Here you can download a free Quadratic Equations questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the Quadratic Equations of answers for the given questions. These questions will help you to make practice and solve the Quadratic Equations questions in the MAH MBA CET exam. Utilize this best PDF practice set<\/strong> which is included answers in detail. Click on the below link to download the Quadratic Equations MCQ<\/strong> PDF for MBA-CET 2022 for free.<\/p>\n Download Quadratic Equations Questions for MAH-CET<\/a><\/p>\n Enroll to MAH-CET 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>I.\u00a0$x^{2}+3x-28=0$ a)\u00a0x > y<\/p>\n b)\u00a0x \u2265 y<\/p>\n c)\u00a0x < y<\/p>\n d)\u00a0x \u2264 y<\/p>\n e)\u00a0x = y or the relation cannot be established.<\/p>\n 1)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$x^{2} + 3x – 28 = 0$<\/p>\n => $x^2 + 7x – 4x – 28 = 0$<\/p>\n => $x (x + 7) – 4 (x + 7) = 0$<\/p>\n => $(x – 4) (x + 7) = 0$<\/p>\n => $x = 4 , -7$<\/p>\n II.$y^{2} – y – 20 = 0$<\/p>\n => $y^2 – 5y + 4y – 20 = 0$<\/p>\n => $y (y – 5) + 4 (y – 5) = 0$<\/p>\n => $(y + 4) (y – 5) = 0$<\/p>\n => $y = -4 , 5$<\/p>\n $\\therefore$ No relation can be established.<\/p>\n Question 2:\u00a0<\/b>I.\u00a0$6x^{2}+29x+35=0$ a)\u00a0x > y<\/p>\n b)\u00a0x \u2265 y<\/p>\n c)\u00a0x < y<\/p>\n d)\u00a0x \u2264 y<\/p>\n e)\u00a0x = y or the relation cannot be established.<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$6x^{2} + 29x + 35 = 0$<\/p>\n => $6x^2 + 15x + 14x + 35 = 0$<\/p>\n => $3x (2x + 5) + 7 (2x + 5) = 0$<\/p>\n => $(2x + 5) (3x + 7) = 0$<\/p>\n => $x = \\frac{-7}{3} , \\frac{-5}{2}$<\/p>\n II.$3y^{2} + 11y + 10 = 0$<\/p>\n => $3y^2 + 6y + 5y + 10 = 0$<\/p>\n => $3y (y + 2) + 5 (y + 2) = 0$<\/p>\n => $(y + 2) (3y + 5) = 0$<\/p>\n => $y = -2 , \\frac{-5}{3}$<\/p>\n Hence $x < y$<\/p>\n Question 3:\u00a0<\/b>\u00a0I.\u00a0$2x^{2}+18x+40=0$ a)\u00a0x > y<\/p>\n b)\u00a0x \u2265 y<\/p>\n c)\u00a0x < y<\/p>\n d)\u00a0x \u2264 y<\/p>\n e)\u00a0x = y or the relation cannot be established.<\/p>\n 3)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$2x^{2} + 18x + 40 = 0$<\/p>\n => $2x^2 + 8x + 10x + 40 = 0$<\/p>\n => $2x (x + 4) + 10 (x + 4) = 0$<\/p>\n => $(x + 4) (2x + 10) = 0$<\/p>\n => $x = -4 , -5$<\/p>\n II.$2y^{2} + 15y + 27 = 0$<\/p>\n => $2y^2 + 6y + 9y + 27 = 0$<\/p>\n => $2y (y + 3) + 9 (y + 3) = 0$<\/p>\n => $(y + 3) (2y + 9) = 0$<\/p>\n => $y = -3 , \\frac{-9}{2}$<\/p>\n $\\therefore$ No relation can be established.<\/p>\n Question 4:\u00a0<\/b>\u00a0I.\u00a0 \u00a0\u00a0$ 2x^{2}+15x+28=0 $ a)\u00a0x > y<\/p>\n b)\u00a0x \u2265 y<\/p>\n c)\u00a0x < y<\/p>\n d)\u00a0x \u2264 y<\/p>\n e)\u00a0x = y or the relation cannot be established.