The Quadratic Equations is an important topic in the Quant section of the SNAP Exam. You can also download this Free Quadratic Equations Questions for SNAP PDF with detailed answers by Cracku. These questions will help you practice and solve the Quadratic Equations questions in the SNAP exam. Utilize this PDF practice set, <\/strong>which is one of the best sources for practicing.<\/p>\n Download Quadratic Equation Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>Four of the following five parts numbered A,B,C,D,E are in the following equation are exactly equal.Which of the part is not equal to the other four.The number of that part is the answer<\/p>\n a)\u00a0$xy^{2}-x^{2}y+2x^{2}y^{2}$<\/p>\n b)\u00a0$xy^{2}(1-2x)+x^{2}y(2y-1)$<\/p>\n c)\u00a0$xy^{2}(1+2x)-x^{2}y(2y+1)$<\/p>\n d)\u00a0$xy[y(1+x)-x(1+y)]$<\/p>\n e)\u00a0$xy[(x+y)^{2}+y(1-y)-x(1+x)-2xy]$<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n (A) :\u00a0$xy^{2}-x^{2}y+2x^{2}y^{2}$<\/p>\n = $xy(y-x+2xy)$<\/p>\n (B) :\u00a0$xy^{2}(1-2x)+x^{2}y(2y-1)$<\/p>\n = $(xy^2-2x^2y^2)+(2x^2y^2-x^2y)$<\/p>\n = $xy^2-x^y = xy(y-x)$<\/p>\n (C) :\u00a0$xy^{2}(1-2x)+x^{2}y(2y-1)$<\/p>\n = $(xy^2-2x^2y^2)+(2x^2y^2-x^2y)$<\/p>\n = $xy^2-x^y = xy(y-x)$<\/p>\n (D) :\u00a0$xy[y(1+x)-x(1+y)]$<\/p>\n = $xy(y+xy-x-xy) = xy(y-x)$<\/p>\n (E) :\u00a0$xy[(x+y)^{2}+y(1-y)-x(1+x)-2xy]$<\/p>\n = $xy[(x^2+y^2+2xy)+(y-y^2)+(-x-x^2)-2xy]$<\/p>\n = $xy(y-x)$<\/p>\n => Ans – (A)<\/p>\n Question 2:\u00a0<\/b>If 2x+y= 15, \u00a02y+z= 25 \u00a0 and \u00a0 2z+x =26 ,what is value of z?<\/p>\n a)\u00a04<\/p>\n b)\u00a07<\/p>\n c)\u00a09<\/p>\n d)\u00a012<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Equation (i) : $2x+y=15$<\/p>\n Equation (ii) : $2y+z=25$<\/p>\n Equation (iii) : $2z+x=26$<\/p>\n Adding the three equations, => $3x+3y+3z=15+25+26$<\/p>\n => $3(x+y+z)=66$<\/p>\n => $x+y+z=\\frac{66}{3}=22$ ————(iv)<\/p>\n Subtracting equation (i) from (iv), => $(z)+(x-2x)+(y-y)=(22-15)$<\/p>\n => $z-x=7$ ————-(v)<\/p>\n Adding equations (v) and (iii), => $3z=26+7=33$<\/p>\n => $z=\\frac{33}{3}=11$<\/p>\n => Ans – (E)<\/p>\n Instructions<\/b><\/p>\n In the following questions two equations numbered I and II are given. You have to solve both the equations and Question 3:\u00a0<\/b>I.\u00a0 $2x^{2}+21x+10=0$ a)\u00a0if x > y<\/p>\n b)\u00a0if x \u2265 y<\/p>\n c)\u00a0if x < y<\/p>\n d)\u00a0if x \u2264 y<\/p>\n e)\u00a0if x = y or the relationship cannot be established.<\/p>\n 3)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$2x^{2} + 21x + 10 = 0$<\/p>\n => $2x^2 + x + 20x + 10 = 0$<\/p>\n => $x (2x + 1) + 10 (2x + 1) = 0$<\/p>\n => $(x + 10) (2x + 1) = 0$<\/p>\n => $x = -10 , \\frac{-1}{2}$<\/p>\n II.$3y^{2} + 13y + 14 = 0$<\/p>\n => $3y^2 + 6y + 7y + 14 = 0$<\/p>\n => $3y (y + 2) + 7 (y + 2) = 0$<\/p>\n => $(y + 2) (3y + 7) = 0$<\/p>\n => $y = -2 , \\frac{-7}{3}$<\/p>\n $\\therefore$\u00a0No relation can be established.<\/p>\n Question 4:\u00a0<\/b>I.\u00a0 $3x^{2}+10x+8=0$ a)\u00a0if x > y<\/p>\n b)\u00a0if x \u2265 y<\/p>\n c)\u00a0if x < y<\/p>\n d)\u00a0if x \u2264 y<\/p>\n e)\u00a0if x = y or the relationship cannot be established.<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$3x^{2} + 10x + 8 = 0$<\/p>\n => $3x^2 + 6x + 4x + 8 = 0$<\/p>\n => $3x (x + 2) + 4 (x + 2) = 0$<\/p>\n => $(x + 2) (3x + 4) = 0$<\/p>\n => $x = -2 , \\frac{-4}{3}$<\/p>\n II.$3y^{2} + 7y + 4 = 0$<\/p>\n => $3y^2 + 3y + 4y + 4 = 0$<\/p>\n => $3y (y + 1) + 4 (y + 1) = 0$<\/p>\n => $(y + 1) (3y + 4) = 0$<\/p>\n => $y = -1 , \\frac{-4}{3}$<\/p>\n $\\therefore x \\leq y$<\/p>\n Question 5:\u00a0<\/b>I.