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Find the quadratic equations whose roots are the reciprocals of the roots of 2×2 + 5x + 3 = 0 ?
A. 3x2 – 5x + 2 = 0
B. 3x2 – 5x – 2 = 0
C. 3x2 + 5x + 2 = 0
D. 3x2 + 5x – 2 = 0The quadratic equation whose roots are reciprocal of 2x2 + 5x + 3 = 0 can be obtained by replacing x by 1/x.
Hence, 2(1/x)2 + 5(1/x) + 3 = 0
=> 3x2 + 5x + 2 = 0(i). a2 – 7a + 12 = 0, (ii). b2 – 3b + 2 = 0 to solve both the equations to find the values of a and b ?
A. if the relationship between a and b cannot be established.
B. if a > b
C. if a < b
D. if a ≤ bExplanation:
I.(a – 3)(a – 4) = 0
=> a = 3, 4
II. (b – 2)(b – 1) = 0
=> b = 1, 2
=> a > b(i). a2 – 9a + 20 = 0, (ii). 2b2 – 5b – 12 = 0 to solve both the equations to find the values of a and b ?
A. If a ≥ b
B. If the relationship between a and b cannot be established
C. If a < b
D. If a ≤ bExplanation:
I. (a – 5)(a – 4) = 0
=> a = 5, 4
II. (2b + 3)(b – 4) = 0
=> b = 4, -3/2 => a ≥ b(i). a2 + 11a + 30 = 0, (ii). b2 + 6b + 5 = 0 to solve both the equations to find the values of a and b ?
A. If a ≤ b
B. If a > b
C. If a < b
D. If the relationship between a and b cannot be establishedExplanation:
I. (a + 6)(a + 5) = 0
=> a = -6, -5
II. (b + 5)(b + 1) = 0
=> b = -5, -1 => a ≤ bI. a2 + 8a + 16 = 0, II. b2 – 4b + 3 = 0 to solve both the equations to find the values of a and b ?
A. If the relationship between a and b cannot be established
B. If a > b
C. If a ≤ b
D. If a < bI. x2 + 5x + 6 = 0, II. y2 + 9y +14 = 0 to solve both the equations to find the values of x and y ?
A. If x > y
B. If x ≤ y
C. If x < y
D. If x = y or the relationship between x and y cannot be established.Explanation:
I. x2 + 3x + 2x + 6 = 0
=> (x + 3)(x + 2) = 0 => x = -3 or -2
II. y2 + 7y + 2y + 14 = 0
=> (y + 7)(y + 2) = 0 => y = -7 or -2
No relationship can be established between x and y.I. x2 – x – 42 = 0, II. y2 – 17y + 72 = 0 to solve both the equations to find the values of x and y ?
A. If x ≥ y
B. If x < y
C. If x > y
D. If x ≤ yExplanation:
I. x2 – 7x + 6x – 42 = 0
=> (x – 7)(x + 6) = 0 => x = 7, -6
II. y2 – 8y – 9y + 72 = 0
=> (y – 8)(y – 9) = 0 => y = 8, 9
=> x < yI. x2 + 9x + 20 = 0,II. y2 + 5y + 6 = 0 to solve both the equations to find the values of x and y ?
A. If x ≤ y
B. If x ≥ y
C. If x < y
D. If x > yI. x2 + 4x + 5x + 20 = 0
=>(x + 4)(x + 5) = 0 => x = -4, -5
II. y2 + 3y + 2y + 6 = 0
=>(y + 3)(y + 2) = 0 => y = -3, -2
= x < y.I. x2 + 3x – 18 = 0, II. y2 + y – 30 = 0 to solve both the equations to find the values of x and y ?
A. If x > y
B. If x ≥ y
C. If x < y
D. If x = y or the relationship between x and y cannot be established.Explanation:
I. x2 + 6x – 3x – 18 = 0
=>(x + 6)(x – 3) = 0 => x = -6, 3
II. y2 + 6y – 5y – 30 = 0
=>(y + 6)(y – 5) = 0 => y = -6, 5
No relationship can be established between x and y.I. x2 + 11x + 30 = 0,II. y2 + 15y + 56 = 0 to solve both the equations to find the values of x and y ?
A. If x ≤ y
B. If x ≥ y
C. If x < y
D. If x > yExplanation:
I. x2 + 6x + 5x + 30 = 0
=>(x + 6)(x + 5) = 0 => x = -6, -5
II. y2 + 8y + 7y + 56 = 0
=>(y + 8)(y + 7) = 0 => y = -8, -7
=> x > y