# Quadratic Equation Questions for IBPS PO PDF

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## Quadratic Equation Questions for IBPS PO PDF

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Instructions

In the following questions two equations numbered I and II are given. You have to solve both the equations and
Give answer a: if x > y
Give answer b: if x ≥ y
Give answer c: if x < y
Give answer d: if x ≤ y
Give answer e: if x = y or the relationship cannot be established.

Question 1: I.  $x^{2}-3x-88=0$
II. $y^{2}+8y-48=0$

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

e) if x = y or the relationship cannot be established.

Question 2: I.  $5x^{2}+29x+20=0$
II. $25y^{2}+25y+6=0$

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

e) if x = y or the relationship cannot be established.

Question 3: I.  $2x^{2}-11x+12=0$
II. $2y^{2}-19y+44=0$

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

e) if x = y or the relationship cannot be established.

Question 4: I.  $3x^{2}+10x+8=0$
II. $3y^{2}+7y+4=0$

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

e) if x = y or the relationship cannot be established.

Question 5: I.  $2x^{2}+21x+10=0$
II. $3y^{2}+13y+14=0$

a) if x > y

b) if x ≥ y

c) if x < y

d) if x ≤ y

e) if x = y or the relationship cannot be established.

Instructions

In the following questions two equations numbered I and II are given. You have to solve both equations and

a. x ˃ y
b. x ≥ y
c. x ˂ y
d. x ≤ y
e. x = y or the relationship cannot be established

Question 6: I. ${x^2}$ – 7x + 10 = 0
II. ${y^2}$ + 11y + 10 = 0

a) x ˃ y

b) x ≥ y

c) x ˂ y

d) x ≤ y

e) x = y or the relationship cannot be established

Question 7: I. ${x^2}$ + 28x + 192 = 0
II. ${y^2}$ + 16y + 48 = 0

a) x ˃ y

b) x ≥ y

c) x ˂ y

d) x ≤ y

e) x = y or the relationship cannot be established

Question 8: I.2x – 3y = – 3.5
II. 3x + 2y = – 6.5

a) x ˃ y

b) x ≥ y

c) x ˂ y

d) x ≤ y

e) x = y or the relationship cannot be established

Question 9: I. ${x^2}$ + 8x + 15 = 0
II. ${y^2}$ + 11y + 30 = 0

a) x ˃ y

b) x ≥ y

c) x ˂ y

d) x ≤ y

e) x = y or the relationship cannot be established

Question 10: I. x = $\sqrt {3136}$
II.${y^2}$ = 3136

a) x ˃ y

b) x ≥ y

c) x ˂ y

d) x ≤ y

e) x = y or the relationship cannot be established

Instructions

In each of these questions two equations numbered I and II are given. You have to solve both the equations and –
Give answer a: if x < y
Give answer b: if x ≤ y
Give answer c: if x > y
Give answer d: if x ≥ y
Give answer e: if x = y or the relationship cannot be established.

Question 11: I.  $x^{2}+13x+42=0$
II. $y^{2} +19y+90=0$

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

e) if x = y or the relationship cannot be established.

Question 12: I.   $x^{2}-15x+56=0$
II. $y^{2} -23y+132=0$

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

e) if x = y or the relationship cannot be established.

Question 13: I. $x^{2}+7x+12=0$
II. $y^{2} +6y+8=0$

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

e) if x = y or the relationship cannot be established.

Question 14: I. $x^{2}-22x+120=0$
II. $y^{2} -26y+168=0$

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

e) if x = y or the relationship cannot be established.

Question 15: I.$x^{2}+12x+32=0$
II. $y^{2} +17y+72=0$

a) if x < y

b) if x ≤ y

c) if x > y

d) if x ≥ y

e) if x = y or the relationship cannot be established.

