Download important Quadratic Equation Questions PDF based on previously asked questions in IBPS Clerk and other MBA Exams. Practice Quadratic Equation Question and Answers for IBPS Clerk Exam.

Download Quadratic Equation Questions For IBPS Clerk PDF

4 Free IBPS Clerk Online Mock Tests (with solutions)

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Go to Free Banking Study Material (15,000 Solved Questions)

**Instructions**

In each of these question two equations I & II with variables a & b are given You have to solve both the equations to find the values of a & b

Mark answer if

a) a<b

</b

<b

</b

<b

b) $a\leq b$

c) relationship between a & b cannot be established

d) a>b

e) $a\geq b$

**Question 1:Â **I.$2a^{2}+a-1=0$

II.$12b^{2}-17b+6=0$

a)Â $a<b$

b)Â $a\leq b$

c)Â Relationship between $a$ & $b$ cannot be established

d)Â $a>b$

e)Â $a\geq b$

**Question 2:Â **I.$a^{2}-5a+6=0$

II. $2b^{2}-13b+21=0$

a)Â $a<b$

b)Â $a\leq b$

c)Â Relationship between $a$ & $b$ cannot be established

d)Â $a>b$

e)Â $a\geq b$

**Question 3:Â **I.$a^{2}+5a+6=0$

II.$b^{2}+7b+12=0$

a)Â $a<b$

b)Â $a \leq b$

c)Â Relationship between $a$ & $b$ cannot be established

d)Â $a>b$

e)Â $a \geq b$

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**Question 4:Â **I.$16a^{2}=1$

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

a)Â $a<b$

b)Â $a\leq b$

c)Â Relationship between $a$ & $b$ cannot be established

d)Â $a>b$

e)Â $a\geq b$

**Question 5:Â **I.$a^{2}+2a+1=0$

II.$b^{2}=\pm4$

a)Â $a<b$

b)Â $a \leq b$

c)Â Relationship between $a$ & $b$ cannot be established

d)Â $a>b$

e)Â $a \geq b$

**Instructions**

In each of the following question two equations are given you have to solve them and

Give answer (a)if p<q

</q

Give answer (b)if p>q

Give answer (c)if $p\leq$q

Give answer(d)if $p\geq$q

Give answer (e)if p=q

**Question 6:Â **I.$p^{2}-7p=-12$

II.$q^{2}-3q+2=0$

a)Â if p<q

b)Â if p>q

c)Â if $p\leq$q

d)Â if $p\geq$q

e)Â if p=q

**Question 7:Â **I. $12p^{2}-7p=-1$

II. $6q^{2}-7q+2=0$

a)Â if $p < q$

b)Â if $p > q$

c)Â if $p\leq q$

d)Â if $p\geq q$

e)Â if $p = q$

**Question 8:Â **I.$p^{2}+12p+35=0$

II.$2q^{2}+22q+56=0$

a)Â if p < q

b)Â if p>q

c)Â if $p \leq q$

d)Â if $p\geq q$

e)Â if p=q or no relationship can be established

**Question 9:Â **I.$p^{2}-8p+15=0$

II.$q^{2}-5q=-6$

a)Â if p < q

b)Â if p>q

c)Â if $p\leq q$

d)Â if $p\geq q$

e)Â if p=q

IBPS Clerk Important Questions PDF

**Question 10:Â **I.$2p^{2}+20p+50=0$

II.$q^{2}=25$

a)Â if p<q

b)Â if p>q

c)Â if $p \leq q$

d)Â if $p\geq q $

e)Â if p = q

**Instructions**

For the two given equations I and II—-

**Question 11:Â **I. $6p^{2}+5p+1=0$

II. $20q^{2}+9q=-1$

a)Â Give answer (A) if p is greater than q.

b)Â Give answer (B) if p is smaller than q.

c)Â Give answer (C) if p is equal to q.

d)Â Give answer (D) if p is either equal to or greater than q.

e)Â Give answer (E) if p is either equal to or smaller than q.

**Question 12:Â **I. $3p^{2}+2p-1=0$ II. $2q^{2}+7q+6=0$

a)Â Give answer (A) if p is greater than q.

b)Â Give answer (B) if p is smaller than q.

c)Â Give answer (C) if p is equal to q.

d)Â Give answer (D) if p is either equal to or greater than q.

e)Â Give answer (E) if p is either equal to or smaller than q.

