Exercise 5.3 — Completing the Square

Solving quadratic equations by completing the square.

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Exercise 5.3 — Quadratic Equations: Completing the Square & Quadratic Formula

Exercise 5.3 is the most important exercise in Chapter 5, Quadratic Equations, of Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus). It covers two powerful methods for solving any quadratic equation — Completing the Square and the famous Quadratic Formula. These methods work even when factorisation fails, making them essential tools for every Class 10 board exam.

This exercise contains 13 problems spread across four question types: finding roots by completing the square, verifying using the quadratic formula, solving fractional and rational equations, and applying quadratic equations to real-life word problems involving age, speed, area, and geometry.

Completing the Square Quadratic Formula Word Problems Discriminant (b²−4ac) Real-life Applications

💡 Why this exercise matters: The Quadratic Formula derived in this lesson can solve any quadratic equation. It appears every year in Telangana SSC and AP SSC board exams, usually as a 4-mark or 5-mark question.

Deriving the Quadratic Formula by Completing the Square

Before solving problems, it is crucial to understand why the quadratic formula works. This derivation starts from the general equation ax² + bx + c = 0 and transforms it step-by-step into the formula. Every board exam student should know this derivation — it is frequently asked as a "Prove that…" question.

Starting equation: ax² + bx + c = 0 (where a ≠ 0)

Multiply both sides by 4a — this clears fractions and prepares for the perfect square.

4a²x² + 4abx + 4ac = 0

Add b² to both sides — this sets up the left side as a perfect square trinomial.

4a²x² + 4abx + b² = b² − 4ac

Recognise the perfect square — the left side fits the pattern (P + Q)² = P² + 2PQ + Q².

(2ax + b)² = b² − 4ac

Take square root on both sides — remember to include both ± since squaring loses sign information.

2ax + b = ±√(b² − 4ac)

Isolate x — shift b to the right and divide everything by 2a.

x = (−b ± √(b² − 4ac)) / 2a

The Quadratic Formula:
x = (−b ± √(b² − 4ac)) / 2a
Valid when b² − 4ac ≥ 0 | If b² − 4ac < 0 → No real roots

📌 The Discriminant (D = b² − 4ac) tells you the nature of roots before you solve:
• D > 0 → Two distinct real roots | D = 0 → Two equal (repeated) roots | D < 0 → No real roots

Question 1 — Finding Roots by Completing the Square

In Question 1, each equation is solved using the Completing the Square method directly — without using the formula. The technique involves multiplying by 4a, adding b² to both sides, forming a perfect square on the left, and then taking the square root. This method builds deep algebraic understanding and is important for board exam proofs.

Q1 (i)

2x² + x − 4 = 0

Here a = 2, b = 1, c = −4. So 4a = 8 and b² = 1.

Step 1 Multiply by 8: 16x² + 8x − 32 = 0 Step 2 Move constant: 16x² + 8x = 32 Step 3 Add b² = 1 to both sides: 16x² + 8x + 1 = 33 Step 4 Perfect square: (4x + 1)² = 33 Step 5 Square root: 4x + 1 = ±√33 Step 6 Solve: x = (−1 ± √33) / 4 ∴ Roots = (−1 + √33)/4 and (−1 − √33)/4

x = (−1 + √33)/4 , (−1 − √33)/4

Q1 (ii)

4x² + 4√3 x + 3 = 0

Here a = 4, b = 4√3, c = 3. Notice that this is already a perfect square — the left side equals (2x + √3)².

Step 1 Recognise pattern: (2x)² + 2(2x)(√3) + (√3)² = 0 Step 2 Perfect square: (2x + √3)² = 0 Step 3 Square root: 2x + √3 = 0 Step 4 Solve: x = −√3/2 ∴ Equal roots: x = −√3/2 (repeated)

✅ Key observation: When D = b² − 4ac = 48 − 48 = 0, the equation has equal roots. Both roots are the same: −√3/2.

x = −√3/2 , −√3/2

Q1 (iii)

5x² − 7x − 6 = 0

Here a = 5, b = −7, c = −6. So 4a = 20 and b² = 49.

Step 1 Multiply by 20: 100x² − 140x − 120 = 0 Step 2 Move constant: 100x² − 140x = 120 Step 3 Add b² = 49: 100x² − 140x + 49 = 169 Step 4 Perfect square: (10x − 7)² = 169 Step 5 Square root: 10x − 7 = ±13 Step 6 x = (7+13)/10 = 20/10 = 2 OR x = (7−13)/10 = −6/10 = −3/5 ∴ Roots = 2 and −3/5

x = 2 , −3/5

Q1 (iv)

x² + 5 = −6x → Rearrange first: x² + 6x + 5 = 0

Always rearrange to standard form before identifying a, b, c. Here a = 1, b = 6, c = 5. So 4a = 4 and b² = 36.

