{"id":5109,"date":"2026-09-26T11:47:00","date_gmt":"2026-09-26T06:17:00","guid":{"rendered":"https:\/\/www.catmock.com\/blog\/?p=5109"},"modified":"2026-09-26T11:47:00","modified_gmt":"2026-09-26T06:17:00","slug":"cat-quadratic-equations","status":"publish","type":"post","link":"https:\/\/www.catmock.com\/blog\/cat-quadratic-equations\/","title":{"rendered":"Top 50 CAT Quadratic Equations Questions with Solutions (CAT 2026 Edition)"},"content":{"rendered":"\n

Quadratic equations are one of the most scoring topics inside CAT Algebra \u2014 yet most aspirants lose easy marks here because they rush the factorisation step or forget the sum-and-product shortcut. This guide fixes that. You get 10 verified CAT previous year questions (1997\u20132025)<\/strong> with year and slot mentioned, followed by 40 CAT-pattern practice questions<\/strong>, all solved step-by-step \u2014 organised the way an actual CAT Quant paper is structured, along with the formulas and shortcuts examiners test.<\/p>\n\n\n\n

Quick answer:<\/strong> A quadratic equation is any equation of the form ax\u00b2 + bx + c = 0<\/strong> (a \u2260 0). In CAT, it usually shows up as direct root-finding, sum\/product-of-roots questions, nature-of-roots (discriminant) questions, or as a hidden step inside Number Systems, Inequalities and Functions questions. Based on the confirmed CAT papers referenced in this article, quadratic equations have appeared in almost every CAT year from 1997 to 2025 \u2014 sometimes directly, sometimes as the final step of a Functions or Number Systems question.<\/p>\n\n\n\n

A note on how this article was built:<\/strong> The 10 previous-year questions were cross-checked with multiple CAT-preparation sources and re-solved to verify their year, slot, and answers. They are presented in original wording while preserving the mathematical conditions and answers. The remaining 40 questions are original CAT-pattern practice questions and are clearly labelled as such.<\/p>\n\n\n\n

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Why Quadratic Equations Matter in CAT Quant<\/h2>\n\n\n\n

Algebra is the single largest scoring area inside CAT Quantitative Ability, and quadratic equations form its foundation. Even when a question is officially tagged under Functions, Inequalities or Number Systems, solving it often collapses down to a quadratic equation in the last step.<\/p>\n\n\n\n

Quant Topic Area<\/th>Approx. Weightage (Questions)<\/th>Where Quadratic Equations Help<\/th><\/tr><\/thead>
Algebra (overall)<\/td>6\u20138 out of 22 QA questions<\/td>Core topic<\/td><\/tr>
Quadratic Equations (direct)<\/td>0\u20132 questions<\/td>Direct application<\/td><\/tr>
Number Systems<\/td>3\u20135 questions<\/td>Integer-root and factorisation logic<\/td><\/tr>
Inequalities & Modulus<\/td>2\u20133 questions<\/td>Sign-scheme and interval logic<\/td><\/tr>
Functions & Graphs<\/td>1\u20132 questions<\/td>Roots as x-intercepts<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

As per the official CAT exam pattern on iimcat.ac.in<\/mark><\/a>, the Quantitative Ability section carries 22 of the exam’s 68 total questions, to be solved in 40 minutes, with +3 marks for every correct answer and \u22121 for a wrong MCQ response (no negative marking on TITA questions). That timing pressure is exactly why speed-based quadratic equation tricks matter more than textbook methods on exam day.<\/p>\n\n\n\n

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CAT Quadratic Equations Formulas You Must Know<\/h2>\n\n\n\n

Before attempting CAT quadratic equations questions with solutions, lock these formulas into memory. Almost every question in this article \u2014 and in the actual CAT paper \u2014 is built on one of these.<\/p>\n\n\n\n

Concept<\/th>Formula<\/th><\/tr><\/thead>
Standard form<\/td>ax\u00b2 + bx + c = 0, where a \u2260 0<\/td><\/tr>
Roots (Quadratic formula)<\/td>x = [\u2212b \u00b1 \u221a(b\u00b2 \u2212 4ac)] \/ 2a<\/td><\/tr>
Discriminant<\/td>D = b\u00b2 \u2212 4ac<\/td><\/tr>
Nature of roots \u2014 real & distinct<\/td>D > 0<\/td><\/tr>
Nature of roots \u2014 real & equal<\/td>D = 0<\/td><\/tr>
Nature of roots \u2014 no real roots (complex)<\/td>D < 0<\/td><\/tr>
Sum of roots<\/td>\u03b1 + \u03b2 = \u2212b\/a<\/td><\/tr>
Product of roots<\/td>\u03b1\u03b2 = c\/a<\/td><\/tr>
Equation from given roots<\/td>x\u00b2 \u2212 (\u03b1 + \u03b2)x + \u03b1\u03b2 = 0<\/td><\/tr>
\u03b1\u00b2 + \u03b2\u00b2<\/td>(\u03b1 + \u03b2)\u00b2 \u2212 2\u03b1\u03b2<\/td><\/tr>
\u03b1\u00b3 + \u03b2\u00b3<\/td>(\u03b1 + \u03b2)\u00b3 \u2212 3\u03b1\u03b2(\u03b1 + \u03b2)<\/td><\/tr>
|\u03b1 \u2212 \u03b2|<\/td>\u221a[(\u03b1 + \u03b2)\u00b2 \u2212 4\u03b1\u03b2]<\/td><\/tr>
Roots in ratio p : q<\/td>b\u00b2\/ac = (p + q)\u00b2\/pq<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
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Top Tricks & Shortcuts for CAT Quadratic Equations<\/h2>\n\n\n\n

