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We are providing the most important **Quadratic Equations (Inequalities) for SBI PO 2019, SBI Clerk 2019** and all **other competitive bank and insurance exams**. These questions have very high chances to be asked in **SBI PO 2019, SBI Clerk 2019.
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Directions:1-10) ** In the following questions, two equations I and II are given. You have to solve both the equations and g****ive Answer as,**

1.

I.√(1225)x + √(1089) =0

II.(625)^{1/4} y + (343)^{1/3} = 0

2.

I. 25x^{2} – 45x + 14 =0

II.48y^{2} -176y – 64 =0

3.

I. 72x^{2} -177x +108 =0

II. 5y^{2} + 24y + 27 =0

4.

I. x^{2} -11x +30 =0

II. y^{2} -13y + 42 =0

5.

I. 25x^{2} +32x -36 =0

II. 57y^{2} -25y-42 =0

6.

I.2x^2 + 19x + 45 = 0

II.2y^2 + 11y + 12 = 0

7.

I.3x^2 – 13x + 12 = 0

II.2y^2 – 15y + 28 = 0

8.

I.x^2 = 16

II.2y^2 – 17y + 36 = 0

9.

I.6x^2 + 19x + 15 = 0

II.3y^2 + 11y + 10 = 0

10.

I.2x^2 – 11x + 15 = 0

II.2y^2 – 11y + 14 = 0

**Check your Answers below:**

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

##### 1. Question

I.√(1225)x + √(1089) =0

II.(625)^{1/4} y + (343)^{1/3} = 0

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. Question

I. 25x^{2} – 45x + 14 =0

II.48y^{2} -176y – 64 =0

I.25x

^{2}– 45x + 14 =0

25x

^{2}– 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.48y^{2} -176y – 64 =0

48y^{2} -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. Question

I. 72x^{2} -177x +108 =0

II. 5y^{2} + 24y + 27 =0

I.72x

^{2}-177x +108 =0

72x

^{2}-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.5y^{2} + 24y + 27 =0

5y^{2} + 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. Question

I. x^{2} -11x +30 =0

II. y^{2} -13y + 42 =0

I.x

^{2}-11x +30 =0

x

^{2}-5x -6x +30 =0

x(x-5)-6 (x-5) =0

(x-6) (x-5) =0

X=6, 5

II.y^{2} -13y + 42 =0

y^{2} – 6y – 7y + 42 =0

y(y-6)-7(y-6) =0

(y-7)(y-6) =0

Y= 7, 6

x ≤ y

##### 5. Question

I. 25x^{2} +32x -36 =0

II. 57y^{2} -25y-42 =0

I.25x

^{2}+32x -36 =0

25x

^{2}-18x + 50x -36 =0

X(25x-18)+2 (25x-18)=0

(x+2) (25x-18) = 0

X =-2, 18/25 = -2, 0.72

II.57y^{2} -25y-42 =0

57y^{2} +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. Question

I.2x^2 + 19x + 45 = 0

II.2y^2 + 11y + 12 = 0

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. Question

I.3x^2 – 13x + 12 = 0

II.2y^2 – 15y + 28 = 0

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. Question

I.x^2 = 16

II.2y^2 – 17y + 36 = 0

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. Question

I.6x^2 + 19x + 15 = 0

II.3y^2 + 11y + 10 = 0

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. Question

I.2x^2 – 11x + 15 = 0

II.2y^2 – 11y + 14 = 0

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