Exercise 5.4 — Nature of Roots

Nature of roots of a quadratic equation based on its discriminant.

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Exercise 5.4 — Nature of Roots of a Quadratic Equation

Exercise 5.4 is the final and most conceptually rich exercise in Chapter 5, Quadratic Equations, of Class 10 Mathematics (CBSE, Telangana & AP Board). It focuses entirely on the Nature of Roots — using a single expression called the Discriminant (D = b²−4ac) to determine whether a quadratic equation has two distinct real roots, two equal real roots, or no real roots at all, without actually solving it.

This concept is immensely powerful: instead of going through the full quadratic formula every time, you compute just one value — the discriminant — and immediately know the character of the solution. Exercise 5.4 also includes reverse problems where the value of an unknown constant k is found given that roots are equal, and real-life geometry problems where the discriminant is used to judge feasibility.

Discriminant D = b²−4ac Two Distinct Roots (D > 0) Equal Roots (D = 0) No Real Roots (D < 0) Finding k for Equal Roots

💡 Why the Discriminant matters: In Telangana & AP SSC board exams, a 2-mark or 4-mark question asking "Find the nature of roots" or "Find k so that roots are equal" appears in almost every paper. Mastering Exercise 5.4 is one of the fastest ways to secure full marks in the Quadratic Equations section.

Understanding the Nature of Roots — The Discriminant

Recall from Exercise 5.3 that the roots of ax² + bx + c = 0 are given by the quadratic formula. The key expression inside the square root — b² − 4ac — is what controls the nature of the roots. It is called the discriminant because it "discriminates" (distinguishes) between the three possible cases.

Discriminant: D = b² − 4ac
for the quadratic equation ax² + bx + c = 0 (where a ≠ 0)

✅

D > 0

Two Distinct
Real Roots

Graph cuts x-axis at two points

🟡

D = 0

Two Equal
Real Roots

Graph touches x-axis at one point

❌

D < 0

No Real Roots
(Imaginary)

Graph does not touch x-axis

Graphical Interpretation — Parabola and the X-Axis

The parabola y = ax² + bx + c tells the same story visually. Where the parabola meets the x-axis (y = 0) gives the real roots of the equation. The shape of the intersection depends directly on the discriminant.

Two distinct roots

Parabola cuts x-axis at x₁ and x₂

Two equal roots

Parabola just touches x-axis at vertex

No real roots

Parabola stays above the x-axis

📌 Concrete examples from the PDF slides:
• x² − 5x + 6 = 0 → D = 1 > 0 → roots are 2 and 3 (two distinct real roots)
• x² − 6x + 9 = 0 → D = 0 → roots are 3 and 3 (two equal real roots)
• x² − 2x + 3 = 0 → D = −8 < 0 → no real roots (imaginary)

Question 1 — Find the Nature of Roots (and Roots if they Exist)

For each equation below, we first compute D = b² − 4ac to classify the nature, then — if D ≥ 0 — apply the quadratic formula to find the actual roots.

Q1 (i)

2x² − 3x + 5 = 0

Identify: a = 2, b = −3, c = 5

Discriminant D = b² − 4ac = (−3)² − 4(2)(5) = 9 − 40 = −31 Since D = −31 < 0 → No real roots (imaginary roots)

❌ D < 0 → The equation 2x² − 3x + 5 = 0 has no real roots. The parabola lies entirely above the x-axis.

No real roots (D = −31 < 0)

Q1 (ii)

3x² − 4√3 x + 4 = 0

Identify: a = 3, b = −4√3, c = 4

Discriminant D = (−4√3)² − 4(3)(4) = 48 − 48 = 0 Since D = 0 → Two equal real roots Roots x = (−b ± 0) / 2a = 4√3 / (2×3) = 4√3 / 6 = 2√3/3 = 2/√3 ∴ Both roots = 2/√3

🟡 D = 0 → Two equal real roots. The parabola touches the x-axis at exactly one point (the vertex). Both roots are 2/√3 (or equivalently 2√3/3).

Equal roots: 2/√3, 2/√3 (D = 0)

Q1 (iii)

2x² − 6x + 3 = 0

Identify: a = 2, b = −6, c = 3

Discriminant D = (−6)² − 4(2)(3) = 36 − 24 = 12 Since D = 12 > 0 → Two distinct real roots Roots x = (6 ± √12) / (2×2) = (6 ± 2√3) / 4 = 2(3 ± √3) / 4 = (3 ± √3) / 2 ∴ x = (3 + √3)/2 and (3 − √3)/2

✅ D = 12 > 0 → Two distinct real roots. The parabola crosses the x-axis at two different points. Note how √12 simplifies to 2√3 — always simplify surds in board exam answers.