Use these CAT quadratic equations tricks to shave seconds off every question \u2014 critical when you have roughly 109 seconds per QA question.<\/p>\n\n\n\n

    \n
  1. Split the middle term by inspection first.<\/strong> Check if b\u00b2 \u2212 4ac is a perfect square before reaching for the formula \u2014 if it is, the equation factorises cleanly.<\/li>\n\n\n\n
  2. Use sum-and-product instead of solving for roots.<\/strong> Most CAT questions ask for \u03b1\u00b2 + \u03b2\u00b2, \u03b1\u00b3 + \u03b2\u00b3, or 1\/\u03b1 + 1\/\u03b2 \u2014 never solve for the actual roots when a direct identity exists.<\/li>\n\n\n\n
  3. Substitute to reduce higher-degree or exponential equations.<\/strong> Expressions like x\u2074 \u2212 13x\u00b2 + 36 = 0, (x + 1\/x)\u00b2 \u2212 3(x + 1\/x) + 2 = 0, or 9^t \u2212 4\u00b73^(t+1) + 27 = 0 all become simple quadratics the moment you substitute a single variable for the repeating expression.<\/li>\n\n\n\n
  4. Remember how a\u1d47 = 1 splits into three cases.<\/strong> This is the exact logic behind CAT 2020’s “distinct positive integer solutions” question: a\u1d47 = 1 when a = 1, or when b = 0 (a \u2260 0), or when a = \u22121 and b is even. Missing the third case is the single most common mistake on this question type.<\/li>\n\n\n\n
  5. Sign-scheme for inequalities.<\/strong> For ax\u00b2 + bx + c > 0 or < 0, plot the roots on a number line and use the “positive-outside, negative-inside” rule (for a > 0) instead of testing multiple values.<\/li>\n\n\n\n
  6. Watch for extraneous roots.<\/strong> Any time you square both sides (as in radical equations), always verify the final answer in the original equation.<\/li>\n\n\n\n
  7. Remember the “sum of roots = 0” shortcut.<\/strong> If a question says roots are “equal in magnitude but opposite in sign,” you only need the coefficient of x to be zero \u2014 no discriminant work needed.<\/li>\n<\/ol>\n\n\n\n
    \n\n\n\n

    Quick Answer Key \u2014 All 50 Questions<\/h2>\n\n\n\n

    Use this table for a fast self-check. Q1\u2013Q10 are verified CAT PYQs; Q11\u2013Q50 are practice questions.<\/p>\n\n\n\n

    Q. No.<\/th>Answer<\/th>Q. No.<\/th>Answer<\/th>Q. No.<\/th>Answer<\/th><\/tr><\/thead>
    1<\/td>c = \u221215<\/td>18<\/td>Sum = 7, Product = 10<\/td>35<\/td>8 and 11<\/td><\/tr>
    2<\/td>Roots = 6, 1<\/td>19<\/td>29\/4<\/td>36<\/td>x = 5 or 1\/5<\/td><\/tr>
    3<\/td>Min value = 5<\/td>20<\/td>3\/4<\/td>37<\/td>40 km\/h<\/td><\/tr>
    4<\/td>n = 24<\/td>21<\/td>52<\/td>38<\/td>x = \u00b12, \u00b13<\/td><\/tr>
    5<\/td>Min value = 3<\/td>22<\/td>k = 7, other root = 4<\/td>39<\/td>x = \u22121, 1, 2, 4<\/td><\/tr>
    6<\/td>Product = \u221216<\/td>23<\/td>k = 8<\/td>40<\/td>x = 0, 1<\/td><\/tr>
    7<\/td>b\u00b2 + c = 549<\/td>24<\/td>Real & equal<\/td>41<\/td>x = 9 (x = 2 rejected)<\/td><\/tr>
    8<\/td>6 solutions<\/td>25<\/td>No real roots<\/td>42<\/td>x = 1, 4<\/td><\/tr>
    9<\/td>1 distinct real root<\/td>26<\/td>Real, rational, distinct<\/td>43<\/td>2 < x < 3<\/td><\/tr>
    10<\/td>Product = 20<\/td>27<\/td>k = \u00b16<\/td>44<\/td>x \u2264 \u22122 or x \u2265 3<\/td><\/tr>
    11<\/td>x = 2, 3<\/td>28<\/td>k \u2264 1 (k \u2260 0)<\/td>45<\/td>x = 0, 2, 4, 6<\/td><\/tr>
    12<\/td>x = 3, 4<\/td>29<\/td>m = \u22125, 3<\/td>46<\/td>6b\u00b2 = 25ac<\/td><\/tr>
    13<\/td>x = \u22121\/2, 2<\/td>30<\/td>x\u00b2 \u2212 3x + 2 = 0<\/td>47<\/td>a = \u22121<\/td><\/tr>
    14<\/td>x = \u22124, 3<\/td>31<\/td>x\u00b2 \u2212 13x + 36 = 0<\/td>48<\/td>k = 1, 4<\/td><\/tr>
    15<\/td>x = 2\/3, 1<\/td>32<\/td>x\u00b2 \u2212 7x + 1 = 0<\/td>49<\/td>2 real roots<\/td><\/tr>
    16<\/td>x = 4, \u22122<\/td>33<\/td>x\u00b2 \u2212 7x + 12 = 0<\/td>50<\/td>4 integer solutions<\/td><\/tr>
    17<\/td>x = 1\/2, \u22121\/3<\/td>34<\/td>10 and 11<\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
    \n\n\n\n

    Part 1: Verified CAT Previous Year Questions (Q1\u2013Q10)<\/h2>\n\n\n\n

    These 10 questions are confirmed CAT PYQs, each cross-verified for its year and slot. The problem statements are written in original wording; the numbers, conditions and final answers match the actual paper exactly.<\/p>\n\n\n\n