<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$2x^{2} + 15x + 28 = 0$<\/p>\n => $2x^2 + 8x + 7x + 28 = 0$<\/p>\n => $2x (x + 4) + 7 (x + 4) = 0$<\/p>\n => $(x + 4) (2x + 7) = 0$<\/p>\n => $x = -4 , \\frac{-7}{2}$<\/p>\n II.$4y^{2} + 18y + 14 = 0$<\/p>\n => $4y^2 + 4y + 14y + 14 = 0$<\/p>\n => $4y (y + 1) + 14 (y +1) = 0$<\/p>\n => $(y + 1) (4y + 14) = 0$<\/p>\n => $y = -1 , \\frac{-7}{2}$<\/p>\n $\\therefore x \\leq y$<\/p>\n Question 5:\u00a0<\/b>I.\u00a0 $ 2x^{2}-19x+45 =0 $ a)\u00a0x > y<\/p>\n b)\u00a0x \u2265 y<\/p>\n c)\u00a0x < y<\/p>\n d)\u00a0x \u2264 y<\/p>\n e)\u00a0x = y or the relation cannot be established.<\/p>\n 5)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$2x^{2} – 19x + 45 = 0$<\/p>\n => $2x^2 – 10x – 9x + 45 = 0$<\/p>\n => $2x (x – 5) – 9 (x – 5) = 0$<\/p>\n => $(x – 5) (2x – 9) = 0$<\/p>\n => $x = 5 , \\frac{9}{2}$<\/p>\n II.$6y^{2} – 48y + 90 = 0$<\/p>\n => $y^2 – 8y + 15 = 0$<\/p>\n => $y^2 – 3y – 5y + 15 = 0$<\/p>\n => $y (y – 3) – 5 (y – 3) = 0$<\/p>\n => $(y – 3) (y – 5) = 0$<\/p>\n => $y = 3 , 5$<\/p>\n Hence no relation can be established.<\/p>\n Take Free MAH-CET mock tests here<\/a><\/strong><\/span><\/p>\n Enrol to 5 MAH CET Latest Mocks For Just Rs. 299<\/a><\/strong><\/span><\/p>\n Instructions<\/b><\/p>\n In the following questions two equations numbered I and II are given. You have to solve both equations and a. x \u02c3 y Question 6:\u00a0<\/b>I. x = $\\sqrt {3136} $ a)\u00a0x \u02c3 y<\/p>\n b)\u00a0x \u2265 y<\/p>\n c)\u00a0x \u02c2 y<\/p>\n d)\u00a0x \u2264 y<\/p>\n e)\u00a0x = y or the relationship cannot be established<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n I. $x = \\sqrt {3136} $<\/p>\n => $x = 56$<\/p>\n II.$ {y^2} = 3136$<\/p>\n => $y = \\sqrt{3136} = \\pm 56$<\/p>\n $\\therefore x \\geq y$<\/p>\n Question 7:\u00a0<\/b>I. ${x^2}$ + 8x + 15 = 0 a)\u00a0x \u02c3 y<\/p>\n b)\u00a0x \u2265 y<\/p>\n c)\u00a0x \u02c2 y<\/p>\n d)\u00a0x \u2264 y<\/p>\n e)\u00a0x = y or the relationship cannot be established<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$x^{2} + 8x + 15 = 0$<\/p>\n => $x^2 + 5x + 3x + 15 = 0$<\/p>\n => $x (x + 5) + 3 (x + 5) = 0$<\/p>\n => $(x + 5) (x + 3) = 0$<\/p>\n => $x = -5 , -3$<\/p>\n II.$y^{2} + 11y + 30 = 0$<\/p>\n => $y^2 + 5y + 6y + 30 = 0$<\/p>\n => $y (y + 5) + 6 (y + 5) = 0$<\/p>\n => $(y + 6) (y + 5) = 0$<\/p>\n => $y = -6 , -5$<\/p>\n $\\therefore x \\geq y$<\/p>\n Question 8:\u00a0<\/b>I.2x – 3y = – 3.5 a)\u00a0x \u02c3 y<\/p>\n b)\u00a0x \u2265 y<\/p>\n c)\u00a0x \u02c2 y<\/p>\n d)\u00a0x \u2264 y<\/p>\n e)\u00a0x = y or the relationship cannot be established<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n I : $2x – 3y = -3.5$<\/p>\n II : $3x + 2y = -6.5$<\/p>\n Multiplying eqn(I) by 2 and eqn(II) by 3, and then adding both equations, we get\u00a0:<\/p>\n => $(4x + 9x) + (-6y + 6y) = (-7 -19.5)$<\/p>\n => $13x = -26.5$<\/p>\n => $x = \\frac{-26.5}{13} \\approx -2$<\/p>\n => $y = \\frac{3x + 6.5}{2} = 0.25$<\/p>\n Hence $x < y$<\/p>\n Question 9:\u00a0<\/b>I. ${x^2}$ + 28x + 192 = 0 a)\u00a0x \u02c3 y<\/p>\n b)\u00a0x \u2265 y<\/p>\n c)\u00a0x \u02c2 y<\/p>\n d)\u00a0x \u2264 y<\/p>\n e)\u00a0x = y or the relationship cannot be established<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$x^{2} + 28x + 192 = 0$<\/p>\n => $x^2 + 16x + 12x + 192 = 0$<\/p>\n => $x (x + 16) + 12 (x + 16) = 0$<\/p>\n => $(x + 16) (x + 12) = 0$<\/p>\n => $x = -16 , -12$<\/p>\n II.$y^{2} + 16y + 48 = 0$<\/p>\n => $y^2 + 12y + 4y + 48 = 0$<\/p>\n => $y (y + 12) + 4 (y + 12) = 0$<\/p>\n => $(y + 12) (y + 4) = 0$<\/p>\n => $y = -12 , -4$<\/p>\n $\\therefore x \\leq y$<\/p>\n Question 10:\u00a0<\/b>I. ${x^2}$ – 7x + 10 = 0 a)\u00a0x \u02c3 y<\/p>\n b)\u00a0x \u2265 y<\/p>\n c)\u00a0x \u02c2 y<\/p>\n d)\u00a0x \u2264 y<\/p>\n e)\u00a0x = y or the relationship cannot be established<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$x^{2} – 7x + 10 = 0$<\/p>\n => $x^2 – 5x – 2x + 10 = 0$<\/p>\n => $x (x – 5) – 2 (x – 5) = 0$<\/p>\n => $(x – 5) (x – 2) = 0$<\/p>\n => $x = 5 , 2$<\/p>\n II.$y^{2} + 11y + 10 = 0$<\/p>\n => $y^2 + 10y + y + 10 = 0$<\/p>\n => $y (y + 10) + 1 (y + 10) = 0$<\/p>\n => $(y + 10) (y + 1) = 0$<\/p>\n => $y = -10 , -1$<\/p>\n $\\therefore x > y$<\/p>\n Instructions<\/b><\/p>\n In each of these questions two equations numbered I and II are given. You have to solve both the equations and \u2013 Question 11:\u00a0<\/b>I.$x^{2}+12x+32=0$ a)\u00a0if x < y<\/p>\n b)\u00a0if x \u2264 y<\/p>\n c)\u00a0if x > y<\/p>\n d)\u00a0if x \u2265 y<\/p>\n e)\u00a0if x = y or the relationship cannot be established.<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$x^{2} + 12x + 32 = 0$<\/p>\n => $x^2 + 8x + 4x + 32 = 0$<\/p>\n => $x (x + 8) + 4 (x + 8) = 0$<\/p>\n => $(x + 8) (x + 4) = 0$<\/p>\n => $x = -8 , -4$<\/p>\n II.$y^{2} + 17y + 72 = 0$<\/p>\n => $y^2 + 9y + 8y + 72 = 0$<\/p>\n => $y (y + 9) + 8 (y + 9) = 0$<\/p>\n => $(y + 9) (y + 8) = 0$<\/p>\n => $y = -9 , -8$<\/p>\n $\\therefore x \\geq y$<\/p>\n Question 12:\u00a0<\/b>I.\u00a0$x^{2}-22x+120=0$ a)\u00a0if x < y<\/p>\n b)\u00a0if x \u2264 y<\/p>\n c)\u00a0if x > y<\/p>\n d)\u00a0if x \u2265 y<\/p>\n e)\u00a0if x = y or the relationship cannot be established.<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$x^{2} – 22x + 120 = 0$<\/p>\n => $x^2 – 10x – 12x + 120 = 0$<\/p>\n => $x (x – 10) – 12 (x – 10) = 0$<\/p>\n => $(x – 10) (x – 12) = 0$<\/p>\n => $x = 10 , 12$<\/p>\n II.$y^{2} – 26y + 168 = 0$<\/p>\n => $y^2 – 12y – 14y + 168 = 0$<\/p>\n => $y (y – 12) – 14 (y – 12) = 0$<\/p>\n => $(y – 12) (y – 14) = 0$<\/p>\n => $y = 12 , 14$<\/p>\n $\\therefore x \\leq y$<\/p>\n Question 13:\u00a0<\/b>I.