Roots: (3+√3)/2 and (3−√3)/2 (D = 12 > 0)

Question 2 — Find the Value of k so that Roots are Equal

When a problem states "roots are equal," the condition is simply D = 0, i.e., b² − 4ac = 0. Set up this equation, substitute the given a, b (or expression in k), and c, then solve for k. This is a favourite 3-mark question in Telangana and AP board exams.

🔑 Key Rule for "Equal Roots" Problems: Roots are equal ⟺ b² − 4ac = 0. This is the only condition you need. Simply set up the discriminant = 0 and solve.

Q2 (i)

2x² + kx + 3 = 0 — Find k for equal roots

Identify: a = 2, b = k, c = 3. Condition for equal roots: D = 0

Set D = 0 b² − 4ac = 0 k² − 4(2)(3) = 0 k² − 24 = 0 k² = 24 k = ±√24 = ±√(4×6) = ±2√6 ∴ k = +2√6 or k = −2√6

k = ±2√6

Q2 (ii)

kx(x − 2) + 6 = 0, k ≠ 0 — Find k for equal roots

First expand and rearrange into standard form: kx² − 2kx + 6 = 0. Now a = k, b = −2k, c = 6.

Set D = 0 b² − 4ac = 0 (−2k)² − 4(k)(6) = 0 4k² − 24k = 0 4k(k − 6) = 0 k = 0 OR k = 6 But k ≠ 0 is given (otherwise equation is not quadratic) ∴ k = 6

📌 Why reject k = 0? If k = 0, the equation becomes 0·x² − 0·x + 6 = 0, i.e., 6 = 0 — which is not a quadratic at all. The condition k ≠ 0 is stated in the problem precisely for this reason.

k = 6

Question 3 — Is it Possible to Design the Rectangular Mango Grove?

Questions 3, 4, and 5 are "is it possible?" problems — a type unique to Exercise 5.4. The strategy is always the same: model the given situation as a quadratic equation, compute the discriminant, and use D ≥ 0 (for real roots) to judge whether the situation is geometrically or physically feasible.

Rectangular Mango Grove: Length = 2 × Breadth, Area = 800 m²

Given: Let breadth = x m. Then length = 2x m. Area = length × breadth = 800 m².

Form equation 2x × x = 800 → 2x² = 800 → x² − 400 = 0 Identify a = 1, b = 0, c = −400 Discriminant D = 0² − 4(1)(−400) = 1600 D = 1600 > 0 → Two distinct real roots → Situation IS possible Find x x = (0 ± √1600) / 2 = ±40/2 = ±20 x = 20 (reject x = −20, breadth cannot be negative) ∴ Breadth = 20 m, Length = 2×20 = 40 m

✅ Yes, it is possible. A rectangular mango grove with breadth 20 m and length 40 m satisfies all the given conditions.

Breadth = 20 m, Length = 40 m

Question 4 — Is the Ages Situation Possible?

Sum of two friends' ages = 20. Four years ago, product of ages = 48. Possible?

Given: Let one friend's age = x years. Other = (20 − x) years. Four years ago: ages were (x − 4) and (16 − x).

Form equation (x − 4)(16 − x) = 48 16x − x² − 64 + 4x = 48 −x² + 20x − 64 = 48 x² − 20x + 112 = 0 Identify a = 1, b = −20, c = 112 Discriminant D = (−20)² − 4(1)(112) = 400 − 448 = −48 D = −48 < 0 → No real roots → Situation is NOT possible

❌ Not possible. There are no real values of age that satisfy the given conditions simultaneously. The discriminant is negative, confirming this mathematically.

Not possible (D = −48 < 0)

Question 5 — Is it Possible to Design the Rectangular Park?

Rectangular Park: Perimeter = 80 m, Area = 400 m². Possible?

Given: Perimeter = 80 m → 2(length + breadth) = 80 → length + breadth = 40. So length = (40 − x) if breadth = x.

Form equation x(40 − x) = 400 → 40x − x² = 400 x² − 40x + 400 = 0 Identify a = 1, b = −40, c = 400 Discriminant D = (−40)² − 4(1)(400) = 1600 − 1600 = 0 D = 0 → Two equal real roots → Situation IS possible Find x x = (40 ± 0) / 2 = 40/2 = 20 Breadth = 20 m, Length = 40 − 20 = 20 m ∴ Length = Breadth = 20 m → The park is a SQUARE!