    Q1. [CAT 1997]<\/strong> The roots x\u2081 and x\u2082 of the equation x\u00b2 \u2212 2x + c = 0 also satisfy the relation 7x\u2082 \u2212 4x\u2081 = 47. Find the value of c. <\/p>\n\n\n\n

    Solution:<\/strong> From the equation, x\u2081 + x\u2082 = 2 (sum of roots), so x\u2081 = 2 \u2212 x\u2082. Substituting into 7x\u2082 \u2212 4x\u2081 = 47: 7x\u2082 \u2212 4(2 \u2212 x\u2082) = 47 \u2192 7x\u2082 \u2212 8 + 4x\u2082 = 47 \u2192 11x\u2082 = 55 \u2192 x\u2082 = 5, so x\u2081 = \u22123. Since c is the product of the roots (c\/a with a = 1): c = x\u2081 \u00b7 x\u2082 = (\u22123)(5) = \u221215. <\/p>\n\n\n\n

    Answer: c = \u221215<\/strong><\/p>\n\n\n\n

    Q2. [CAT 2001]<\/strong> Two students attempted to solve the same quadratic equation. The first student copied the coefficient of x correctly but made an error in the constant term, and ended up with roots 4 and 3. The second student copied the constant term correctly but made an error in the coefficient of x, and ended up with roots 3 and 2. Find the correct roots of the original equation. <\/p>\n\n\n\n

    Solution:<\/strong> The first student’s equation (roots 4, 3) is x\u00b2 \u2212 7x + 12 = 0 \u2014 since he made the error only in the constant term, his coefficient of x (i.e., \u22127) is correct. The second student’s equation (roots 3, 2) is x\u00b2 \u2212 5x + 6 = 0 \u2014 since he made the error only in the coefficient of x, his constant term (i.e., 6) is correct. Combining the correct coefficient of x (\u22127) with the correct constant (6), the original equation is x\u00b2 \u2212 7x + 6 = 0 \u2192 (x \u2212 6)(x \u2212 1) = 0. Answer: Roots = 6, 1<\/strong><\/p>\n\n\n\n

    Q3. [CAT 2003]<\/strong> Let p and q be the roots of x\u00b2 \u2212 (k \u2212 2)x \u2212 (k + 1) = 0, where k is a real parameter. Find the minimum possible value of p\u00b2 + q\u00b2. <\/p>\n\n\n\n

    Solution:<\/strong> Sum of roots: p + q = k \u2212 2. Product of roots: pq = \u2212(k + 1). Using p\u00b2 + q\u00b2 = (p+q)\u00b2 \u2212 2pq: p\u00b2 + q\u00b2 = (k\u22122)\u00b2 \u2212 2(\u2212(k+1)) = (k\u22122)\u00b2 + 2k + 2 = k\u00b2 \u2212 4k + 4 + 2k + 2 = k\u00b2 \u2212 2k + 6. This is itself a quadratic in k that opens upward, so its minimum occurs at k = \u2212(\u22122)\/(2\u00d71) = 1. Substituting k = 1: 1 \u2212 2 + 6 = 5. <\/p>\n\n\n\n

    Answer: Minimum value = 5<\/strong><\/p>\n\n\n\n

    Q4. [CAT 2017, Slot 1]<\/strong> If f(x) = x\u00b2 + 11x + n and g(x) = x, find the largest positive integer value of n for which f(x) = g(x) has two distinct real roots. <\/p>\n\n\n\n

    Solution:<\/strong> Setting f(x) = g(x): x\u00b2 + 11x + n = x \u2192 x\u00b2 + 10x + n = 0. For two distinct real roots, the discriminant must be positive: 10\u00b2 \u2212 4(1)(n) > 0 \u2192 100 \u2212 4n > 0 \u2192 n < 25. The largest positive integer satisfying this is n = 24. <\/p>\n\n\n\n

    Answer: n = 24<\/strong><\/p>\n\n\n\n

    Q5. [CAT 2017, Slot 2]<\/strong> Find the minimum possible value of the sum of squares of the roots of the equation x\u00b2 + (a + 3)x \u2212 (a + 5) = 0, where a is real. <\/p>\n\n\n\n

    Solution:<\/strong> Sum of roots = \u2212(a + 3), product of roots = \u2212(a + 5). Sum of squares = (sum)\u00b2 \u2212 2(product) = (a+3)\u00b2 \u2212 2(\u2212(a+5)) = (a+3)\u00b2 + 2a + 10 = a\u00b2 + 6a + 9 + 2a + 10 = a\u00b2 + 8a + 19. Completing the square: a\u00b2 + 8a + 19 = (a + 4)\u00b2 + 3. The minimum value of (a+4)\u00b2 is 0 (at a = \u22124), so the minimum of the whole expression is 3. <\/p>\n\n\n\n

    Answer: Minimum value = 3<\/strong><\/p>\n\n\n\n

    Q6. [CAT 2019, Slot 1]<\/strong> Find the product of the distinct roots of |x\u00b2 \u2212 x \u2212 6| = x + 2. <\/p>\n\n\n\n

    Solution:<\/strong> Since the right-hand side must be non-negative, we need x \u2265 \u22122. Case 1: x\u00b2 \u2212 x \u2212 6 = x + 2 \u2192 x\u00b2 \u2212 2x \u2212 8 = 0 \u2192 (x\u22124)(x+2) = 0 \u2192 x = 4 or x = \u22122 (both satisfy x \u2265 \u22122). Case 2: \u2212(x\u00b2 \u2212 x \u2212 6) = x + 2 \u2192 x\u00b2 \u2212 4 = 0 \u2192 x = 2 or x = \u22122 (both satisfy x \u2265 \u22122). Combining and removing the repeated value, the distinct roots are 4, \u22122, and 2. Product = 4 \u00d7 (\u22122) \u00d7 2 = \u221216. <\/p>\n\n\n\n