**Question 13:Â **I. $3p^2+15p=-18$ II. $q^2+7q+12=0$

a)Â Give answer (A) if p is greater than q.

b)Â Give answer (B) if p is smaller than q.

c)Â Give answer (C) if p is equal to q.

d)Â Give answer (D) if p is either equal to or greater than q.

e)Â Give answer (E) if p is either equal to or smaller than q.

**Question 14:Â **I. $p=\frac{\sqrt{4}}{\sqrt{9}}$ II. $9q^{2}-12q+4=0$

a)Â Give answer (A) if p is greater than q.

b)Â Give answer (B) if p is smaller than q.

c)Â Give answer (C) if p is equal to q.

d)Â Give answer (D) if p is either equal to or greater than q.

e)Â Give answer (E) if p is either equal to or smaller than q.

**Question 15:Â **I. $p^{2}+13p+42=0$ II. $q^{2}=36$

a)Â Give answer (A) if p is greater than q.

b)Â Give answer (B) if p is smaller than q.

c)Â Give answer (C) if p is equal to q.

d)Â Give answer (D) if p is either equal to or greater than q.

e)Â Give answer (E) if p is either equal to or smaller than q.

Free Banking Study Material (15,000 Solved Questions)

**Instructions**

In these questions, two equations numbered I and II are given. You have to solve both the equations and select the appropriate option.

**Question 16:Â **I. $2x^{2}+19x+45=0$

II. $2y^{2}+11y+12=0$

a)Â x = y

b)Â x> y

c)Â x < y

d)Â relationship between xand y cannot be determined

e)Â x + y

**Question 17:Â **I. $3x^{2}-13x+12=0$

II. $2y^{2}-15y+28=0$

a)Â x> y

b)Â x= y

c)Â x < y

d)Â relationship between x and y cannot be determined

e)Â xâ‰¤ y

**Question 18:Â **I. $x^{2}=16$

II. $2y^{2}-17y+36=0$

a)Â x > y

b)Â x > y

c)Â x < y

d)Â relationship between x and y cannot be determined

e)Â $x \leq y$

Daily Free Banking Online Tests

**Question 19:Â **I. $6x^{2}+19x+15=0$

II. $3y^{2}+11y+10=0$

a)Â x = y

b)Â x > y

c)Â x < y

d)Â $x \geq y$

e)Â $x \leq y$

**Question 20:Â **I. $2x^{2}-11x+15=0$

II. $2y^{2}-11y+14=0$

a)Â x > y

b)Â x> y

c)Â x < y

d)Â relationship between x and y cannot be determined

e)Â x â‰¤ y

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**Answers & Solutions:**

**1)Â AnswerÂ (A)**

$2a^{2}+a-1=0$

We get the factor as:

a=-1, a=0.5

$12b^{2}-17b+6=0$

Solving, we get the factor as,

b= 1.5, b= .75

Hence, b>a

Option A is correct option.

**2)Â AnswerÂ (B)**

Soving the quadratic equations we get,

$a^{2}-5a+6=0$

i.e (a-2)(a-3)=0

i.e a=2, a=3

$2b^{2}-13b+21=0$

i.e (b-3.5)(b-3)=0

i.e b= 3.5 and b=3

Hence, we can deduce that $a\leq b$

Therefore, option B is correct.

**3)Â AnswerÂ (E)**

$a^{2}+5a+6=0$

i.e (a+2)(a+3)=0

i.e a=-2, a=-3

.$b^{2}+7b+12=0$

i.e (b+4)(b+3)=0

i.e b=-4, b=-3

Hence, we can deduce that $a \geq b$.

Therefore, option E is correct.

**4)Â AnswerÂ (D)**

$16a^{2}=1$

Solving we get, a=-.25, a=+.25

$3b^{2}+7b+2=0$

Solving we get, b= -2. b = -1/3

Hence, a>b. Option D is correct.

**5)Â AnswerÂ (C)**

We can easily solve equation I to get a = -1

But we cannot solve $b^{2}=\pm4$. Square root of negative number is not a real number.

Hence, we cannot find a value of b. Therefore, we cannot establish a relationship between a and b.

