Exercise 4.1 — Quadratic Equations

1. 14 marksNCERT Cl-10 Maths, Ex 4.1, Q1 (eight parts)

   Decide whether each of the following is a quadratic equation: (i) (x+1)² = 2(x−3); (ii) x² − 2x = (−2)(3−x); (iii) (x−2)(x+1) = (x−1)(x+3); (iv) (x−3)(2x+1) = x(x+5); (v) (2x−1)(x−3) = (x+5)(x−1); (vi) x² + 3x + 1 = (x−2)²; (vii) (x+2)³ = 2x(x²−1); (viii) x³ − 4x² − x + 1 = (x−2)³.

   Hint. Expand both sides, bring everything to one side, and then look at what survives. The test is whether the highest power is exactly 2 with a non-zero coefficient.

   An equation is quadratic if, after full simplification, it has the form ax² + bx + c = 0 with a ≠ 0. The trap in every part is that terms cancel — so you cannot judge by appearance.

   (i) (x+1)² = 2(x−3) becomes x² + 2x + 1 = 2x − 6, so x² + 7 = 0. Degree 2. Quadratic.

   (ii) x² − 2x = −6 + 2x, so x² − 4x + 6 = 0. Quadratic.

   (iii) Left: x² − x − 2. Right: x² + 2x − 3. Subtracting, the x² terms cancel: −3x + 1 = 0. Degree 1. Not quadratic.

   (iv) Left: 2x² − 5x − 3. Right: x² + 5x. Subtracting: x² − 10x − 3 = 0. Quadratic.

   (v) Left: 2x² − 7x + 3. Right: x² + 4x − 5. Subtracting: x² − 11x + 8 = 0. Quadratic.

   (vi) Right: x² − 4x + 4. So x² + 3x + 1 = x² − 4x + 4, and the x² terms cancel: 7x − 3 = 0. Not quadratic.

   (vii) Left: x³ + 6x² + 12x + 8. Right: 2x³ − 2x. Subtracting: −x³ + 6x² + 14x + 8 = 0. The cubic term survives. Not quadratic (it is cubic).

   (viii) Right: x³ − 6x² + 12x − 8. Subtracting from the left: 2x² − 13x + 9 = 0. The cubic terms cancel here, leaving degree 2. Quadratic.

   ✦ Quadratic: (i), (ii), (iv), (v), (viii). Not quadratic: (iii), (vi), (vii).

   Where students slip. Calling (vii) quadratic because a cube appears on both sides, and (viii) cubic for the same reason. In (vii) the x³ terms do not cancel; in (viii) they do. Only the simplified form can tell you — always expand first.

   Another way. A quick pre-check: if both sides have the same leading term with the same coefficient, that term will cancel, so look one degree lower to find the real answer.

2. 24 marksNCERT Cl-10 Maths, Ex 4.1, Q2 (four parts)

   Represent each situation as a quadratic equation. (i) A rectangular plot has area 528 m², and its length is one more than twice its breadth. (ii) The product of two consecutive positive integers is 306. (iii) Rohan's mother is 26 years older than him; three years from now the product of their ages will be 360. (iv) A train travels 480 km at a uniform speed; had the speed been 8 km/h less, it would have taken 3 hours longer.

   Hint. Name the unknown, express every other quantity in terms of it, then write the one sentence that gives an equation.

   The examinable skill here is forming the equation — the question does not ask you to solve it.

   (i) The plot. Let the breadth be x metres. Then the length is 2x + 1.

   Area = length × breadth: x(2x + 1) = 528.

   ✦ 2x² + x − 528 = 0

   (ii) Consecutive integers. Let the smaller be x, so the next is x + 1.

   x(x + 1) = 306.

   ✦ x² + x − 306 = 0

   (iii) Rohan and his mother. Let Rohan's present age be x, so his mother is x + 26.

   In three years they will be x + 3 and x + 29. Their product is 360:

   (x + 3)(x + 29) = 360 ⟹ x² + 32x + 87 = 360.

   ✦ x² + 32x − 273 = 0

   (iv) The train. Let the speed be x km/h. Time taken is 480/x hours.

   At speed x − 8 the time is 480/(x − 8), and that is 3 hours more:

   480/(x − 8) − 480/x = 3.

   Multiply through by x(x − 8): 480x − 480(x − 8) = 3x(x − 8), so 3840 = 3x² − 24x.

   Dividing by 3: x² − 8x − 1280 = 0.

   ✦ x² − 8x − 1280 = 0

   Where students slip. In (iv), writing 480/x − 480/(x−8) = 3. Going slower takes *longer*, so the slower time is the bigger one and must come first. Sanity-check the direction before multiplying out.

   Another way. In (iii), a useful check on the expansion: the constant 3 × 29 = 87, and the x-coefficient 3 + 29 = 32 — the sum and product of the two shifts.

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.