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# Quadratic Equations Questions for MAH-CET

Question 1: I. $x^{2}+3x-28=0$
II. $y^{2} -y-20=0$

a) x > y

b) x ≥ y

c) x < y

d) x ≤ y

e) x = y or the relation cannot be established.

Solution:

I.$x^{2} + 3x – 28 = 0$

=> $x^2 + 7x – 4x – 28 = 0$

=> $x (x + 7) – 4 (x + 7) = 0$

=> $(x – 4) (x + 7) = 0$

=> $x = 4 , -7$

II.$y^{2} – y – 20 = 0$

=> $y^2 – 5y + 4y – 20 = 0$

=> $y (y – 5) + 4 (y – 5) = 0$

=> $(y + 4) (y – 5) = 0$

=> $y = -4 , 5$

$\therefore$ No relation can be established.

Question 2: I. $6x^{2}+29x+35=0$
II. $3y^{2} +11y+10=0$

a) x > y

b) x ≥ y

c) x < y

d) x ≤ y

e) x = y or the relation cannot be established.

Solution:

I.$6x^{2} + 29x + 35 = 0$

=> $6x^2 + 15x + 14x + 35 = 0$

=> $3x (2x + 5) + 7 (2x + 5) = 0$

=> $(2x + 5) (3x + 7) = 0$

=> $x = \frac{-7}{3} , \frac{-5}{2}$

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

=> $3y^2 + 6y + 5y + 10 = 0$

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

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

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

Hence $x < y$

Question 3:  I. $2x^{2}+18x+40=0$
II. $2y^{2} +15y+27=0$

a) x > y

b) x ≥ y

c) x < y

d) x ≤ y

e) x = y or the relation cannot be established.

Solution:

I.$2x^{2} + 18x + 40 = 0$

=> $2x^2 + 8x + 10x + 40 = 0$

=> $2x (x + 4) + 10 (x + 4) = 0$

=> $(x + 4) (2x + 10) = 0$

=> $x = -4 , -5$

II.$2y^{2} + 15y + 27 = 0$

=> $2y^2 + 6y + 9y + 27 = 0$

=> $2y (y + 3) + 9 (y + 3) = 0$

=> $(y + 3) (2y + 9) = 0$

=> $y = -3 , \frac{-9}{2}$

$\therefore$ No relation can be established.

Question 4:  I.    $2x^{2}+15x+28=0$
II.    $4y^{2} +18y+14=0$

a) x > y

b) x ≥ y

c) x < y

d) x ≤ y

e) x = y or the relation cannot be established.

Solution:

I.$2x^{2} + 15x + 28 = 0$

=> $2x^2 + 8x + 7x + 28 = 0$

=> $2x (x + 4) + 7 (x + 4) = 0$

=> $(x + 4) (2x + 7) = 0$

=> $x = -4 , \frac{-7}{2}$

II.$4y^{2} + 18y + 14 = 0$

=> $4y^2 + 4y + 14y + 14 = 0$

=> $4y (y + 1) + 14 (y +1) = 0$

=> $(y + 1) (4y + 14) = 0$

=> $y = -1 , \frac{-7}{2}$

$\therefore x \leq y$

Question 5: I.  $2x^{2}-19x+45 =0$
II. $6y^{2}-48y+90=0$

a) x > y

b) x ≥ y

c) x < y

d) x ≤ y

e) x = y or the relation cannot be established.

Solution:

I.$2x^{2} – 19x + 45 = 0$

=> $2x^2 – 10x – 9x + 45 = 0$

=> $2x (x – 5) – 9 (x – 5) = 0$

=> $(x – 5) (2x – 9) = 0$

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

II.$6y^{2} – 48y + 90 = 0$

=> $y^2 – 8y + 15 = 0$

=> $y^2 – 3y – 5y + 15 = 0$

=> $y (y – 3) – 5 (y – 3) = 0$

=> $(y – 3) (y – 5) = 0$

=> $y = 3 , 5$

Hence no relation can be established.

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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 = $\sqrt {3136}$
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

Solution:

I. $x = \sqrt {3136}$

=> $x = 56$

II.${y^2} = 3136$

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

$\therefore x \geq y$

Question 7: 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

Solution:

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

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

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

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

=> $y = -6 , -5$

$\therefore x \geq y$

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

Solution:

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 $x < y$

Question 9: 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

Solution:

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

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

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

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

=> $y = -12 , -4$

$\therefore x \leq y$

Question 10: 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

Solution:

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

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

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

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

=> $y = -10 , -1$

$\therefore x > y$

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

Solution:

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

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

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

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

=> $y = -9 , -8$

$\therefore x \geq y$

Question 12: 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.

Solution:

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.$y^{2} – 26y + 168 = 0$

=> $y^2 – 12y – 14y + 168 = 0$

=> $y (y – 12) – 14 (y – 12) = 0$

=> $(y – 12) (y – 14) = 0$

=> $y = 12 , 14$

$\therefore x \leq y$

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.

Solution:

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.

Question 14: 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.

Solution:

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.$y^{2} – 23y + 132 = 0$

=> $y^2 – 11y – 12y + 132 = 0$

=> $y (y – 11) – 12 (y – 11) = 0$

=> $(y – 11) (y – 12) = 0$

=> $y = 11 , 12$

$\therefore x < y$

Question 15: 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.

Solution:

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

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

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

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

=> $y = -9 , -10$

$\therefore x > y$