Exercise 4.2 — Quadratic Equations

1. 15 marksNCERT Cl-10 Maths, Ex 4.2, Q1 (five parts)

   Find the roots by factorisation: (i) x² − 3x − 10 = 0; (ii) 2x² + x − 6 = 0; (iii) √2x² + 7x + 5√2 = 0; (iv) 2x² − x + 1/8 = 0; (v) 100x² − 20x + 1 = 0.

   Hint. Split the middle term into two numbers whose product is ac and whose sum is b, then factor by grouping.

   (i) x² − 3x − 10 = 0. Need two numbers with product −10 and sum −3: those are −5 and +2.

   x² − 5x + 2x − 10 = x(x − 5) + 2(x − 5) = (x − 5)(x + 2).

   ✦ x = 5 or x = −2

   (ii) 2x² + x − 6 = 0. Here ac = −12 and b = 1, so use +4 and −3.

   2x² + 4x − 3x − 6 = 2x(x + 2) − 3(x + 2) = (x + 2)(2x − 3).

   ✦ x = −2 or x = 3/2

   (iii) √2x² + 7x + 5√2 = 0. Here ac = √2 × 5√2 = 10 and b = 7, so use 5 and 2.

   √2x² + 5x + 2x + 5√2 = x(√2x + 5) + √2(√2x + 5) = (√2x + 5)(x + √2).

   Note the middle step: 2x = √2·√2·x, which is what lets √2 come out as the common factor.

   ✦ x = −5/√2 (= −5√2/2) or x = −√2

   (iv) 2x² − x + 1/8 = 0. Clear the fraction first by multiplying through by 8: 16x² − 8x + 1 = 0.

   That is a perfect square: (4x − 1)².

   ✦ x = 1/4, a repeated root

   (v) 100x² − 20x + 1 = 0. Again a perfect square: (10x − 1)².

   ✦ x = 1/10, a repeated root

   Where students slip. In (iv), forgetting to clear the 1/8 before splitting the middle term. Working with fractions here is possible but invites arithmetic slips — multiply up first.

   Another way. Parts (iv) and (v) both have discriminant zero, which you can check in one line (64 − 64 = 0 and 400 − 400 = 0) before factorising. Spotting a perfect square early saves the search for factors.

2. 24 marksNCERT Cl-10 Maths, Ex 4.2, Q2

   (i) John and Jivanti together have 45 marbles. Each loses 5 marbles, and the product of the numbers they then have is 124. Find how many each started with. (ii) A cottage industry makes toys in a day; the cost in rupees of each toy was 55 minus the number of toys made. The day's total production cost was ₹750. Find the number of toys made.

   Hint. In both parts, name one quantity x and write the other as (total − x).

   (i) The marbles.

   Step 1 — Let John have x marbles, so Jivanti has 45 − x.

   Step 2 — After each loses 5: (x − 5) and (40 − x).

   Step 3 — Their product is 124: (x − 5)(40 − x) = 124.

   Step 4 — Expand: 40x − x² − 200 + 5x = 124, so −x² + 45x − 324 = 0, that is x² − 45x + 324 = 0.

   Step 5 — Factorise: need product 324 and sum 45 — that is 36 and 9.

   (x − 36)(x − 9) = 0, so x = 36 or x = 9.

   Step 6 — Both work, and they describe the same pair from opposite sides.

   ✦ One had 36 marbles and the other 9.

   (ii) The toys.

   Step 1 — Let x toys be made. Each costs (55 − x) rupees.

   Step 2 — Total cost: x(55 − x) = 750.

   Step 3 — Rearranged: x² − 55x + 750 = 0.

   Step 4 — Factorise: product 750, sum 55 — that is 30 and 25.

   (x − 30)(x − 25) = 0, so x = 30 or x = 25.

   Step 5 — Check both are sensible: 30 toys at ₹25 each = ₹750 ✓, and 25 toys at ₹30 each = ₹750 ✓. Both are genuine answers.

   ✦ Either 25 or 30 toys were made.

   Where students slip. Discarding one root out of habit. Reject a root only when it is impossible in context — negative counts, negative lengths, ages below zero. Here both roots describe real situations, so both must be reported.

   Another way. For (i), note the two answers must add to 45 (36 + 9 = 45) — that is a fast structural check on your factorisation.

3. 33 marksNCERT Cl-10 Maths, Ex 4.2, Q3

   Find two numbers whose sum is 27 and whose product is 182.

   Hint. Call one number x and the other 27 − x.

   Step 1 — Let the numbers be x and 27 − x.

   Step 2 — Their product is 182: x(27 − x) = 182.

   Step 3 — Rearrange into standard form: 27x − x² = 182, so x² − 27x + 182 = 0.

   Step 4 — Split the middle term. We need two numbers with product 182 and sum 27. Since 182 = 2 × 7 × 13, the pair 13 and 14 works.

   x² − 13x − 14x + 182 = x(x − 13) − 14(x − 13) = (x − 13)(x − 14).