    Answer: Product = \u221216<\/strong><\/p>\n\n\n\n

    Q7. [CAT 2019, Slot 2]<\/strong> The equation x\u00b2 + bx + c = 0 has two roots, 4a and 3a, where a is an integer. Find a possible value of b\u00b2 + c. <\/p>\n\n\n\n

    Solution:<\/strong> Sum of roots: 4a + 3a = 7a = \u2212b \u2192 b = \u22127a. Product of roots: (4a)(3a) = 12a\u00b2 = c. So b\u00b2 + c = 49a\u00b2 + 12a\u00b2 = 61a\u00b2. This must equal one of the given answer choices for some integer a. Testing a = 3: 61(3)\u00b2 = 61 \u00d7 9 = 549, and a = 3 is indeed an integer, so this works. <\/p>\n\n\n\n

    Answer: b\u00b2 + c = 549 (when a = 3)<\/strong><\/p>\n\n\n\n

    Q8. [CAT 2020, Slot 1]<\/strong> How many distinct positive integer-valued solutions exist for the equation (x\u00b2 \u2212 7x + 11)^(x\u00b2 \u2212 13x + 42) = 1? <\/p>\n\n\n\n

    Solution:<\/strong> An expression of the form (base)^(exponent) equals 1 in exactly three scenarios: (i) base = 1, for any exponent; (ii) exponent = 0, provided base \u2260 0; (iii) base = \u22121, provided the exponent is even.<\/p>\n\n\n\n

    Case (i): x\u00b2 \u2212 7x + 11 = 1 \u2192 x\u00b2 \u2212 7x + 10 = 0 \u2192 (x\u22122)(x\u22125) = 0 \u2192 x = 2, 5. <\/p>\n\n\n\n

    Case (ii): x\u00b2 \u2212 13x + 42 = 0 \u2192 (x\u22126)(x\u22127) = 0 \u2192 x = 6, 7 (checking the base is non-zero at both points \u2014 it is).<\/p>\n\n\n\n

    Case (iii): x\u00b2 \u2212 7x + 11 = \u22121 \u2192 x\u00b2 \u2212 7x + 12 = 0 \u2192 (x\u22123)(x\u22124) = 0 \u2192 x = 3, 4. Checking the exponent is even at both: at x = 3, exponent = 9\u221239+42 = 12 (even) \u2713; at x = 4, exponent = 16\u221252+42 = 6 (even) \u2713. Collecting all distinct values: {2, 3, 4, 5, 6, 7} \u2014 six values in total. <\/p>\n\n\n\n

    Answer: 6 solutions<\/strong><\/p>\n\n\n\n

    Q9. [CAT 2020, Slot 1]<\/strong> Find the number of distinct real roots of the equation (x + 1\/x)\u00b2 \u2212 3(x + 1\/x) + 2 = 0. <\/p>\n\n\n\n

    Solution:<\/strong> Substitute y = x + 1\/x: y\u00b2 \u2212 3y + 2 = 0 \u2192 (y\u22121)(y\u22122) = 0 \u2192 y = 1 or y = 2. For y = 1: x + 1\/x = 1 \u2192 x\u00b2 \u2212 x + 1 = 0 \u2192 discriminant = 1 \u2212 4 = \u22123 < 0, so no real roots here. For y = 2: x + 1\/x = 2 \u2192 x\u00b2 \u2212 2x + 1 = 0 \u2192 (x\u22121)\u00b2 = 0 \u2192 x = 1 (a single repeated value). So across both cases, there is exactly one distinct real value of x. <\/p>\n\n\n\n

    Answer: 1 distinct real root<\/strong><\/p>\n\n\n\n

    Q10. [CAT 2025, Slot 2]<\/strong> If 9^(x\u00b2 + 2x \u2212 3) \u2212 4\u00b73^(x\u00b2 + 2x \u2212 2) + 27 = 0, find the product of all possible values of x. <\/p>\n\n\n\n

    Solution:<\/strong> Let t = x\u00b2 + 2x \u2212 3. Then 9^t = 3^(2t), and the middle term 3^(x\u00b2+2x\u22122) = 3^(t+1) = 3\u00b73^t. Substituting y = 3^t, the equation becomes y\u00b2 \u2212 4\u00b73\u00b7y + 27 = 0 \u2192 y\u00b2 \u2212 12y + 27 = 0 \u2192 (y\u22123)(y\u22129) = 0 \u2192 y = 3 or y = 9, giving 3^t = 3 \u2192 t = 1, or 3^t = 9 \u2192 t = 2. For t = 1: x\u00b2 + 2x \u2212 3 = 1 \u2192 x\u00b2 + 2x \u2212 4 = 0. Discriminant = 4 + 16 = 20 > 0 (real roots); product of these two roots = \u22124\/1 = \u22124. For t = 2: x\u00b2 + 2x \u2212 3 = 2 \u2192 x\u00b2 + 2x \u2212 5 = 0. Discriminant = 4 + 20 = 24 > 0 (real roots); product of these two roots = \u22125\/1 = \u22125. Product of all four real values of x = (\u22124) \u00d7 (\u22125) = 20. <\/p>\n\n\n\n

    Answer: Product = 20<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    Part 2: CAT-Pattern Quadratic Equations Practice Questions (Q11\u2013Q50)<\/h2>\n\n\n\n

    These 40 questions are original, written in CAT style and difficulty to build the same skills the PYQs above test. They are practice material, not attributed to any specific CAT year.<\/p>\n\n\n\n

