Exercise 4.3 — Quadratic Equations

1. 13 marksNCERT Cl-10 Maths, Ex 4.3, Q1 (three parts)

   Find the nature of the roots of each equation, and where the roots are real, find them: (i) 2x² − 3x + 5 = 0; (ii) 3x² − 4√3x + 4 = 0; (iii) 2x² − 6x + 3 = 0.

   Hint. Compute D = b² − 4ac first. Only if D ≥ 0 do you go on to find the roots.

   The discriminant D = b² − 4ac answers the question before any root-finding starts:

   • D > 0 — two distinct real roots
   • D = 0 — two equal real roots
   • D < 0 — no real roots

   (i) 2x² − 3x + 5 = 0. Here a = 2, b = −3, c = 5.

   D = (−3)² − 4(2)(5) = 9 − 40 = −31.

   Since D < 0, there is nothing further to find.

   ✦ No real roots.

   (ii) 3x² − 4√3x + 4 = 0. Here a = 3, b = −4√3, c = 4.

   D = (−4√3)² − 4(3)(4) = 16 × 3 − 48 = 48 − 48 = 0.

   Take care with (−4√3)²: square both the 4 and the √3, giving 16 × 3 = 48, not 4√3 × 2.

   With D = 0 the two roots coincide at x = −b/2a = 4√3/6 = 2√3/3 = 2/√3.

   ✦ Two equal real roots, both x = 2/√3 (≈ 1.155).

   (iii) 2x² − 6x + 3 = 0. Here a = 2, b = −6, c = 3.

   D = 36 − 24 = 12 > 0, so two distinct real roots.

   x = (6 ± √12)/4 = (6 ± 2√3)/4 = (3 ± √3)/2.

   ✦ Two distinct real roots: x = (3 + √3)/2 and x = (3 − √3)/2 (≈ 2.366 and 0.634).

   Where students slip. Mis-squaring a surd coefficient in (ii). (−4√3)² means 16 × 3 = 48. Students often write 4 × 3 = 12 or leave a √ behind.

   Another way. In (iii), simplifying √12 to 2√3 before dividing lets you cancel the 2 cleanly. Leaving it as √12 and dividing by 4 gives the same value but an unsimplified answer, which can lose a mark.

2. 23 marksNCERT Cl-10 Maths, Ex 4.3, Q2

   Find the value of k for each equation so that it has two equal roots: (i) 2x² + kx + 3 = 0; (ii) kx(x − 2) + 6 = 0.

   Hint. Two equal roots means D = 0 exactly. Set b² − 4ac = 0 and solve for k.

   (i) 2x² + kx + 3 = 0. Here a = 2, b = k, c = 3.

   Set D = 0: k² − 4(2)(3) = 0, so k² = 24.

   k = ±√24 = ±2√6.

   Both signs are valid — the equation has equal roots for either.

   ✦ k = 2√6 or k = −2√6

   (ii) kx(x − 2) + 6 = 0. First expand into standard form: kx² − 2kx + 6 = 0, so a = k, b = −2k, c = 6.

   Set D = 0: (−2k)² − 4(k)(6) = 0, that is 4k² − 24k = 0.

   Factorise: 4k(k − 6) = 0, so k = 0 or k = 6.

   Now the important step. If k = 0 the equation collapses to 6 = 0, which is not an equation in x at all — certainly not a quadratic with equal roots. So k = 0 must be rejected.

   ✦ k = 6

   Where students slip. Reporting k = 0 in part (ii). The coefficient a must be non-zero for the equation to be quadratic, so any value of k that kills the x² term is inadmissible — say so explicitly, it carries a mark.

   Another way. For (ii) you can check the answer directly: with k = 6 the equation is 6x² − 12x + 6 = 0, that is 6(x − 1)², whose repeated root is x = 1.

3. 33 marksNCERT Cl-10 Maths, Ex 4.3, Q3

   Is it possible to design a rectangular mango grove whose length is twice its breadth and whose area is 800 m²? If so, find its dimensions.