I.$x^{2} – 3x – 88 = 0$

=> $x^2 + 8x – 11x – 88 = 0$

=> $x (x + 8) – 11 (x + 8) = 0$

=> $(x + 8) (x – 11) = 0$

=> $x = -8 , 11$

II.$y^{2} + 8y – 48 = 0$

=> $y^2 + 12y – 4y – 48 = 0$

=> $y (y + 12) – 4 (y + 12) = 0$

=> $(y + 12) (y – 4) = 0$

=> $y = -12 , 4$

$\therefore$ No relation can be established.

I.$5x^{2} + 29x + 20 = 0$

=> $5x^2 + 25x + 4x + 20 = 0$

=> $5x (x + 5) + 4 (x + 5) = 0$

=> $(x + 5) (5x + 4) = 0$

=> $x = -5 , \frac{-4}{5}$

II.$25y^{2} + 25y + 6 = 0$

=> $25y^2 + 10y + 15y + 6 = 0$

=> $5y (5y + 2) + 3 (5y + 2) = 0$

=> $(5y + 3) (5y + 2) = 0$

=> $y = \frac{-3}{5} , \frac{-2}{5}$

Therefore $x < y$

I.$2x^{2} – 11x + 12 = 0$

=> $2x^2 – 8x – 3x + 12 = 0$

=> $2x (x – 4) – 3 (x – 4) = 0$

=> $(x – 4) (2x – 3) = 0$

=> $x = 4 , \frac{3}{2}$

II.$2y^{2} – 19y + 44 = 0$

=> $2y^2 – 8y – 11y + 44 = 0$

=> $2y (y – 4) – 11 (y – 4) = 0$

=> $(y – 4) (2y – 11) = 0$

=> $y = 4 , \frac{11}{2}$

$\therefore x \leq y$

I.$3x^{2} + 10x + 8 = 0$

=> $3x^2 + 6x + 4x + 8 = 0$

=> $3x (x + 2) + 4 (x + 2) = 0$

=> $(x + 2) (3x + 4) = 0$

=> $x = -2 , \frac{-4}{3}$

II.$3y^{2} + 7y + 4 = 0$

=> $3y^2 + 3y + 4y + 4 = 0$

=> $3y (y + 1) + 4 (y + 1) = 0$

=> $(y + 1) (3y + 4) = 0$

=> $y = -1 , \frac{-4}{3}$

$\therefore x \leq y$

I.$2x^{2} + 21x + 10 = 0$

=> $2x^2 + x + 20x + 10 = 0$

=> $x (2x + 1) + 10 (2x + 1) = 0$

=> $(x + 10) (2x + 1) = 0$

=> $x = -10 , \frac{-1}{2}$

II.$3y^{2} + 13y + 14 = 0$

=> $3y^2 + 6y + 7y + 14 = 0$

=> $3y (y + 2) + 7 (y + 2) = 0$

=> $(y + 2) (3y + 7) = 0$

=> $y = -2 , \frac{-7}{3}$

$\therefore$ No relation can be established.

I.$x^{2} – 7x + 10 = 0$

=> $x^2 – 5x – 2x + 10 = 0$

=> $x (x – 5) – 2 (x – 5) = 0$

=> $(x – 5) (x – 2) = 0$

=> $x = 5 , 2$

II.$y^{2} + 11y + 10 = 0$

=> $y^2 + 10y + y + 10 = 0$

=> $y (y + 10) + 1 (y + 10) = 0$

=> $(y + 10) (y + 1) = 0$

=> $y = -10 , -1$

$\therefore x > y$

I.$x^{2} + 28x + 192 = 0$

=> $x^2 + 16x + 12x + 192 = 0$

=> $x (x + 16) + 12 (x + 16) = 0$

=> $(x + 16) (x + 12) = 0$

=> $x = -16 , -12$

II.$y^{2} + 16y + 48 = 0$

=> $y^2 + 12y + 4y + 48 = 0$

=> $y (y + 12) + 4 (y + 12) = 0$

=> $(y + 12) (y + 4) = 0$

=> $y = -12 , -4$

$\therefore x \leq y$

I : $2x – 3y = -3.5$

II : $3x + 2y = -6.5$

Multiplying eqn(I) by 2 and eqn(II) by 3, and then adding both equations, we get :