    Section A: Basic Factorisation (Q11\u2013Q17)<\/h3>\n\n\n\n

    Q11.<\/strong> Solve for x: x\u00b2 \u2212 5x + 6 = 0 <\/p>\n\n\n\n

    Solution:<\/strong> Split the middle term: x\u00b2 \u2212 2x \u2212 3x + 6 = 0 \u2192 x(x\u22122) \u2212 3(x\u22122) = 0 \u2192 (x\u22122)(x\u22123) = 0. <\/p>\n\n\n\n

    Answer: x = 2, 3<\/strong><\/p>\n\n\n\n

    Q12.<\/strong> Solve for x: x\u00b2 \u2212 7x + 12 = 0<\/p>\n\n\n\n

    Solution:<\/strong> (x \u2212 3)(x \u2212 4) = 0. <\/p>\n\n\n\n

    Answer: x = 3, 4<\/strong><\/p>\n\n\n\n

    Q13.<\/strong> Solve for x: 2x\u00b2 \u2212 3x \u2212 2 = 0 <\/p>\n\n\n\n

    Solution:<\/strong> Split: 2x\u00b2 + x \u2212 4x \u2212 2 = 0 \u2192 x(2x+1) \u2212 2(2x+1) = 0 \u2192 (2x+1)(x\u22122) = 0. <\/p>\n\n\n\n

    Answer: x = \u22121\/2, 2<\/strong><\/p>\n\n\n\n

    Q14.<\/strong> Solve for x: x\u00b2 + x \u2212 12 = 0 <\/p>\n\n\n\n

    Solution:<\/strong> (x + 4)(x \u2212 3) = 0. <\/p>\n\n\n\n

    Answer: x = \u22124, 3<\/strong><\/p>\n\n\n\n

    Q15.<\/strong> Solve for x: 3x\u00b2 \u2212 5x + 2 = 0 <\/p>\n\n\n\n

    Solution:<\/strong> 3x\u00b2 \u2212 3x \u2212 2x + 2 = 0 \u2192 3x(x\u22121) \u2212 2(x\u22121) = 0 \u2192 (3x\u22122)(x\u22121) = 0. <\/p>\n\n\n\n

    Answer: x = 2\/3, 1<\/strong><\/p>\n\n\n\n

    Q16.<\/strong> Solve for x: x\u00b2 \u2212 2x \u2212 8 = 0 <\/p>\n\n\n\n

    Solution:<\/strong> (x \u2212 4)(x + 2) = 0. <\/p>\n\n\n\n

    Answer: x = 4, \u22122<\/strong><\/p>\n\n\n\n

    Q17.<\/strong> Solve for x: 6x\u00b2 \u2212 x \u2212 1 = 0 <\/p>\n\n\n\n

    Solution:<\/strong> Discriminant = 1 + 24 = 25. x = (1 \u00b1 5)\/12. <\/p>\n\n\n\n

    Answer: x = 1\/2, \u22121\/3<\/strong><\/p>\n\n\n\n

    Section B: Sum & Product of Roots (Q18\u2013Q23)<\/h3>\n\n\n\n

    Q18.<\/strong> Find the sum and product of the roots of x\u00b2 \u2212 7x + 10 = 0. <\/p>\n\n\n\n

    Solution:<\/strong> Using \u03b1 + \u03b2 = \u2212b\/a and \u03b1\u03b2 = c\/a directly. <\/p>\n\n\n\n

    Answer: Sum = 7, Product = 10<\/strong><\/p>\n\n\n\n

    Q19.<\/strong> If \u03b1, \u03b2 are the roots of 2x\u00b2 \u2212 3x \u2212 5 = 0, find \u03b1\u00b2 + \u03b2\u00b2. <\/p>\n\n\n\n

    Solution:<\/strong> \u03b1 + \u03b2 = 3\/2, \u03b1\u03b2 = \u22125\/2. \u03b1\u00b2 + \u03b2\u00b2 = (\u03b1+\u03b2)\u00b2 \u2212 2\u03b1\u03b2 = 9\/4 + 5 = 29\/4. <\/p>\n\n\n\n

    Answer: 29\/4<\/strong><\/p>\n\n\n\n

    Q20.<\/strong> If \u03b1, \u03b2 are the roots of x\u00b2 \u2212 6x + 8 = 0, find 1\/\u03b1 + 1\/\u03b2. <\/p>\n\n\n\n

    Solution:<\/strong> 1\/\u03b1 + 1\/\u03b2 = (\u03b1+\u03b2)\/\u03b1\u03b2 = 6\/8 = 3\/4. <\/p>\n\n\n\n

    Answer: 3\/4<\/strong><\/p>\n\n\n\n

    Q21.<\/strong> If \u03b1, \u03b2 are the roots of x\u00b2 \u2212 4x + 1 = 0, find \u03b1\u00b3 + \u03b2\u00b3. <\/p>\n\n\n\n

    Solution:<\/strong> \u03b1 + \u03b2 = 4, \u03b1\u03b2 = 1. \u03b1\u00b3 + \u03b2\u00b3 = (\u03b1+\u03b2)\u00b3 \u2212 3\u03b1\u03b2(\u03b1+\u03b2) = 64 \u2212 12 = 52. <\/p>\n\n\n\n

    Answer: 52<\/strong><\/p>\n\n\n\n

    Q22.<\/strong> If one root of x\u00b2 \u2212 kx + 12 = 0 is 3, find k and the other root. <\/p>\n\n\n\n

    Solution:<\/strong> Substituting x = 3: 9 \u2212 3k + 12 = 0 \u2192 k = 7. Since product of roots = 12, the other root = 12\/3 = 4. <\/p>\n\n\n\n