✅ Yes, it is possible — but only if the "rectangle" is actually a square with side 20 m. D = 0 means there is exactly one solution (equal roots), which corresponds to the unique square that satisfies both conditions.

Length = Breadth = 20 m (Square) (D = 0)

Quick Reference — All Answers at a Glance

Q#	Equation / Problem	a, b, c	D = b²−4ac	Nature	Roots / Answer
Q1(i)	2x² − 3x + 5 = 0	2, −3, 5	−31	No real roots	Imaginary
Q1(ii)	3x² − 4√3 x + 4 = 0	3, −4√3, 4	0	Equal roots	2/√3, 2/√3
Q1(iii)	2x² − 6x + 3 = 0	2, −6, 3	12	Two distinct roots	(3±√3)/2
Q2(i)	2x² + kx + 3 = 0 (equal roots)	2, k, 3	Set = 0	Equal roots	k = ±2√6
Q2(ii)	kx(x−2)+6=0 (equal roots, k≠0)	k, −2k, 6	Set = 0	Equal roots	k = 6
Q3	Mango grove: length=2×breadth, area=800 m²	1, 0, −400	1600	Possible ✓	20 m × 40 m
Q4	Two friends: sum=20, product 4yrs ago=48	1, −20, 112	−48	Not possible ✗	No real solution
Q5	Rectangular park: perimeter=80, area=400 m²	1, −40, 400	0	Possible ✓ (Square)	20 m × 20 m

Common Mistakes to Avoid

Computing D correctly when b is negative: For b = −3, b² = (−3)² = +9, not −9. Squaring always gives a positive result. This is the single most common arithmetic error in discriminant problems.
Forgetting to expand before identifying a, b, c: In Q2(ii), the equation kx(x−2)+6 = 0 must first be expanded to kx²−2kx+6 = 0. Treating it as-is gives wrong values of a, b, c.
Not rejecting k = 0 in Q2(ii): The factoring gives k = 0 or k = 6, but k = 0 makes the equation non-quadratic. Always check whether any value must be rejected based on the problem constraints.
Saying "roots exist" when D = 0 without finding them: When D ≥ 0, the question says "If real roots exist, find them." Students often stop after classifying the nature without actually computing the roots — losing half the marks.
Incorrect conclusion in "Is it possible?" questions: D > 0 or D = 0 → possible; D < 0 → not possible. Some students confuse this: they say "not possible" even when D = 0 (equal roots still means real roots exist, so the situation IS possible).
Not simplifying surds: √12 should be simplified to 2√3, and 2√3/3 is the simplified form of 4√3/6. Board examiners expect fully simplified answers.

⚠️ Board Exam Alert (Telangana & AP SSC): A common 4-mark exam question combines both parts of Exercise 5.4: "Find the nature of roots of [equation]. If real roots exist, find them." You must (1) compute D, (2) state the nature, and (3) find the roots using the formula — all three steps earn marks separately.

The Discriminant — Complete Reference

Condition	Value of D	Nature of Roots	Formula Result	Graph Behaviour
D > 0	Positive	Two distinct real roots	x = (−b + √D)/2a and (−b − √D)/2a	Cuts x-axis at 2 points
D = 0	Zero	Two equal real roots	x = −b/2a (both roots the same)	Touches x-axis at 1 point
D < 0	Negative	No real roots (imaginary)	√D is not real — roots are complex	Never meets the x-axis

What This Exercise Prepares You For

Exercise 5.4 completes Chapter 5 on Quadratic Equations — the most important chapter of Class 10 Mathematics for both CBSE and state boards. The discriminant concept reappears throughout higher mathematics: in Class 11, it determines the number of roots of polynomial equations; in coordinate geometry, it is used to determine whether a line is tangent to a conic section.

Within Class 10, the techniques from this exercise directly support Chapter 6 (Real Numbers) and Chapter 10 (Circles), where the condition for equal roots corresponds to a line being tangent to a circle. For students building on this topic, see also Exercise 5.1 for standard form, Exercise 5.2 for factorisation, and Exercise 5.3 for the quadratic formula.

📐 Board Exam Tip (Telangana & AP SSC): Learn the three discriminant cases as a single "D rule" — write it on top of your answer before starting any nature-of-roots problem. Examiners award a dedicated mark just for correctly stating "since D [>/=/<] 0, the roots are [distinct/equal/imaginary]" as a conclusion line. Never skip this statement even if the calculation is obvious.