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# Quadratic Equation Questions For IBPS Clerk PDF

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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
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: I.$2a^{2}+a-1=0$
II.$12b^{2}-17b+6=0$

a) $a<b$

b) $a\leq b$

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

d) $a>b$

e) $a\geq b$

Question 2: I.$a^{2}-5a+6=0$
II. $2b^{2}-13b+21=0$

a) $a<b$

b) $a\leq b$

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

d) $a>b$

e) $a\geq b$

Question 3: I.$a^{2}+5a+6=0$
II.$b^{2}+7b+12=0$

a) $a<b$

b) $a \leq b$

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

d) $a>b$

e) $a \geq b$

Question 4: I.$16a^{2}=1$
II.$3b^{2}+7b+2=0$

a) $a<b$

b) $a\leq b$

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

d) $a>b$

e) $a\geq b$

Question 5: I.$a^{2}+2a+1=0$
II.$b^{2}=\pm4$

a) $a<b$

b) $a \leq b$

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

d) $a>b$

e) $a \geq b$

Instructions

In each of the following question two equations are given you have to solve them and
</q
Give answer (c)if $p\leq$q
Give answer(d)if $p\geq$q

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

a) if p<q

b) if p>q

c) if $p\leq$q

d) if $p\geq$q

e) if p=q

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

a) if $p < q$

b) if $p > q$

c) if $p\leq q$

d) if $p\geq q$

e) if $p = q$

Question 8: I.$p^{2}+12p+35=0$
II.$2q^{2}+22q+56=0$

a) if p < q

b) if p>q

c) if $p \leq q$

d) if $p\geq q$

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

Question 9: I.$p^{2}-8p+15=0$
II.$q^{2}-5q=-6$

a) if p < q

b) if p>q

c) if $p\leq q$

d) if $p\geq q$

e) if p=q

Question 10: I.$2p^{2}+20p+50=0$
II.$q^{2}=25$

a) if p<q

b) if p>q

c) if $p \leq q$

d) if $p\geq q$

e) if p = q

Instructions

For the two given equations I and II—-

Question 11: I. $6p^{2}+5p+1=0$
II. $20q^{2}+9q=-1$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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: I. $2x^{2}+19x+45=0$
II. $2y^{2}+11y+12=0$

a) x = y

b) x> y

c) x < y

d) relationship between xand y cannot be determined

e) x + y

Question 17: I. $3x^{2}-13x+12=0$
II. $2y^{2}-15y+28=0$

a) x> y

b) x= y

c) x < y

d) relationship between x and y cannot be determined

e) x≤ y

Question 18: I. $x^{2}=16$
II. $2y^{2}-17y+36=0$

a) x > y

b) x > y

c) x < y

d) relationship between x and y cannot be determined

e) $x \leq y$

Question 19: I. $6x^{2}+19x+15=0$
II. $3y^{2}+11y+10=0$

a) x = y

b) x > y

c) x < y

d) $x \geq y$

e) $x \leq y$

Question 20: I. $2x^{2}-11x+15=0$
II. $2y^{2}-11y+14=0$

a) x > y

b) x> y

c) x < y

d) relationship between x and y cannot be determined

e) x ≤ y

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

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.

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

$16a^{2}=1$
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.

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.

$p^2-7p+12 = 0$
$(p-3)(p-4) = 0$
$p = 3, 4$

$q^2-3q+2 = 0$
$(q-1)(q-2) = 0$
$q = 1, 2$

$\therefore p > q$

$12p^2-7p+1 = 0$
$(4p-1)(3p-1) = 0$
$p = \frac{1}{3}, \frac{1}{4}$

$6q^2-7q+2 = 0$
$(2q-1)(3q-2) = 0$
$q = \frac{1}{2}, \frac{2}{3}$

$\therefore p < q$

$p^2+12p+35 = 0$
$(p+5)(p+7) = 0$
$p = -5, -7$

$2q^2+22q+56 = 0$
$q^2+11q+28 = 0$
$(q+4)(q+7) = 0$
$q = -4, -7$

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.

$p^2-8p+15 = 0$
$(p-3)(p-5) = 0$
$p = 3, 5$

$q^2-5q+6 = 0$
$(q-2)(q-3) = 0$
$q = 2, 3$

$p\geq q$

$2p^2+20p+50 = 0$
$p^2+10p+25 = 0$
$(p+5)^2 = 0$
$p = -5$

$q^2 = 25$
$q = 5, -5$

$p\leq q$

$6p^2+5p+1 = 0$
$(2p+1)(3p+1) = 0$
$p = -\frac{1}{2}, -\frac{1}{3}$

$20q^2+9q+1 = 0$
$(4q+1)(5q+1) = 0$
$q = -\frac{1}{4}, -\frac{1}{5}$

$p < q$

$3p^2+2p-1 = 0$
$(3p-1)(p+1) = 0$
$p = -1, \frac{1}{3}$

$2q^2+7q+6 = 0$
$(2q+3)(q+2) = 0$
$q = -2, -\frac{3}{2}$

p > q

$3p^2+15p+18 = 0$
$p^2+5p+6 = 0$
$(p+2)(p+3) = 0$
$p = -3, -2$

$q^2+7q+12 = 0$
$(q+4)(q+3) = 0$
$q = -4, -3$

$p\geq q$

$p = \frac{\sqrt{4}}{\sqrt{9}}$
$p = \frac{2}{3}$

$9q^2-12q+4 = 0$
${(3q-2)}^2 = 0$
$q = \frac{2}{3}$

p = q

$p^2+13p+42 = 0$
$(p+6)(p+7) = 0$
$p = -6, -7$

$q^2 = 36$
$q = -6, 6$

$p\leq q$

$2x^2+19x+45 = 0$
$(2x+9)(x+5) = 0$
$x = -5, -\frac{9}{2}$

$2y^2+11y+12 = 0$
$(2y+3)(y+4) = 0$
$y = -4, -\frac{3}{2}$

x < y

$3x^2-13x+12 = 0$
$(3x-4)(x-3) = 0$
$x = \frac{4}{3}, 3$

$2y^2-15y+28 = 0$
$(2y-7)(y-4) = 0$
$y = \frac{7}{2}, 4$

x < y

$x^2 = 16$
$x = 4, -4$

$2y^2-17y+36 = 0$
$(2y-9)(y-4) = 0$
$y = \frac{9}{2}, 4$

$x \leq y$

$6x^2+19x+15 = 0$
$(3x+5)(2x+3) = 0$
$x = -\frac{5}{3}, -\frac{3}{2}$

$3y^2+11y+10 = 0$
$(3y+5)(y+2) = 0$
$y = -\frac{5}{3}, -2$

$x\geq y$

$2x^2-11x+15 = 0$
$(2x-5)(x-3) = 0$
$x = 3, \frac{5}{2}$
$2y^2-11y+14 = 0$
$(2y-7)(y-2) = 0$
$y = 2, \frac{7}{2}$