Chapter 4 of 19

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Andhra Pradesh (BIEAP)Class 10 Mathematics★ AP Board SSC Mathematics · 8–10 marks · Paper: 4-mark word problem + 4-mark formula/discriminant + 2-mark nature of roots

Quadratic Equations

Master QUADRATIC EQUATIONS — the quadratic formula, discriminant, nature of roots, and solving word problems. The most important algebra topic in AP SSC Mathematics.

⏱ 12 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Recognise standard form ax² + bx + c = 0 and rearrange any equation to it
2Solve by factorisation (splitting the middle term)
3Solve by completing the square
4Apply the quadratic formula x = [−b ± √(b²−4ac)] / 2a
5Use discriminant Δ = b²−4ac to determine nature of roots: Δ>0 (two real), Δ=0 (equal), Δ<0 (no real)
6Find k when roots satisfy given conditions (equal roots, real roots, one root given)
7Set up and solve word problems (numbers, ages, speed-distance-time, area, work)

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Why this chapter matters

Quadratic Equations is one of the most tested chapters in AP SSC Mathematics — expect 6–10 marks every paper. The quadratic formula is the most powerful algebraic tool in Class 10. The discriminant test for nature of roots is a near-certain 2-mark question. Word problems (numbers, ages, speed, area) are predictable in structure and reward students who practise systematically.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Polynomials — zeroes and coefficients (Class 10)

Sum and product of zeroes directly apply here

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Linear equations (Class 9)

Equation rearrangement and basic algebra

Quadratic Equations

"The quadratic formula solves ANY quadratic. It is your most powerful algebra tool."

Standard Form: ax² + bx + c = 0 (a ≠ 0)

Solving Methods

1. Factorisation

Find two numbers whose PRODUCT = ac and SUM = b. Split the middle term. Factor by grouping. Only works when roots are RATIONAL.

2. Quadratic Formula: x = [−b ± √(b²−4ac)] / 2a

Works for ANY quadratic. Root 1 uses (+). Root 2 uses (−).

The Discriminant — Δ = b² − 4ac

Δ > 0: TWO distinct real roots.
Δ = 0: ONE real root (equal/repeated).
Δ < 0: NO real roots (complex).
For RATIONAL roots: Δ must be a PERFECT SQUARE.

Sum and Product of Roots

If α and β are roots: α + β = −b/a. αβ = c/a.

Common Word Problem Types

Type	Approach
Numbers	Let number = x. Frame using conditions.
Speed-Distance-Time	Speed = Distance/Time.
Area	Use formulas. When dimensions change, form quadratic.

Common Mistakes

Forgetting the ±: The formula gives TWO roots.
Not setting the equation to ZERO first: All terms must be on ONE side.

The Discriminant and Nature of Roots

Δ = b² − 4ac

Δ	Nature of Roots	Example
Δ > 0, perfect square	Two distinct RATIONAL roots	x² − 5x + 6 = 0, Δ = 1
Δ > 0, not a perfect square	Two distinct IRRATIONAL roots	x² − 4x + 2 = 0, Δ = 8
Δ = 0	One real root (REPEATED / equal)	x² − 4x + 4 = 0, Δ = 0
Δ < 0	NO real roots (complex)	x² + x + 1 = 0, Δ = −3

Example 1: Find the nature of roots of 2x² − 5x + 3 = 0. Δ = (−5)² − 4×2×3 = 25 − 24 = 1 > 0 and perfect square → Two distinct rational roots. Check: 2x² − 5x + 3 = (2x−3)(x−1) = 0 → x = 1, x = 3/2.

Example 2: Find k so that x² − 6x + k = 0 has equal roots. For equal roots: Δ = 0 → b² − 4ac = 0 → 36 − 4(1)(k) = 0 → 4k = 36 → k = 9. When k = 9: x² − 6x + 9 = (x−3)² = 0 → x = 3 (repeated).

Completing the Square Method

Convert ax² + bx + c = 0 to (x + p)² = q and solve.

Steps:

Keep x² and x terms on LHS, constant on RHS.
Divide both sides by a (if a ≠ 1).
Add (b/2a)² to both sides to complete the square.
Write LHS as (x + b/2a)².
Take square root and solve.