    Answer: k = 7, other root = 4<\/strong><\/p>\n\n\n\n

    Q23.<\/strong> If the roots of 3x\u00b2 + (2k \u2212 1)x + (k \u2212 5) = 0 are reciprocals of each other, find k. <\/p>\n\n\n\n

    Solution:<\/strong> Roots are reciprocal \u27f9 product of roots = 1 \u27f9 (k \u2212 5)\/3 = 1 \u27f9 k = 8. <\/p>\n\n\n\n

    Answer: k = 8<\/strong><\/p>\n\n\n\n

    Section C: Nature of Roots \u2014 Discriminant (Q24\u2013Q29)<\/h3>\n\n\n\n

    Q24.<\/strong> Determine the nature of the roots of x\u00b2 \u2212 4x + 4 = 0. <\/p>\n\n\n\n

    Solution:<\/strong> D = 16 \u2212 16 = 0. <\/p>\n\n\n\n

    Answer: Real and equal roots<\/strong><\/p>\n\n\n\n

    Q25.<\/strong> Determine the nature of the roots of x\u00b2 + 2x + 5 = 0. <\/p>\n\n\n\n

    Solution:<\/strong> D = 4 \u2212 20 = \u221216 < 0. <\/p>\n\n\n\n

    Answer: No real roots (complex\/imaginary roots)<\/strong><\/p>\n\n\n\n

    Q26.<\/strong> Determine the nature of the roots of 2x\u00b2 \u2212 7x + 3 = 0. <\/p>\n\n\n\n

    Solution:<\/strong> D = 49 \u2212 24 = 25, a perfect square. <\/p>\n\n\n\n

    Answer: Real, rational and distinct roots<\/strong><\/p>\n\n\n\n

    Q27.<\/strong> Find the value(s) of k for which x\u00b2 \u2212 kx + 9 = 0 has equal roots. <\/p>\n\n\n\n

    Solution:<\/strong> D = 0 \u27f9 k\u00b2 \u2212 36 = 0 \u27f9 k = \u00b16. <\/p>\n\n\n\n

    Answer: k = \u00b16<\/strong><\/p>\n\n\n\n

    Q28.<\/strong> For what values of k does kx\u00b2 + 2x + 1 = 0 have two real roots? <\/p>\n\n\n\n

    Solution:<\/strong> For a genuine quadratic, k \u2260 0. D \u2265 0 \u27f9 4 \u2212 4k \u2265 0 \u27f9 k \u2264 1. <\/p>\n\n\n\n

    Answer: k \u2264 1, k \u2260 0<\/strong><\/p>\n\n\n\n

    Q29.<\/strong> For what value(s) of m does x\u00b2 \u2212 (m + 3)x + (m + 6) = 0 have equal roots? <\/p>\n\n\n\n

    Solution:<\/strong> D = (m + 3)\u00b2 \u2212 4(m + 6) = 0 \u27f9 m\u00b2 + 2m \u2212 15 = 0 \u27f9 (m + 5)(m \u2212 3) = 0. <\/p>\n\n\n\n

    Answer: m = \u22125 or m = 3<\/strong><\/p>\n\n\n\n

    Section D: Forming Quadratic Equations (Q30\u2013Q33)<\/h3>\n\n\n\n

    Q30.<\/strong> Form the quadratic equation whose roots are the reciprocals of the roots of 2x\u00b2 \u2212 3x + 1 = 0. <\/p>\n\n\n\n

    Solution:<\/strong> Original sum = 3\/2, product = 1\/2. New sum = (\u03b1+\u03b2)\/(\u03b1\u03b2) = 3, new product = 1\/(\u03b1\u03b2) = 2. <\/p>\n\n\n\n

    Answer: x\u00b2 \u2212 3x + 2 = 0<\/strong><\/p>\n\n\n\n

    Q31.<\/strong> Form the quadratic equation whose roots are the squares of the roots of x\u00b2 \u2212 5x + 6 = 0. <\/p>\n\n\n\n

    Solution:<\/strong> Roots of original equation: 2, 3. Squares: 4, 9. Sum = 13, product = 36. <\/p>\n\n\n\n

    Answer: x\u00b2 \u2212 13x + 36 = 0<\/strong><\/p>\n\n\n\n

    Q32.<\/strong> If \u03b1, \u03b2 are the roots of x\u00b2 + 3x + 1 = 0, form the equation whose roots are \u03b1\/\u03b2 and \u03b2\/\u03b1. <\/p>\n\n\n\n

    Solution:<\/strong> \u03b1 + \u03b2 = \u22123, \u03b1\u03b2 = 1. \u03b1\/\u03b2 + \u03b2\/\u03b1 = (\u03b1\u00b2 + \u03b2\u00b2)\/\u03b1\u03b2 = (9 \u2212 2)\/1 = 7. Product = 1. <\/p>\n\n\n\n

    Answer: x\u00b2 \u2212 7x + 1 = 0<\/strong><\/p>\n\n\n\n

    Q33.<\/strong> Form the quadratic equation whose roots exceed the roots of x\u00b2 \u2212 3x + 2 = 0 by 2. <\/p>\n\n\n\n

    Solution:<\/strong> Roots of original equation: 1, 2. New roots: 3, 4. Sum = 7, product = 12. <\/p>\n\n\n\n

    Answer: x\u00b2 \u2212 7x + 12 = 0<\/strong><\/p>\n\n\n\n

    Section E: CAT-Style Word Problems (Q34\u2013Q37)<\/h3>\n\n\n\n

    Q34.<\/strong> The product of two consecutive positive integers exceeds their sum by 89. Find the integers. <\/p>\n\n\n\n

    Solution:<\/strong> Let integers be n and n+1. n(n+1) \u2212 (2n+1) = 89 \u27f9 n\u00b2 \u2212 n \u2212 90 = 0 \u27f9 (n\u221210)(n+9) = 0 \u27f9 n = 10. <\/p>\n\n\n\n