**6)Â AnswerÂ (B)**

$p^2-7p+12 = 0$

$(p-3)(p-4) = 0$

$p = 3, 4$

$q^2-3q+2 = 0$

$(q-1)(q-2) = 0$

$q = 1, 2$

$\therefore p > q$

**7)Â AnswerÂ (A)**

$12p^2-7p+1 = 0$

$(4p-1)(3p-1) = 0$

$p = \frac{1}{3}, \frac{1}{4}$

$6q^2-7q+2 = 0$

$(2q-1)(3q-2) = 0$

$q = \frac{1}{2}, \frac{2}{3}$

$\therefore p < q$

**8)Â AnswerÂ (E)**

$p^2+12p+35 = 0$

$(p+5)(p+7) = 0$

$p = -5, -7$

$2q^2+22q+56 = 0$

$q^2+11q+28 = 0$

$(q+4)(q+7) = 0$

$q = -4, -7$

As we can see $p$ can be greater than, less than or equal to $q$. No relationship can be established between $p$ and $q$ and hence, option E is the right answer.

**9)Â AnswerÂ (D)**

$p^2-8p+15 = 0$

$(p-3)(p-5) = 0$

$p = 3, 5$

$q^2-5q+6 = 0$

$(q-2)(q-3) = 0$

$q = 2, 3$

$p\geq q$

**10)Â AnswerÂ (C)**

$2p^2+20p+50 = 0$

$p^2+10p+25 = 0$

$(p+5)^2 = 0$

$p = -5$

$q^2 = 25$

$q = 5, -5$

$p\leq q$

**11)Â AnswerÂ (B)**

$6p^2+5p+1 = 0$

$(2p+1)(3p+1) = 0$

$p = -\frac{1}{2}, -\frac{1}{3}$

$20q^2+9q+1 = 0$

$(4q+1)(5q+1) = 0$

$q = -\frac{1}{4}, -\frac{1}{5}$

$p < q$

**12)Â AnswerÂ (A)**

$3p^2+2p-1 = 0$

$(3p-1)(p+1) = 0$

$p = -1, \frac{1}{3}$

$2q^2+7q+6 = 0$

$(2q+3)(q+2) = 0$

$q = -2, -\frac{3}{2}$

p > q

**13)Â AnswerÂ (D)**

$3p^2+15p+18 = 0$

$p^2+5p+6 = 0$

$(p+2)(p+3) = 0$

$p = -3, -2$

$q^2+7q+12 = 0$

$(q+4)(q+3) = 0$

$q = -4, -3$

$p\geq q$

**14)Â AnswerÂ (C)**

$p = \frac{\sqrt{4}}{\sqrt{9}}$

$p = \frac{2}{3}$

$9q^2-12q+4 = 0$

${(3q-2)}^2 = 0$

$q = \frac{2}{3}$

p = q

**15)Â AnswerÂ (E)**

$p^2+13p+42 = 0$

$(p+6)(p+7) = 0$

$p = -6, -7$

$q^2 = 36$

$q = -6, 6$

$p\leq q$

**16)Â AnswerÂ (C)**

$2x^2+19x+45 = 0$

$(2x+9)(x+5) = 0$

$x = -5, -\frac{9}{2}$

$2y^2+11y+12 = 0$

$(2y+3)(y+4) = 0$

$y = -4, -\frac{3}{2}$

x < y

**17)Â AnswerÂ (C)**

$3x^2-13x+12 = 0$

$(3x-4)(x-3) = 0$

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

$2y^2-15y+28 = 0$

$(2y-7)(y-4) = 0$

$y = \frac{7}{2}, 4$

x < y

**18)Â AnswerÂ (E)**

$x^2 = 16$

$x = 4, -4$

$2y^2-17y+36 = 0$

$(2y-9)(y-4) = 0$

$y = \frac{9}{2}, 4$

$x \leq y$

**19)Â AnswerÂ (D)**

$6x^2+19x+15 = 0$

$(3x+5)(2x+3) = 0$

$x = -\frac{5}{3}, -\frac{3}{2}$

$3y^2+11y+10 = 0$

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

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

$x\geq y$

**20)Â AnswerÂ (D)**

$2x^2-11x+15 = 0$

$(2x-5)(x-3) = 0$

$x = 3, \frac{5}{2}$

$2y^2-11y+14 = 0$

$(2y-7)(y-2) = 0$

$y = 2, \frac{7}{2}$

relationship between x and y cannot be established