Step 1 Multiply by 4: 4x² + 24x + 20 = 0 Step 2 Move constant: 4x² + 24x = −20 Step 3 Add b² = 36: 4x² + 24x + 36 = 16 Step 4 Perfect square: (2x + 6)² = 16 Step 5 Square root: 2x + 6 = ±4 Step 6 2x = −6+4 = −2 OR 2x = −6−4 = −10 ∴ Roots = −1 and −5

x = −1 , −5

Question 2 — Same Equations, Now Using the Quadratic Formula

Question 2 solves the same four equations from Q1 by directly applying x = (−b ± √(b²−4ac)) / 2a. This cross-checks the answers and shows that both methods give identical results. In board exams, you may use either method unless the question specifies one.

🔢 Completing the Square

Builds deep understanding
Required when "Prove the formula" is asked
Slightly more steps
Useful for deriving the discriminant

⚡ Quadratic Formula

Fastest method in exams
Works for all quadratic equations
Just substitute a, b, c — done!
Best when factorisation is unclear

Q2 (i)

2x² + x − 4 = 0 | a=2, b=1, c=−4

x = (−1 ± √(1² − 4×2×(−4))) / (2×2) = (−1 ± √(1 + 32)) / 4 = (−1 ± √33) / 4 ∴ x = (−1 + √33)/4 and (−1 − √33)/4 ✓

Same as Q1(i) — confirmed ✓

Q2 (ii)

4x² + 4√3 x + 3 = 0 | a=4, b=4√3, c=3

x = (−4√3 ± √((4√3)² − 4×4×3)) / (2×4) = (−4√3 ± √(48 − 48)) / 8 = (−4√3 ± √0) / 8 = −4√3 / 8 ∴ x = −√3/2 (equal roots) ✓

Same as Q1(ii) — D = 0, equal roots ✓

Q2 (iii)

5x² − 7x − 6 = 0 | a=5, b=−7, c=−6

x = (7 ± √(49 − 4×5×(−6))) / (2×5) = (7 ± √(49 + 120)) / 10 = (7 ± √169) / 10 = (7 ± 13) / 10 x = (7+13)/10 = 2 OR x = (7−13)/10 = −3/5 ∴ x = 2 and −3/5 ✓

Same as Q1(iii) — confirmed ✓

Q2 (iv)

x² + 6x + 5 = 0 | a=1, b=6, c=5

x = (−6 ± √(36 − 20)) / 2 = (−6 ± √16) / 2 = (−6 ± 4) / 2 x = (−6+4)/2 = −1 OR x = (−6−4)/2 = −5 ∴ x = −1 and −5 ✓

Same as Q1(iv) — confirmed ✓

Question 3 — Solving Equations That Reduce to Quadratics

Question 3 involves equations that look different from standard quadratics but can be converted into one by cross-multiplication and simplification. These are commonly asked in boards to test whether students can reduce and solve.

Q3 (i)

x − 1/x = 3, x ≠ 0

First multiply through by x to eliminate the denominator.

Convert x(x − 1/x) = 3x → x² − 1 = 3x → x² − 3x − 1 = 0 Values a = 1, b = −3, c = −1 Formula x = (3 ± √(9 + 4)) / 2 = (3 ± √13) / 2 ∴ x = (3 + √13)/2 and (3 − √13)/2

x = (3 + √13)/2 , (3 − √13)/2

Q3 (ii)

1/(x+4) − 1/(x−7) = 11/30, x ≠ −4, 7

Cross-multiply and simplify the left side first.

Step 1 LHS = (x−7 − x−4) / (x+4)(x−7) = −11 / (x²−3x−28) Step 2 So: −11/(x²−3x−28) = 11/30 Step 3 Cross-multiply: −11×30 = 11×(x²−3x−28) Step 4 −30 = x²−3x−28 → x²−3x+2 = 0 Formula x = (3 ± √(9−8)) / 2 = (3 ± 1) / 2 x = 4/2 = 2 OR x = 2/2 = 1 ∴ x = 2 and x = 1

x = 2 , 1

Questions 4 to 13 — Real-Life Word Problems

The remaining problems apply quadratic equations to practical situations. These word problems are a key part of the Telangana SSC and AP SSC board exam pattern — a 5-mark word problem from this section appears almost every year. The key skill is forming the correct equation from the given information, then solving it.

Rehman's Age Problem

Given: The sum of reciprocals of Rehman's age 3 years ago and 5 years from now is 1/3. Find his present age.

