β = - 2 / a, ɑβ = 3a / a ɑ +... View Article

If α and β are the roots of the quadratic equation x² + x + 1 = 0, then the equation, whose roots are α¹⁹ and β⁷, is

1) x2 - x + 1 = 0 2) x2 - x - 1 = 0 3) x2 + x - 1 = 0 4) x2 + x + 1 = 0 Solution: (4) x2 + x + 1 = 0 The roots of x2 + x + 1 = 0 are x = [-... View Article

If the roots of the equation x²2

1) - 1 2) 0 3) 1 4) 2 5) 3 Solution: (3) 1 Let α, α + 1 are consecutive integer ∴ (x + α) (x + α + 1) = x2 - bx + c Comparing both sides -... View Article

If a and b are the roots of the equation x² + ax + b = 0, a ≠ 0, b ≠ 0, the values of a and b are respectively

1) 2 and -2 2) 2 and -1 3) 1 and - 2 4) 1 and 2 5) - 1 and 2 Solution: (3) 1 and - 2 Since a + b = - a; ab = b 2a + b = 0; ab - b = 0 2a =... View Article

If x² + px + q = 0 has the roots α and β, then the value of (α – β)² is

1) p2 - 4q 2) (p2 - 4q)2 3) p2 + 4q 4) (p2 + 4q)2 Solution: (1) p2 - 4q ɑ, β are roots of equation, x2 + px + q = 0 ɑ + β = p, ɑβ = q (ɑ -... View Article

If one root of the equation x² + px + q = 0 is 2 + √3, then the values of p and q are respectively

1) - 4, 1 2) 4, -1 3) 2,√3 4) - 2, -√3 Solution: (1) - 4, 1 If 2 + √3 is a root and 2 - √3 is its conjugate root. (2 + √3 + 2 - √3) = - p p... View Article

If α and β are the roots of the equation ax² + bx + c = 0 (c ≠ 0), then the equation, whose roots are 1 / [aα + b] and 1 / [aβ + b], is

1) acx2 - bx + 1 = 0 2) x2 - acx + bc + 1 = 0 3) acx2 + bx - 1 = 0 4) x2 + acx + 11 = 0 Solution: (1) acx2 - bx + 1 = 0 ɑ, β are roots of... View Article

If α and β are the roots of the equation x² – x + 1 = 0, then α²⁰⁰⁹ + β²⁰⁰⁹ is equal to

1) - 2 2) - 1 3) 1 4) 2 Solution: (3) 1 x2 - x + 1 = 0 x3 + 1 = (x + 1) (x2 - x + 1) = 0 x3 = - 1 If ɑ, β are roots then ɑ + β = 1, ɑβ = 1... View Article

Let p and q be real numbers such that p ≠ 0, p³ ≠ q and p³ ≠ – q. If α and β are non – zero complex numbers satisfying α + β = – p and α³ + β³ = q, then a quadratic equation having α / β and α / β as its roots is

1) (p3 + q)x2 - (p3 + 2q)x + (p3 + q ) = 0 2) (p3 + q)x2 - (p3 - 2q)x + (p3 + q ) = 0 3) (p3 - q)x2 - (5p3 - 2q)x + (p3 - q ) = 0 4) (p3 -... View Article

If α, β and γ are the roots of x³ – 2x + 1 = 0 then the value of ∑ [1] / [α + β – γ] is

1) -1 / 2 2) -1 3) 0 4) 1/2 Solution: (2) -1 x3 - x - x + 1 = 0 x2 (x2 - 1) - (x - 1) = 0 x = 1, x2 + x - 1 = 0 α = 1, β = [- 1 + √5] / 2, γ... View Article

If α and β are the roots of the equation x² – 2x + 4 = 0, then the value of αⁿ + βⁿ will be

1) i2n+1 sin (nπ/3 ) 2) 2n+1 cos (nπ/3 ) 3) i 2n-1 sin (nπ/3 ) 4) 2n-1 cos (nπ/3 ) Solution: (2) 2n+1 cos (nπ / 3) Since α and β are the... View Article

If the product of the roots of the equation x² – 2 √2kx + 2e^2logk – 1 = 0 is 31, then the roots of the equation are real for k, is equal to

1) - 4 2) 1 3) 4 4) 0 Solution: (3) 4 Product = 31 2e2logk - 1 = 31 2e2logk = 32 k2 = 16 k = ± 4 But log k > 0, k ≠ - 4, k = 4... View Article

If (α + √β ) and (α -√β ) are the roots of the equation where x² + px + q = 0, where α, β p and q are real, then the roots of the equation (p² – 4q) (p²x² + 4px) – 16q = 0 are

1) [(1 / ɑ) + (1 / √β)] and [(1 / ɑ) - (1 / √β)] 2) [(1 / √ɑ) + (1 / β)] and [(1 / √ɑ) - (1 / β)] 3) [(1 / √ɑ) + (1 / √β)] and [(1 / √ɑ) - (1 /... View Article

If α , β and γ are the roots of the equation x³ + 4x + 2 = 0, then α³ + β³ + γ³ is equal to

1) 2 2) 6 3) - 2 4) - 6 Solution: (4) - 6 x3 + 4x + 2 = 0 ∑ɑ = 0, ∑ɑβ = 4 / 1 = 4, ɑβγ = - 2 / 1 = - 2 α3 + β3 + γ3 = 3ɑβγ = 3 * - 2 = - 6... View Article

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