   Hint. Form the equation, then let the discriminant decide whether a design exists before you solve.

   Questions phrased 'is it possible' are really discriminant questions. Form the equation, test D, and only then answer.

   Step 1 — Let the breadth be x metres, so the length is 2x metres.

   Step 2 — Area: x × 2x = 800, that is 2x² = 800, so x² − 400 = 0.

   Step 3 — Here a = 1, b = 0, c = −400, so D = 0 − 4(1)(−400) = 1600.

   Step 4 — D > 0, so real roots exist and the design is possible.

   Step 5 — Solve: x² = 400, so x = ±20. A breadth cannot be negative, so x = 20.

   Step 6 — Length = 2(20) = 40 m.

   Step 7 — Check: 40 × 20 = 800 m² ✓

   ✦ Yes — the grove is 40 m long and 20 m broad.

   Where students slip. Jumping straight to x = 20 without stating the discriminant. The question asks whether it is *possible*, so the D > 0 line is the part that actually answers it.

   Another way. You can reason without the formula: 2x² = 800 gives x² = 400, and a positive number always has a real square root — so a solution must exist.

4. 43 marksNCERT Cl-10 Maths, Ex 4.3, Q4

   Is the following situation possible? The sum of the ages of two friends is 20 years, and four years ago the product of their ages was 48. If it is possible, find their present ages.

   Hint. Set up the equation, then compute the discriminant — this one is designed to come out negative.

   Step 1 — Let one friend be x years old now, so the other is 20 − x.

   Step 2 — Four years ago they were x − 4 and 16 − x.

   Step 3 — Their product then was 48: (x − 4)(16 − x) = 48.

   Step 4 — Expand: 16x − x² − 64 + 4x = 48, so −x² + 20x − 112 = 0, that is x² − 20x + 112 = 0.

   Step 5 — Compute the discriminant: D = (−20)² − 4(1)(112) = 400 − 448 = −48.

   Step 6 — D < 0, so the equation has no real roots. There is no pair of real ages meeting both conditions.

   ✦ No — the situation is not possible.

   Where students slip. Trying to force an answer by taking the square root of a negative discriminant, or by 'fixing' a sign until the numbers work. A negative D here is the correct answer, not a mistake to be corrected.

   Another way. Intuition worth having: with a fixed sum of 20, four years ago the ages summed to 12, and a pair summing to 12 has maximum product 6 × 6 = 36 — comfortably below 48. So impossibility is visible before any algebra.

5. 54 marksNCERT Cl-10 Maths, Ex 4.3, Q5

   Is it possible to design a rectangular park with perimeter 80 m and area 400 m²? If so, find its length and breadth.

   Hint. Perimeter 80 means length + breadth = 40. Use that to write the area equation in one unknown.

   Step 1 — Let the length be x metres. Since the perimeter is 80, we have 2(length + breadth) = 80, so length + breadth = 40 and the breadth is 40 − x.

   Step 2 — Area: x(40 − x) = 400.

   Step 3 — Rearrange: 40x − x² = 400, so x² − 40x + 400 = 0.

   Step 4 — Discriminant: D = (−40)² − 4(1)(400) = 1600 − 1600 = 0.

   Step 5 — D = 0, so real (and equal) roots exist. The design is possible, and there is exactly one shape that works.

   Step 6 — The repeated root is x = −b/2a = 40/2 = 20. Then the breadth is 40 − 20 = 20.

   Step 7 — Check: perimeter 2(20 + 20) = 80 ✓ and area 20 × 20 = 400 ✓

   ✦ Yes — and the park must be a 20 m × 20 m square.

   The D = 0 is telling you something real: among all rectangles of perimeter 80, the square has the largest possible area, and 400 m² is exactly that maximum. Ask for any more area and the discriminant would go negative.

   Where students slip. Writing 2(x + breadth) = 80 and then using 80 as the sum of the sides. Halve the perimeter first: length + breadth is 40, not 80.

   Another way. x² − 40x + 400 is a perfect square, (x − 20)², which shows the repeated root immediately without the discriminant formula.

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.