=> $(4x + 9x) + (-6y + 6y) = (-7 -19.5)$

=> $13x = -26.5$

=> $x = \frac{-26.5}{13} \approx -2$

=> $y = \frac{3x + 6.5}{2} = 0.25$

Hence $x < y$

I.$x^{2} + 8x + 15 = 0$

=> $x^2 + 5x + 3x + 15 = 0$

=> $x (x + 5) + 3 (x + 5) = 0$

=> $(x + 5) (x + 3) = 0$

=> $x = -5 , -3$

II.$y^{2} + 11y + 30 = 0$

=> $y^2 + 5y + 6y + 30 = 0$

=> $y (y + 5) + 6 (y + 5) = 0$

=> $(y + 6) (y + 5) = 0$

=> $y = -6 , -5$

$\therefore x \geq y$

I. $x = \sqrt {3136}$

=> $x = 56$

II.${y^2} = 3136$

=> $y = \sqrt{3136} = \pm 56$

$\therefore x \geq y$

I.$x^{2} + 13x + 42 = 0$

=> $x^2 + 7x + 6x + 42 = 0$

=> $x (x + 7) + 6 (x + 7) = 0$

=> $(x + 7) (x + 6) = 0$

=> $x = -7 , -6$

II.$y^{2} + 19y + 90 = 0$

=> $y^2 + 9y + 10y + 90 = 0$

=> $y (y + 9) + 10 (y + 9) = 0$

=> $(y + 9) (y + 10) = 0$

=> $y = -9 , -10$

$\therefore x > y$

I.$x^{2} – 15x + 56 = 0$

=> $x^2 – 8x – 7x + 56 = 0$

=> $x (x – 8) – 7 (x – 8) = 0$

=> $(x – 8) (x – 7) = 0$

=> $x = 8 , 7$

II.$y^{2} – 23y + 132 = 0$

=> $y^2 – 11y – 12y + 132 = 0$

=> $y (y – 11) – 12 (y – 11) = 0$

=> $(y – 11) (y – 12) = 0$

=> $y = 11 , 12$

$\therefore x < y$

I.$x^{2} + 7x + 12 = 0$

=> $x^2 + 3x + 4x + 12 = 0$

=> $x (x + 3) + 4 (x + 3) = 0$

=> $(x + 3) (x + 4) = 0$

=> $x = -3 , -4$

II.$y^{2} + 6y + 8 = 0$

=> $y^2 + 4y + 2y + 8 = 0$

=> $y (y + 4) + 2 (y + 4) = 0$

=> $(y + 4) (y + 2) = 0$

=> $y = -4 , -2$

Because $-2 > -4$ and $-3 > -4$

Therefore, no relation can be established.

I.$x^{2} – 22x + 120 = 0$

=> $x^2 – 10x – 12x + 120 = 0$

=> $x (x – 10) – 12 (x – 10) = 0$

=> $(x – 10) (x – 12) = 0$

=> $x = 10 , 12$

II.$y^{2} – 26y + 168 = 0$

=> $y^2 – 12y – 14y + 168 = 0$

=> $y (y – 12) – 14 (y – 12) = 0$

=> $(y – 12) (y – 14) = 0$

=> $y = 12 , 14$

$\therefore x \leq y$

I.$x^{2} + 12x + 32 = 0$

=> $x^2 + 8x + 4x + 32 = 0$

=> $x (x + 8) + 4 (x + 8) = 0$

=> $(x + 8) (x + 4) = 0$

=> $x = -8 , -4$

II.$y^{2} + 17y + 72 = 0$

=> $y^2 + 9y + 8y + 72 = 0$

=> $y (y + 9) + 8 (y + 9) = 0$

=> $(y + 9) (y + 8) = 0$

=> $y = -9 , -8$

$\therefore x \geq y$

We hope this Quadratic Equation Questions for IBPS PO  will be highly useful for your preparation.