Set up Let present age = x years 3 years ago: (x−3), After 5 years: (x+5) Equation: 1/(x−3) + 1/(x+5) = 1/3 Simplify (2x+2)/(x²+2x−15) = 1/3 3(2x+2) = x²+2x−15 → x²−4x−21 = 0 Formula x = (4 ± √(16+84))/2 = (4 ± 10)/2 x = 7 OR x = −3 (age cannot be negative) ∴ Rehman's present age = 7 years

Present age = 7 years

Moulika's Marks Problem

Given: Sum of Maths and English marks = 30. If she got 2 more in Maths and 3 less in English, product = 210.

Set up Let Maths marks = x. English = 30−x Modified Maths = x+2, Modified English = 27−x Equation: (x+2)(27−x) = 210 Expand 27x−x²+54−2x = 210 → x²−25x+156 = 0 Formula x = (25 ± √(625−624))/2 = (25 ± 1)/2 x = 13 → Maths=13, English=17 OR x=12 → Maths=12, English=18 ∴ Two solutions: (13, 17) or (12, 18)

Maths=13, English=17 OR Maths=12, English=18

Rectangular Field — Diagonal Problem

Given: Diagonal = shorter side + 60 m. Longer side = shorter side + 30 m. Find both sides.

Set up Let shorter side = x m. Longer side = x+30. Diagonal = x+60. Pythagoras x² + (x+30)² = (x+60)² x² + x²+60x+900 = x²+120x+3600 x²−60x−2700 = 0 → (x−90)(x+30) = 0 x = 90 OR x = −30 (length cannot be negative) ∴ Shorter side = 90 m, Longer side = 120 m

Shorter side = 90 m, Longer side = 120 m

Difference of Squares of Two Numbers = 180

Given: Difference of squares of two numbers = 180. Square of the smaller = 8 × larger. Find both numbers.

Set up Let larger number = x. Square of smaller = 8x. Difference of squares: x² − 8x = 180 x²−8x−180 = 0 → (x−18)(x+10) = 0 x = 18 OR x = −10 (x must be positive) Smaller² = 8×18 = 144 → smaller = √144 = 12 ∴ The two numbers are 18 and 12

18 and 12

Train Speed Problem

Given: A train travels 360 km at uniform speed. If speed was 5 km/h more, it would take 1 hour less. Find the speed.

Set up Let speed = x km/h. Time = 360/x hours. At speed (x+5): time = 360/(x+5) Equation: 360/x − 360/(x+5) = 1 360×5 / x(x+5) = 1 → x²+5x−1800 = 0 x = (−5 ± √(25+7200))/2 = (−5 ± √7225)/2 = (−5 ± 85)/2 x = 40 OR x = −45 (speed cannot be negative) ∴ Speed of train = 40 km/h

Speed = 40 km/h

Two Water Taps Filling a Tank

Given: Both taps together fill a tank in 9⅜ hours = 75/8 hours. Larger tap takes 10 hours less than smaller. Find each tap's time.

Set up Let smaller tap alone take x hours. Larger = x−10 hours. Part filled per hour: 1/x + 1/(x−10) = 8/75 75(2x−10) = 8x(x−10) 150x−750 = 8x²−80x → 4x²−115x+375 = 0 4x(x−25)−15(x−25) = 0 → (x−25)(4x−15) = 0 x = 25 OR x = 15/4 (x=15/4 gives negative time for larger tap — invalid) ∴ Smaller tap = 25 hours, Larger tap = 15 hours

Smaller tap: 25 hrs, Larger tap: 15 hrs

Q10

Express vs Passenger Train (Mysore–Bengaluru)

Given: Express takes 1 hour less than passenger to cover 132 km. Express speed = passenger speed + 11 km/h.

Set up Let passenger speed = x km/h. Express = x+11 km/h. 132/x − 132/(x+11) = 1 132×11 = x(x+11) → x²+11x−1452 = 0 x(x+44)−33(x+44) = 0 → (x+44)(x−33) = 0 x = 33 OR x = −44 (speed cannot be negative) ∴ Passenger train = 33 km/h, Express train = 44 km/h

Passenger: 33 km/h, Express: 44 km/h

Q11

Sum of Areas of Two Squares = 468 m²

Given: Sum of areas of two squares = 468 m². Difference of perimeters = 24 m. Find the sides.

Set up Let side of first square = x m. Perimeter = 4x. Second perimeter = 4x+24 → second side = x+6 m Equation: x² + (x+6)² = 468 2x²+12x+36 = 468 → x²+6x−216 = 0 (x+18)(x−12) = 0 → x = 12 or x = −18 (rejected) ∴ First square = 12 m, Second square = 18 m

Sides: 12 m and 18 m

Q12

Object Thrown Upwards — Time to Hit Ground

Given: Height from ground: S = 96 + 80t − 16t². Find t when S = 0.

