Download Quadratic Equations Questions for XAT PDF \u2013 XAT Quadratic Equations questions pdf by Cracku. Practice XAT solved Quadratic Equations Questions paper tests, and these are the practice question to have a firm grasp on the Quadratic Equations topic in the XAT exam. Top 20 very Important Quadratic Equations Questions for XAT based on asked questions in previous exam papers. \u00a0The XAT question papers contain actual questions asked with answers and solutions.<\/p>\n
Download Quadratic Equations Questions for XAT<\/a><\/p>\n Enroll to XAT 2023 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>If $a^3 – b^3 = 899\u00a0 and\u00a0 a – b = 31$, then $(a – b)^2 + 3ab$ is equal to:<\/p>\n a)\u00a031<\/p>\n b)\u00a035<\/p>\n c)\u00a016<\/p>\n d)\u00a029<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0 $a^3-b^3=899$\u00a0 and\u00a0$a-b=31$<\/p>\n $=$> \u00a0$\\left(a-b\\right)\\left(a^2+ab+b^2\\right)=899$<\/p>\n $=$> \u00a0$31\\left(a^2+ab+b^2-2ab+2ab\\right)=899$<\/p>\n $=$> \u00a0$\\left(a-b\\right)^2+3ab=\\frac{899}{31}$<\/p>\n $=$> \u00a0$\\left(a-b\\right)^2+3ab=29$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 2:\u00a0<\/b>If $(x – 7)^3 + (2x + 8)^3 + (2x – 3)^3 = 3 (x \u2014 7) (2x + 8) (2x \u2014 3)$, then what is the value of $x$ ?<\/p>\n a)\u00a01.6<\/p>\n b)\u00a00.4<\/p>\n c)\u00a02.4<\/p>\n d)\u00a01.2<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, $(x – 7)^3 + (2x + 8)^3 + (2x – 3)^3 = 3 (x \u2014 7) (2x + 8) (2x \u2014 3)$<\/p>\n $=$> $(x – 7)^3 + (2x + 8)^3 + (2x – 3)^3 – 3 (x \u2014 7) (2x + 8) (2x \u2014 3)=0$<\/p>\n We know that if $a^3+b^3+c^3-3abc=0$ then $a+b+c=0$<\/p>\n $=$> $\\left(x-7\\right)+\\left(2x+8\\right)+\\left(2x-3\\right)=0$<\/p>\n $=$> $5x-2=0$<\/p>\n $=$> $x=\\frac{2}{5}$<\/p>\n $=$> $x=0.4$<\/p>\n Hence, the correct option is Option B<\/p>\n Question 3:\u00a0<\/b>If $(x – 8)^3 + (2x + 16)^3 + (2x – 13)^3 = 3 (x – 8) (2x + 16) (2x – 13 )$, then what is the value of $x$ ?<\/p>\n a)\u00a01<\/p>\n b)\u00a0-1<\/p>\n c)\u00a00<\/p>\n d)\u00a00.7<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, $(x-8)^3+(2x+16)^3+(2x-13)^3 = 3(x-8)(2x+16)(2x-13)$<\/p>\n $=$> $(x-8)^3+(2x+16)^3+(2x-13)^3-3(x-8)(2x+16)(2x-13)=0$<\/p>\n We know that if $a^3+b^3+c^3-3abc=0$ then $a+b+c=0$<\/p>\n $=$> $\\left(x-8\\right)+\\left(2x+16\\right)+\\left(2x-13\\right)=0$<\/p>\n $=$> $5x-5=0$<\/p>\n $=$> $x=\\frac{5}{5}$<\/p>\n $=$> $x=1$<\/p>\n Hence, the correct option is Option A<\/p>\n Question 4:\u00a0<\/b>If $(x + 7)^3 + (2x + 8)^3 + (2x + 3)^3 = 3 (x + 7) (2x + 8) (2x + 3)$, then what is the value of $x$ ?<\/p>\n a)\u00a03.6<\/p>\n b)\u00a0-3.6<\/p>\n c)\u00a02.4<\/p>\n d)\u00a0-2.4<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0$(x+7)^3+(2x+8)^3+(2x+3)^3 = 3(x+7)(2x+8)(2x+3)$<\/p>\n $=$> \u00a0$(x+7)^3+(2x+8)^3+(2x+3)^3-3(x+7)(2x+8)(2x+3)=0$<\/p>\n We know that if\u00a0$a^3+b^3+c^3-3abc=0$ then $a+b+c=0$<\/p>\n $=$> \u00a0$\\left(x+7\\right)+\\left(2x+8\\right)+\\left(2x+3\\right)=0$<\/p>\n $=$> \u00a0$5x+18=0$<\/p>\n $=$> \u00a0$x=\\frac{-18}{5}$<\/p>\n $=$> \u00a0$x=-3.6$<\/p>\n Hence, the correct option is Option B<\/p>\n Question 5:\u00a0<\/b>If $(x – 3)^2 + (2x – 5)^3 + (x – 4)^3 = (3x – 9)(2x – 5)(x – 4)$ , then what is the value of $x$ ?<\/p>\n a)\u00a02<\/p>\n b)\u00a05<\/p>\n c)\u00a04<\/p>\n d)\u00a03<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The best way to solve such questions is by substituting the options given : Take XAT 2023 Mock Tests<\/a><\/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 6:\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 6)\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 7:\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 7)\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 8:\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 8)\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 Question 9:\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 9)\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 10:\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 10)\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
\nOn substituting x=3
\nwe get LHS = 0
\nAnd RHS =0
\nso we get x=3<\/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