    Answer: 10 and 11<\/strong><\/p>\n\n\n\n

    Q35.<\/strong> Two numbers differ by 3 and their product is 88. Find the numbers. <\/p>\n\n\n\n

    Solution:<\/strong> Let numbers be x and x+3. x(x+3) = 88 \u27f9 x\u00b2 + 3x \u2212 88 = 0 \u27f9 D = 361 = 19\u00b2 \u27f9 x = 8. <\/p>\n\n\n\n

    Answer: 8 and 11<\/strong><\/p>\n\n\n\n

    Q36.<\/strong> The sum of a positive number and its reciprocal is 26\/5. Find the number. <\/p>\n\n\n\n

    Solution:<\/strong> x + 1\/x = 26\/5 \u27f9 5x\u00b2 \u2212 26x + 5 = 0 \u27f9 D = 576 = 24\u00b2 \u27f9 x = (26 \u00b1 24)\/10. <\/p>\n\n\n\n

    Answer: x = 5 or x = 1\/5<\/strong><\/p>\n\n\n\n

    Q37.<\/strong> A train travels 360 km at a uniform speed. If the speed had been 5 km\/h more, it would have taken 1 hour less for the same journey. Find the original speed. <\/p>\n\n\n\n

    Solution:<\/strong> Let speed = x km\/h. 360\/x \u2212 360\/(x+5) = 1 \u27f9 1800 = x\u00b2 + 5x \u27f9 x\u00b2 + 5x \u2212 1800 = 0 \u27f9 D = 7225 = 85\u00b2 \u27f9 x = 40. <\/p>\n\n\n\n

    Answer: 40 km\/h<\/strong><\/p>\n\n\n\n

    Section F: Higher-Degree Equations Reducible to Quadratic (Q38\u2013Q41)<\/h3>\n\n\n\n

    Q38.<\/strong> Solve for x: x\u2074 \u2212 13x\u00b2 + 36 = 0 <\/p>\n\n\n\n

    Solution:<\/strong> Substitute y = x\u00b2: y\u00b2 \u2212 13y + 36 = 0 \u27f9 (y\u22124)(y\u22129) = 0 \u27f9 y = 4, 9. <\/p>\n\n\n\n

    Answer: x = \u00b12, \u00b13<\/strong><\/p>\n\n\n\n

    Q39.<\/strong> Solve for x: (x\u00b2 \u2212 3x)\u00b2 \u2212 2(x\u00b2 \u2212 3x) \u2212 8 = 0 <\/p>\n\n\n\n

    Solution:<\/strong> Substitute y = x\u00b2 \u2212 3x: y\u00b2 \u2212 2y \u2212 8 = 0 \u27f9 (y\u22124)(y+2) = 0 \u27f9 y = 4 or y = \u22122. For y = 4: x\u00b2 \u2212 3x \u2212 4 = 0 \u27f9 x = 4, \u22121. For y = \u22122: x\u00b2 \u2212 3x + 2 = 0 \u27f9 x = 1, 2. <\/p>\n\n\n\n

    Answer: x = \u22121, 1, 2, 4<\/strong><\/p>\n\n\n\n

    Q40.<\/strong> Solve for x: 2\u00b2\u02e3 \u2212 3\u00b72\u02e3 + 2 = 0 <\/p>\n\n\n\n

    Solution:<\/strong> Substitute y = 2\u02e3: y\u00b2 \u2212 3y + 2 = 0 \u27f9 (y\u22121)(y\u22122) = 0 \u27f9 y = 1 or y = 2. <\/p>\n\n\n\n

    Answer: x = 0, 1<\/strong><\/p>\n\n\n\n

    Q41.<\/strong> Solve for x: \u221a(x + 7) = x \u2212 5 <\/p>\n\n\n\n

    Solution:<\/strong> Squaring: x + 7 = x\u00b2 \u2212 10x + 25 \u27f9 x\u00b2 \u2212 11x + 18 = 0 \u27f9 (x\u22129)(x\u22122) = 0 \u27f9 x = 9 or 2. Checking x = 2: \u221a9 = 3 \u2260 (2\u22125) = \u22123, rejected. Checking x = 9: \u221a16 = 4 = (9\u22125) \u2713. <\/p>\n\n\n\n

    Answer: x = 9 (x = 2 is an extraneous root)<\/strong><\/p>\n\n\n\n

    Section G: Quadratic Equations with Modulus & Inequalities (Q42\u2013Q45)<\/h3>\n\n\n\n

    Q42.<\/strong> Solve for x: |x\u00b2 \u2212 5x + 6| = 2 <\/p>\n\n\n\n

    Solution:<\/strong> Case 1: x\u00b2 \u2212 5x + 6 = 2 \u27f9 x\u00b2 \u2212 5x + 4 = 0 \u27f9 x = 1, 4. Case 2: x\u00b2 \u2212 5x + 6 = \u22122 \u27f9 x\u00b2 \u2212 5x + 8 = 0 \u27f9 D = \u22127 < 0, no real solutions. Answer: x = 1, 4<\/strong><\/p>\n\n\n\n

    Q43.<\/strong> Solve the inequality: x\u00b2 \u2212 5x + 6 < 0 <\/p>\n\n\n\n

    Solution:<\/strong> Roots are 2 and 3. Since the coefficient of x\u00b2 is positive, the expression is negative between the roots. <\/p>\n\n\n\n

    Answer: 2 < x < 3<\/strong><\/p>\n\n\n\n

    Q44.<\/strong> Solve the inequality: x\u00b2 \u2212 x \u2212 6 \u2265 0 <\/p>\n\n\n\n

