Quadratic Equations for SBI PO 2017

Quadratic Equations for SBI PO 2017

Directions (1 – 10): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and given answer:

a) If x< y

b) If x ≤ y

c) If x = y, or no relation can be established between x and y.

d) If x > y

e) If x ≥ y

1. I. 6x ² + 13x + 5 =0     II.3y² + 11y + 10 =0

2. I. 2x² + 3x – 5 =0          II.2y² + 11y + 15 =0

3.  I. (x^1/4 / 16)² = 144 / x^3/2  II. y^1/3  × y^2/3  × 3014 = 16 × y²

4. I. 16x^2 – 40x – 39 = 0 II. 12y^2 – 113y + 255 = 0

5. I. 63x -194 √x +143 = 0 II. 99y - 255 √y +150 = 0

6. I. x² – x – 56 = 0 II. y=³√729

7. I. x² + 7x + 10=0  II. y² + 13y + 42 =0

8. I. 14x² - 55x + 50 = 0 II. 2y² – 13y + 20 = 0

9. I. 7x + 8y = 5    II. 6x-5y = 28

10. I. 5x² – 52x + 96 = 0     II. 5y² + 3y-36 = 0

Solutions:

1. E) I. 6x² + 13x +5 =0

Or, 6x² + 3x +10x + 5=0

Or, 3x (2x +1)+ 5(2x + 1)= 0

Or, (3x + 5) (2x +1 ) = 0

Therefore, X = - 5/3, -1/2

II.3y²+11y+10 = 0

Or, 3y² + 6y + 5y+10 = 0

Or, 3y(y + 2) + 5(y + 2) = 0

Or, (3y + 5) (y + 2) = 0

Therefore, y= - 5/3, -2

Hence, x ≥ y

2. D) I. 2x² + 3x – 5 =0

Or, x = 1, - 5/2

II. 2y² + 11y + 15 = 0

Or, 2y² + 6y + 5y+ 15 = 0

Or, 2y(y + 3)+ 5(y + 3) = 0

Or,(y + 3) (2y + 5) = 0

Or, y = -3, - 5/2

Hence, x ≥ y

3. C) I. (x^1/4 / 16)² = (144 / x^3/2 )  = (x^1/2 / 256) = (144 / x^3/2 )

(x^1/2 )  × (x^3/2 ) = 256 × 144

x² = (256 × 144)

x = √(256 × 144)

x = ± (16 × 12) = ±192

II. y^1/3 × y^2/3 × 3104 = 16y²

 y × 3104 = 16y²

3104 = 16y

Y = 3104 / 16 = 194

Clearly, x > y

4. B) I. 16x^2 - 40x - 39 = 0

or 16x^2 - 52x + 12x - 39 = 0

or (4x- 13) (4x + 3)

x = 13/4, -3/4

II. 12y^2 - 113y + 255 = 0

or 12y^2 - 45y - 68y + 255 = 0

or (4y - 15) (3y - 17) = 0

y = 15/4, 17/3

Therefore y > x or, x < y

5. E) I. 63x -194 √x +143 = 0

or63x -117 √x - 77√x +143 = 0

or (7 √x -13)(9 √x -11) = 0

x = 169/49, 121/81

II. 99y - 225 √y +150 = 0

or 99y - 90 √y -165 √y +150 = 0

or (11 √y -10)(9√y -15) = 0

y = 100/121, 225/81

Therefore relation cannot be established between x and y.

6. C) I . x² – x – 56 = 0

Or,x² + 7x – 8x-56 = 0

Or, x(x+7) -8(x+7) =0

Or,(x-8)(x+7)=0

Or, x= 8,-7

II. y=³√729

Or, y=9

Hence, x < y

7. A) I. x² + 7x + 10 = 0

Or,x² + 2x +5x + 10 = 0

Or, x(x+2)+ 5(x+2)=0

Or,(x+5)(x+2) = 0

or,x= -2,-5

II.y² + 13y + 42 =0

Or,y² + 6y+7y+42 = 0

Or, y(y+6)+7(y+6) = 0

Or,(y+6)(y+7)=0

Y = -6,-7

Hence, x > y

8. D) I. 14x² – 55x+50 = 0

Or,14x² -35x-20x+50 = 0

Or,7x(2x-5)-10(2x-5)=0

Or,(7x-10)(2x-5) = 0

 X=10/7,5/2

 II.2y²-13y + 20 = 0

Or,2y²-8y-5y+20 =0

Or,2y(y-4)-5(y-4) = 0

Or,(y-4)(2y-5) = 0

Or, y=4,5/2

Hence, x ≤ y

9. A) I. 7x + 8y = 5                            …..(i)

II.6x - 5y = 28                           …..(ii)

Now, eqn(i) × 5 + eqn (ii) × 8

35x + 40 y =25

48x – 40y = 224

Simplify above 2 eqns we get à83x = 249

X=3 and y=-2

Hence, x > y

10. B) I. 5x²-52x + 96 = 0

Or,5x²-40x-12x + 96 =0

Or,5x(x-8)-12(x-8)=0

Or,(x-8)(5x-12)=0

Or,x=8,12/5

II. 5y² + 3y -36=0

Or,5y² +15y -12y -36 =0

Or,5y(y + 3)-12(y+3) =0

Or,(5y-12)(y+3) = 0

Or,y=-3,12/5

Hence,x ≥ y