Set up When object hits ground, S = 0: −16t² + 80t + 96 = 0 → −16(t²−5t−6) = 0 t²−5t−6 = 0 → (t−6)(t+1) = 0 t = 6 OR t = −1 (time cannot be negative) ∴ Object hits ground after 6 seconds

t = 6 seconds

Q13

Number of Diagonals in a Polygon

Given: A polygon with n sides has n(n−3)/2 diagonals. (a) Find n if diagonals = 65. (b) Is there a polygon with 50 diagonals?

Part a n(n−3)/2 = 65 → n²−3n−130 = 0 n = (3 ± √(9+520))/2 = (3 ± √529)/2 = (3 ± 23)/2 n = 13 OR n = −10 (rejected — sides must be positive integer) ∴ Polygon with 65 diagonals has 13 sides Part b n(n−3)/2 = 50 → n²−3n−100 = 0 n = (3 ± √409)/2 √409 is not a whole number → n is not an integer ∴ No polygon can have exactly 50 diagonals

Part a: 13 sides | Part b: No such polygon exists

Quick Reference — All Answers at a Glance

Q#	Equation / Problem	Method	Answer / Roots
Q1(i)	2x² + x − 4 = 0	Completing Square	(−1±√33)/4
Q1(ii)	4x² + 4√3 x + 3 = 0	Completing Square	−√3/2, −√3/2
Q1(iii)	5x² − 7x − 6 = 0	Completing Square	2, −3/5
Q1(iv)	x² + 5 = −6x	Completing Square	−1, −5
Q2(i–iv)	Same as Q1 using formula	Quadratic Formula	Same results ✓
Q3(i)	x − 1/x = 3	Reduce + Formula	(3±√13)/2
Q3(ii)	1/(x+4) − 1/(x−7) = 11/30	Cross-multiply + Formula	2, 1
Q4	Rehman's age (reciprocals sum = 1/3)	Word problem	7 years
Q5	Moulika's marks (sum=30, product=210)	Word problem	(13,17) or (12,18)
Q6	Rectangular field sides	Pythagoras + QE	90 m, 120 m
Q7	Difference of squares = 180	Word problem	18 and 12
Q8	Train, 360 km, 5 km/h faster = 1 hr less	Speed-Time	40 km/h
Q9	Two taps fill tank in 75/8 hrs	Work-Rate	25 hrs, 15 hrs
Q10	Express vs passenger, 132 km	Speed-Time	33 km/h, 44 km/h
Q11	Two squares, areas sum = 468 m²	Area + QE	12 m, 18 m
Q12	S = 96 + 80t − 16t², find t at S=0	Projectile Motion	6 seconds
Q13a	Polygon with 65 diagonals	Diagonal formula	13 sides
Q13b	Polygon with 50 diagonals?	Check discriminant	Not possible

Common Mistakes to Avoid

Forgetting to rearrange first: Always convert to standard form ax² + bx + c = 0 before identifying a, b, and c. In Q1(iv), failing to rearrange x² + 5 = −6x leads to wrong values of b and c.
Wrong sign for b in the formula: The formula uses −b, not b. When b is already negative (like b = −7 in Q1(iii)), −b = +7. This sign error is the most frequent mistake in board exams.
Forgetting ± when taking square root: √(x²) = ±x, not just +x. Missing the ± sign gives only one root instead of two.
Accepting negative values in context: In word problems involving age, speed, time, or length — always reject the negative root with a clear statement: "since age/speed/length cannot be negative."
Incomplete perfect square in completing-the-square method: You must add b² (not b²/4) when you have first multiplied the equation by 4a. Students sometimes add b²/4 without doing the 4a multiplication step, leading to wrong answers.
Assuming non-integer discriminant means error: In Q13(b), √409 is not a whole number — that is the intended result, proving no polygon with 50 diagonals can exist.

⚠️ Board Exam Alert (Telangana & AP SSC): The most frequently asked 5-mark question from this exercise is the train speed problem (Q8) or a work-rate problem (Q9). Practice setting up the equation from scratch — not just the algebraic solving part. Marks are often awarded for the correct equation formation even if the final answer has an arithmetic slip.

What This Exercise Prepares You For

Exercise 5.3 builds the core skill of solving any quadratic equation — a tool you will use in every remaining chapter of Class 10 mathematics. The quadratic formula reappears directly in Chapter 8 (Exercise 5.4 — Nature of Roots) and in trigonometric identities involving quadratic relationships.

Word problem techniques from Q8–Q11 (forming equations from speed-time and area contexts) directly prepare you for problems in the Real Numbers and Statistics chapters. For students revising from Class 9, the foundation of forming equations was built in Chapter 5 Introduction and Exercise 5.1.

📐 Board Exam Tip (Telangana & AP SSC): Know the derivation of the quadratic formula by heart — it is frequently asked as a 2-mark "derive" or "prove" question. Write all 5 steps clearly: multiply by 4a → add b² → form perfect square → take square root → isolate x. Each step earns partial marks.