    Solution:<\/strong> Roots are \u22122 and 3. Since the parabola opens upward, the expression is non-negative outside the roots. <\/p>\n\n\n\n

    Answer: x \u2264 \u22122 or x \u2265 3<\/strong><\/p>\n\n\n\n

    Q45.<\/strong> Solve for x: |x \u2212 3|\u00b2 \u2212 4|x \u2212 3| + 3 = 0 <\/p>\n\n\n\n

    Solution:<\/strong> Substitute y = |x\u22123|: y\u00b2 \u2212 4y + 3 = 0 \u27f9 (y\u22121)(y\u22123) = 0 \u27f9 y = 1 or 3. y=1 \u27f9 x = 4, 2. y=3 \u27f9 x = 6, 0. <\/p>\n\n\n\n

    Answer: x = 0, 2, 4, 6<\/strong><\/p>\n\n\n\n

    Section H: Advanced \/ TITA-Style Questions (Q46\u2013Q50)<\/h3>\n\n\n\n

    Q46.<\/strong> If the roots of ax\u00b2 + bx + c = 0 are in the ratio 2 : 3, prove the relation connecting a, b and c. <\/p>\n\n\n\n

    Solution:<\/strong> Let roots be 2k and 3k. Sum = 5k = \u2212b\/a \u27f9 k = \u2212b\/5a. Product = 6k\u00b2 = c\/a. Substituting k: 6b\u00b2\/25a\u00b2 = c\/a. <\/p>\n\n\n\n

    Answer: 6b\u00b2 = 25ac<\/strong><\/p>\n\n\n\n

    Q47.<\/strong> If the equation x\u00b2 \u2212 (a+1)x + (a\u22121) = 0 has roots that are equal in magnitude but opposite in sign, find a. <\/p>\n\n\n\n

    Solution:<\/strong> Roots equal in magnitude but opposite in sign \u27f9 sum of roots = 0 \u27f9 a + 1 = 0. <\/p>\n\n\n\n

    Answer: a = \u22121<\/strong><\/p>\n\n\n\n

    Q48.<\/strong> Find the value(s) of k for which x\u00b2 + 2(k+2)x + 9k = 0 has equal roots. <\/p>\n\n\n\n

    Solution:<\/strong> D = 0 \u27f9 4(k+2)\u00b2 \u2212 36k = 0 \u27f9 (k+2)\u00b2 \u2212 9k = 0 \u27f9 k\u00b2 \u2212 5k + 4 = 0 \u27f9 (k\u22121)(k\u22124) = 0. <\/p>\n\n\n\n

    Answer: k = 1, 4<\/strong><\/p>\n\n\n\n

    Q49.<\/strong> Find the number of real roots of (x\u00b2 + x + 1)(x\u00b2 + x + 2) = 12. <\/p>\n\n\n\n

    Solution:<\/strong> Let y = x\u00b2 + x. (y+1)(y+2) = 12 \u27f9 y\u00b2 + 3y \u2212 10 = 0 \u27f9 (y+5)(y\u22122) = 0 \u27f9 y = \u22125 or y = 2. For y = 2: x\u00b2 + x \u2212 2 = 0 \u27f9 x = 1, \u22122 (2 real roots). For y = \u22125: x\u00b2 + x + 5 = 0 \u27f9 D = \u221219 < 0 (no real roots). <\/p>\n\n\n\n

    Answer: 2 real roots<\/strong><\/p>\n\n\n\n

    Q50.<\/strong> Find the number of integer values of x satisfying x\u00b2 \u2212 5|x| + 6 = 0. <\/p>\n\n\n\n

    Solution:<\/strong> Substitute t = |x|, t \u2265 0: t\u00b2 \u2212 5t + 6 = 0 \u27f9 (t\u22122)(t\u22123) = 0 \u27f9 t = 2 or 3. So |x| = 2 gives x = \u00b12, and |x| = 3 gives x = \u00b13. <\/p>\n\n\n\n

    Answer: 4 integer solutions (x = \u22123, \u22122, 2, 3)<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    Common Mistakes Students Make in CAT Quadratic Equations<\/h2>\n\n\n\n
    Mistake<\/th>Why It Costs Marks<\/th>Fix<\/th><\/tr><\/thead>
    Forgetting the third case of a\u1d47 = 1 (a = \u22121, b even)<\/td>Directly causes wrong answers on questions like CAT 2020’s Q8 above<\/td>Always check all three cases: base = 1, exponent = 0, or base = \u22121 with even exponent<\/td><\/tr>
    Not verifying roots after squaring radical equations<\/td>Leads to selecting an extraneous root as the final answer<\/td>Always substitute the final answer back into the original equation<\/td><\/tr>
    Solving for actual roots when only \u03b1\u00b2 + \u03b2\u00b2 or \u03b1\u00b3 + \u03b2\u00b3 is asked<\/td>Slower and more error-prone<\/td>Use sum-and-product identities directly<\/td><\/tr>
    Ignoring the case a = 0 in “for what value of k” questions<\/td>Missing valid cases where the equation becomes linear<\/td>Always state a \u2260 0 as a condition, or check the linear case separately<\/td><\/tr>
    Mixing up “roots differ by n” with “roots in ratio n”<\/td>Leads to setting up the wrong two equations<\/td>Read word problems twice; identify whether it’s a difference or a ratio condition<\/td><\/tr>
    Sign errors in the modulus cases<\/td>Missing one branch of the solution set<\/td>Always write both Case 1 (positive) and Case 2 (negative) explicitly<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
    \n\n\n\n

    How to Practice Quadratic Equations for CAT 2026<\/h2>\n\n\n\n

    Reading solved questions builds understanding, but CAT rewards speed under exam-like pressure. Once you’re comfortable with all 50 questions above:<\/p>\n\n\n\n