   Step 5 — So x = 13 or x = 14, giving the other number as 14 or 13 respectively — the same pair either way.

   Step 6 — Check: 13 + 14 = 27 ✓ and 13 × 14 = 182 ✓

   ✦ The numbers are 13 and 14.

   Where students slip. Treating x = 13 and x = 14 as two different answers to report separately. They name the same pair; the question asks for the two numbers, so give them once.

   Another way. Because you are looking for numbers with a known sum and product, this is exactly the 'build a quadratic from its zeroes' idea from Polynomials: x² − 27x + 182 is the polynomial whose zeroes are the two numbers.

4. 43 marksNCERT Cl-10 Maths, Ex 4.2, Q4

   Find two consecutive positive integers the sum of whose squares is 365.

   Hint. Consecutive integers differ by 1, so call them x and x + 1.

   Step 1 — Let the integers be x and x + 1.

   Step 2 — The sum of their squares is 365: x² + (x + 1)² = 365.

   Step 3 — Expand: x² + x² + 2x + 1 = 365, so 2x² + 2x − 364 = 0.

   Step 4 — Divide through by 2 to keep the numbers small: x² + x − 182 = 0.

   Step 5 — Factorise: product −182, sum 1 — that is 14 and −13.

   x² + 14x − 13x − 182 = x(x + 14) − 13(x + 14) = (x + 14)(x − 13).

   Step 6 — So x = −14 or x = 13. The question says positive integers, so x = −14 is rejected.

   Step 7 — Check: 13² + 14² = 169 + 196 = 365 ✓

   ✦ The integers are 13 and 14.

   Where students slip. Keeping x = −14 as a second answer. It does satisfy the equation (−14 and −13 have squares summing to 365), but the question restricts to positive integers — always re-read the constraint before finalising.

   Another way. Dividing by 2 at Step 4 is worth the habit: factorising x² + x − 182 is much easier than 2x² + 2x − 364.

5. 53 marksNCERT Cl-10 Maths, Ex 4.2, Q5

   The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

   Hint. Let the base be x, write the altitude in terms of x, then apply Pythagoras.

   Step 1 — Let the base be x cm. Then the altitude is x − 7 cm.

   Step 2 — By Pythagoras, base² + altitude² = hypotenuse²:

   x² + (x − 7)² = 13².

   Step 3 — Expand: x² + x² − 14x + 49 = 169, so 2x² − 14x − 120 = 0.

   Step 4 — Divide by 2: x² − 7x − 60 = 0.

   Step 5 — Factorise: product −60, sum −7 — that is −12 and +5.

   (x − 12)(x + 5) = 0, so x = 12 or x = −5.

   Step 6 — A length cannot be negative, so x = 12. Then the altitude is 12 − 7 = 5.

   Step 7 — Check: 12² + 5² = 144 + 25 = 169 = 13² ✓

   ✦ The base is 12 cm and the altitude is 5 cm.

   Where students slip. Accepting x = −5 and reporting an altitude of −12 cm. Whenever the unknown is a physical length, state explicitly that the negative root is rejected — examiners look for that line.

   Another way. Recognising 5-12-13 as a standard Pythagorean triple gives you the answer instantly, which makes a good check on the algebra.

6. 63 marksNCERT Cl-10 Maths, Ex 4.2, Q6

   A cottage industry produces a certain number of pottery articles in a day. On a particular day the cost of production of each article in rupees was 3 more than twice the number of articles produced, and the total cost was ₹90. Find the number of articles and the cost of each.

   Hint. Let the number of articles be x, express the per-article cost in terms of x, then multiply.

   Step 1 — Let x articles be produced. The cost of each is 2x + 3 rupees.

   Step 2 — Total cost = number × cost each: x(2x + 3) = 90.

   Step 3 — Rearrange: 2x² + 3x − 90 = 0.

   Step 4 — Split the middle term. Here ac = −180 and b = 3, so we need 15 and −12.

   2x² + 15x − 12x − 90 = x(2x + 15) − 6(2x + 15) = (2x + 15)(x − 6).

   Step 5 — So x = 6 or x = −15/2. A count of articles cannot be negative or fractional, so x = 6.

   Step 6 — Cost of each article: 2(6) + 3 = ₹15.

   Step 7 — Check: 6 × 15 = ₹90 ✓

   ✦ 6 articles were produced, each costing ₹15.

   Where students slip. Reading '3 more than twice the number' as 2(x + 3) instead of 2x + 3. Translate the phrase piece by piece: 'twice the number' is 2x, then '3 more' adds 3.

   Another way. Since 90 = 6 × 15 and the numbers are small, you could also test the factor pairs of 90 directly for one fitting cost = 2 × count + 3.

Solutions written by the tuition.in editorial team and checked against the rationalised NCERT Class 10 Mathematics textbook and the CBSE 2026-27 syllabus (solving by completing the square is no longer part of this chapter; the nature-of-roots exercise is now Exercise 4.3). Questions are referenced from the NCERT textbook for identification.