Quadratic Equation - gkquestionsguru.com

Directions:1-10) In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,

1.
I.√(1225)x + √(1089) =0
II.(625)^1/4 y + (343)^1/3 = 0

If x > y
If x ≥ y
If x < y
If x ≤ y
If x = y or the relation cannot be established

Ans:
I.√(1225)x + √(1089) =0
35x + 33 = 0
35x = -33
X= -33/35 = -0.94

II.(625)^1/4 y + (343)^1/3 = 0
5y+7 =0
Y= -7/5 = -1.4
x>y

2.
I. 25x² – 45x + 14 =0
II.48y² -176y – 64 =0

If x > y
If x ≥ y
If x < y
If x ≤ y
If x = y or the relation cannot be established

Ans:
I.25x² – 45x + 14 =0
25x² – 10x – 35x + 14 =0
5x(5x-2) -7(5x-2) =0
(5x-7) (5x-2) = 0
X= 7/5 , 2/5 =1.4, 0.4

II.48y² -176y – 64 =0
48y² -192y +16y – 64 =0
48y(y-4)+16 (y-4)=0
(48y + 16) (y-4) =0
Y= – 16/48, 4 = -0.33, 4
Can’t be determined.

3.
I. 72x² -177x +108 =0
II. 5y² + 24y + 27 =0

If x > y
If x ≥ y
If x < y
If x ≤ y
If x = y or the relation cannot be established

Ans:
I.72x² -177x +108 =0
72x² -81x -96x +108 =0
9x(8x-9)-12(8x-9) =0
(9x-12)(8x-9)=0
X=12/9 , 9/8 =1.33, 1.125

II.5y² + 24y + 27 =0
5y² + 15y + 9y + 27 =0
5y(y+3)+9(y+3) =0
(5y+9) (y+3) =0
Y= -9/5 , -3 = -1.8, -3
X > y

4.
I. x² -11x +30 =0
II. y² -13y + 42 =0

If x > y
If x ≥ y
If x < y
If x ≤ y
If x = y or the relation cannot be established

Ans:
I.x² -11x +30 =0
x² -5x -6x +30 =0
x(x-5)-6 (x-5) =0
(x-6) (x-5) =0
X=6, 5

II.y² -13y + 42 =0
y² – 6y – 7y + 42 =0
y(y-6)-7(y-6) =0
(y-7)(y-6) =0
Y= 7, 6
x ≤ y

5.
I. 25x² +32x -36 =0
II. 57y² -25y-42 =0

If x > y
If x ≥ y
If x < y
If x ≤ y
If x = y or the relation cannot be established

Ans:
I.25x² +32x -36 =0
25x² -18x + 50x -36 =0
X(25x-18)+2 (25x-18)=0
(x+2) (25x-18) = 0
X =-2, 18/25 = -2, 0.72

II.57y² -25y-42 =0
57y² +38y-63y -42 =0
19y (3y+2)- 21(3y+2) =0
(19y-21) (3y+2) =0
y= 21/19, -2/3 = 1.105, -0.66
Can’t be determined.

6.
I.2x^2 + 19x + 45 = 0
II.2y^2 + 11y + 12 = 0

x> y
x ≥ y
x< y
Relationship between x and y cannot be determined
x ≤ y

Ans:
I.2x^2 + 19x + 45 = 0
2 × 45 = 90 = (10 × 9)
(10 + 9 = 19) x = (-10/2), (-9/2) (dividing by co efficient of x^2 and changing signs)
x = -5, -4.5

II.2y^2 + 11y + 12 = 0
2 × 12 = 24 (8 × 3 = 24) (8 + 3 = 11)
y = (-8/2), (-3/2) ) (dividing by co efficient of y^2 and changing signs)
y = -4, -1.5
Hence x < y

7.
I.3x^2 – 13x + 12 = 0
II.2y^2 – 15y + 28 = 0

x ≥ y
Relationship between x and y cannot be determined
x> y
x< y
x ≤ y

Ans:
I.3x^2 – 13x + 12 = 0
12 × 3 = 36 (-9 × -4 = 36) (-9 – 4 = -13)
x = 9/3, 4/3 (dividing by co efficient of x^2 and changing signs)
x = 3, 4/3

II.2y^2 – 15y + 28 = 0
2 × 28 = 56 (-8 × -7 = 56)
(-8 -7 = -15) (Dividing by co efficient of y^2 and changing signs)
y = 4, 3.5
Hence x< y

8.
I.x^2 = 16
II.2y^2 – 17y + 36 = 0

x< y
Relationship between x and y cannot be determined
x> y
x ≥ y
x ≤ y

Ans:
I.x^2 = 16
x = ±4

II.2y^2 – 17y + 36 = 0
2 × 36 = 72 (-9 × -8 = 72)
(-9 -8 = -17) y = 9/2, 8/2(dividing by co efficient of y^2 and changing signs)
y = 4.5, 4
Hence x ≤ y

9.
I.6x^2 + 19x + 15 = 0
II.3y^2 + 11y + 10 = 0

x ≤ y
x< y
x ≥ y
x> y
Relationship between x and y cannot be determined

Ans:
I.6x^2 + 19x + 15 = 0
6 × 15 = 90 (10 × 9 = 90) (10 + 9 = 19)
x = -10/6, -9/6 (dividing by co efficient of x^2 and changing signs)
x = -5/3, -3/2 = -1.66, -1.5

II.3y^2 + 11y + 10 = 0
3 × 10 = 30 (6 × 5 = 30) (6 + 5 = 11)
y = -6/3, -5/2 (dividing by co efficient of y^2 and changing signs)
y = -1.6, -2
Hence x ≥ y

10.
I.2x^2 – 11x + 15 = 0
II.2y^2 – 11y + 14 = 0

x ≤ y
x> y
x ≥ y
x< y
Relationship between x and y cannot be determined

Ans:
I.2x^2 – 11x + 15 = 0
2 × 15 = 30 (-6 × -5 = 30) (-6 – 5 = -11)
x = -6/2, -5/2 (dividing by co efficient of x^2 and changing signs)
x = 3, 2.5

II.2y^2 – 11y + 14 = 0
2 × 14 = 28 (-7 × -4 = 28) (-7 -4 = -11)
y = 7/2, 4/2 (dividing by co efficient of y^2 and changing signs)
y = 3.5, 2
Hence relationship cannot be established.

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11 thoughts on “Quadratic Equation”

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