\u00a0 $2x^{2}-11x+12=0$ a)\u00a0if x > y<\/p>\n b)\u00a0if x \u2265 y<\/p>\n c)\u00a0if x < y<\/p>\n d)\u00a0if x \u2264 y<\/p>\n e)\u00a0if x = y or the relationship cannot be established.<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$2x^{2} – 11x + 12 = 0$<\/p>\n => $2x^2 – 8x – 3x + 12 = 0$<\/p>\n => $2x (x – 4) – 3 (x – 4) = 0$<\/p>\n => $(x – 4) (2x – 3) = 0$<\/p>\n => $x = 4 , \\frac{3}{2}$<\/p>\n II.$2y^{2} – 19y + 44 = 0$<\/p>\n => $2y^2 – 8y – 11y + 44 = 0$<\/p>\n => $2y (y – 4) – 11 (y – 4) = 0$<\/p>\n => $(y – 4) (2y – 11) = 0$<\/p>\n => $y = 4 , \\frac{11}{2}$<\/p>\n $\\therefore x \\leq y$<\/p>\n Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>I.\u00a0 $5x^{2}+29x+20=0$ a)\u00a0if x > y<\/p>\n b)\u00a0if x \u2265 y<\/p>\n c)\u00a0if x < y<\/p>\n d)\u00a0if x \u2264 y<\/p>\n e)\u00a0if x = y or the relationship cannot be established.<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$5x^{2} + 29x + 20 = 0$<\/p>\n => $5x^2 + 25x + 4x + 20 = 0$<\/p>\n => $5x (x + 5) + 4 (x + 5) = 0$<\/p>\n => $(x + 5) (5x + 4) = 0$<\/p>\n => $x = -5 , \\frac{-4}{5}$<\/p>\n II.$25y^{2} + 25y + 6 = 0$<\/p>\n => $25y^2 + 10y + 15y + 6 = 0$<\/p>\n => $5y (5y + 2) + 3 (5y + 2) = 0$<\/p>\n => $(5y + 3) (5y + 2) = 0$<\/p>\n => $y = \\frac{-3}{5} , \\frac{-2}{5}$<\/p>\n Therefore $x < y$<\/p>\n Question 7:\u00a0<\/b>I.\u00a0 $x^{2}-3x-88=0$ a)\u00a0if x > y<\/p>\n b)\u00a0if x \u2265 y<\/p>\n c)\u00a0if x < y<\/p>\n d)\u00a0if x \u2264 y<\/p>\n e)\u00a0if x = y or the relationship cannot be established.<\/p>\n 7)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n I.$x^{2} – 3x – 88 = 0$<\/p>\n => $x^2 + 8x – 11x – 88 = 0$<\/p>\n => $x (x + 8) – 11 (x + 8) = 0$<\/p>\n => $(x + 8) (x – 11) = 0$<\/p>\n => $x = -8 , 11$<\/p>\n II.$y^{2} + 8y – 48 = 0$<\/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 $\\therefore$\u00a0No relation can be established.<\/p>\n Question 8:\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 8)\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 9:\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 9)\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 10:\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 10)\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 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 Question 16:\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 16)\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 17:\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 17)\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 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 18:\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 18)\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 19:\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 19)\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 20:\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 20)\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 21:\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 21)\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 22:\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 22)\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
\nGive answer a: if x > y
\nGive answer b: if x \u2265 y
\nGive answer c: if x < y
\nGive answer d: if x \u2264 y
\nGive answer e: if x = y or the relationship cannot be established.<\/p>\n
\nII.\u00a0$3y^{2}+13y+14=0$<\/p>\n
\nII.\u00a0$3y^{2}+7y+4=0$<\/p>\n
\nII.\u00a0$2y^{2}-19y+44=0$<\/p>\n
\nII.\u00a0$25y^{2}+25y+6=0$<\/p>\n
\nII.\u00a0$y^{2}+8y-48=0$<\/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
\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
\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