\u00a0$x^{2}+7x+12=0$ a)\u00a0if x < y<\/p>\n b)\u00a0if x \u2264 y<\/p>\n c)\u00a0if x > y<\/p>\n d)\u00a0if x \u2265 y<\/p>\n e)\u00a0if x = y or the relationship cannot be established.<\/p>\n 13)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$x^{2} + 7x + 12 = 0$<\/p>\n => $x^2 + 3x + 4x + 12 = 0$<\/p>\n => $x (x + 3) + 4 (x + 3) = 0$<\/p>\n => $(x + 3) (x + 4) = 0$<\/p>\n => $x = -3 , -4$<\/p>\n II.$y^{2} + 6y + 8 = 0$<\/p>\n => $y^2 + 4y + 2y + 8 = 0$<\/p>\n => $y (y + 4) + 2 (y + 4) = 0$<\/p>\n => $(y + 4) (y + 2) = 0$<\/p>\n => $y = -4 , -2$<\/p>\n Because $-2 > -4$ and $-3 > -4$<\/p>\n Therefore, no relation can be established.<\/p>\n Question 14:\u00a0<\/b>I.\u00a0 \u00a0$x^{2}-15x+56=0$ a)\u00a0if x < y<\/p>\n b)\u00a0if x \u2264 y<\/p>\n c)\u00a0if x > y<\/p>\n d)\u00a0if x \u2265 y<\/p>\n e)\u00a0if x = y or the relationship cannot be established.<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$x^{2} – 15x + 56 = 0$<\/p>\n => $x^2 – 8x – 7x + 56 = 0$<\/p>\n => $x (x – 8) – 7 (x – 8) = 0$<\/p>\n => $(x – 8) (x – 7) = 0$<\/p>\n => $x = 8 , 7$<\/p>\n II.$y^{2} – 23y + 132 = 0$<\/p>\n => $y^2 – 11y – 12y + 132 = 0$<\/p>\n => $y (y – 11) – 12 (y – 11) = 0$<\/p>\n => $(y – 11) (y – 12) = 0$<\/p>\n => $y = 11 , 12$<\/p>\n $\\therefore x < y$<\/p>\n Question 15:\u00a0<\/b>I.\u00a0 $x^{2}+13x+42=0$ a)\u00a0if x < y<\/p>\n b)\u00a0if x \u2264 y<\/p>\n c)\u00a0if x > y<\/p>\n d)\u00a0if x \u2265 y<\/p>\n e)\u00a0if x = y or the relationship cannot be established.<\/p>\n 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$x^{2} + 13x + 42 = 0$<\/p>\n => $x^2 + 7x + 6x + 42 = 0$<\/p>\n => $x (x + 7) + 6 (x + 7) = 0$<\/p>\n => $(x + 7) (x + 6) = 0$<\/p>\n => $x = -7 , -6$<\/p>\n II.$y^{2} + 19y + 90 = 0$<\/p>\n => $y^2 + 9y + 10y + 90 = 0$<\/p>\n => $y (y + 9) + 10 (y + 9) = 0$<\/p>\n => $(y + 9) (y + 10) = 0$<\/p>\n => $y = -9 , -10$<\/p>\n $\\therefore x > y$<\/p>\n
\nII.\u00a0$y^{2} -y-20=0$<\/p>\n
\nII.\u00a0$3y^{2} +11y+10=0$<\/p>\n
\nII.\u00a0$2y^{2} +15y+27=0$<\/p>\n
\nII.\u00a0 \u00a0\u00a0$4y^{2} +18y+14=0 $<\/p>\n
\nII.\u00a0$ 6y^{2}-48y+90=0\u00a0 $<\/p>\n
\nGive answer If<\/p>\n
\nb. x \u2265 y
\nc. x \u02c2 y
\nd. x \u2264 y
\ne. x = y or the relationship cannot be established<\/p>\n
\nII.$ {y^2}$ = 3136<\/p>\n
\nII. ${y^2}$ + 11y + 30 = 0<\/p>\n
\nII. 3x + 2y = – 6.5<\/p>\n
\nII. ${y^2}$ + 16y + 48 = 0<\/p>\n
\nII. ${y^2}$ + 11y + 10 = 0<\/p>\n
\nGive answer a: if x < y
\nGive answer b: if x \u2264 y
\nGive answer c: if x > y
\nGive answer d: if x \u2265 y
\nGive answer e: if x = y or the relationship cannot be established.<\/p>\n
\nII.\u00a0$y^{2} +17y+72=0$<\/p>\n
\nII.\u00a0$y^{2} -26y+168=0$<\/p>\n
\nII.\u00a0$y^{2} +6y+8=0$<\/p>\n
\nII.\u00a0$y^{2} -23y+132=0$<\/p>\n
\nII.\u00a0$y^{2} +19y+90=0$<\/p>\n