Example 3: Solve x² + 6x − 7 = 0 by completing the square. x² + 6x = 7. Half of 6 = 3, square = 9. x² + 6x + 9 = 7 + 9 = 16. (x + 3)² = 16 → x + 3 = ±4 → x = 1 or x = −7.

Example 4: Solve 2x² − 8x + 5 = 0 by completing the square. 2(x² − 4x) = −5. x² − 4x = −5/2. Half of 4 = 2, square = 4. x² − 4x + 4 = −5/2 + 4 = 3/2. (x − 2)² = 3/2 → x − 2 = ±√(3/2) → x = 2 ± √(3/2) = 2 ± √6/2 = (4 ± √6)/2.

Quadratic Formula Method

x = [−b ± √(b²−4ac)] / 2a

This formula works for ALL quadratic equations. 'Memorise this formula. It is your MOST POWERFUL tool for quadratics.'

Example 5: Solve 2x² + 3x − 5 = 0 using the formula. a = 2, b = 3, c = −5. Δ = 9 − 4(2)(−5) = 9 + 40 = 49. x = [−3 ± √49] / 4 = [−3 ± 7] / 4. x = 4/4 = 1 or x = −10/4 = −5/2.

Example 6: Solve 3x² − 5x + 1 = 0 (roots are irrational). Δ = 25 − 12 = 13. x = [5 ± √13] / 6. Roots: (5+√13)/6 and (5−√13)/6.

Sum and Product of Roots

If α and β are roots of ax² + bx + c = 0:

α + β = −b/a
αβ = c/a

Example 7: Find the sum and product of roots of 3x² − 6x + 2 = 0. α+β = −(−6)/3 = 2. αβ = 2/3.

Example 8: Find k if one root of x² − 5x + k = 0 is 2. Since x = 2 is a root: (2)² − 5(2) + k = 0 → 4 − 10 + k = 0 → k = 6. Check: x² − 5x + 6 = (x−2)(x−3) = 0 → roots are 2 and 3 ✓.

Word Problems — Detailed

Number Problems

Example 9: Find two consecutive positive integers whose product is 306. Let integers be x and x+1. x(x+1) = 306 → x² + x − 306 = 0. Δ = 1 + 1224 = 1225 = 35². x = [−1 ± 35] / 2. x = 17 or x = −18 (reject, positive required). Integers: 17 and 18. Check: 17×18 = 306 ✓.

Example 10: The sum of the squares of two consecutive odd numbers is 290. Find the numbers. Let numbers be 2x+1 and 2x+3. (2x+1)² + (2x+3)² = 290. 4x²+4x+1 + 4x²+12x+9 = 290 → 8x²+16x+10 = 290 → 8x²+16x−280 = 0. Divide by 8: x²+2x−35 = 0 → (x+7)(x−5) = 0 → x = 5 (positive). Numbers: 11 and 13. Check: 121+169 = 290 ✓.

Age Problems

Example 11: The product of the ages of a man and his son is 240. Four years hence, the man's age will be twice the son's age. Find their present ages. Let son's age = x. Man's age = 240/x. After 4 years: 240/x + 4 = 2(x+4) → 240/x + 4 = 2x + 8 → 240/x = 2x + 4. Multiply: 240 = 2x² + 4x → 2x² + 4x − 240 = 0 → x² + 2x − 120 = 0. (x+12)(x−10) = 0 → x = 10 (positive). Man's age = 240/10 = 24. Son = 10, man = 24. Check: 4 years hence: 28 = 2×14 ✓.

Speed-Distance Problems

Example 12: A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, the journey would have taken 1 hour less. Find the speed of the train. Let speed = x km/h. Time = 360/x hours. New speed = x+5 km/h. New time = 360/(x+5). 360/x − 360/(x+5) = 1. 360(x+5) − 360x = x(x+5) → 360x+1800−360x = x²+5x → x²+5x−1800 = 0. (x+45)(x−40) = 0 → x = 40 (positive). Speed = 40 km/h. Check: Original time = 360/40 = 9 h. New time = 360/45 = 8 h. Difference = 1 h ✓.

Work Problems

Example 13: Two pipes together can fill a tank in 6 hours. The larger pipe alone takes 5 hours less than the smaller pipe. Find the time each pipe takes alone. Let smaller pipe take x hours. Larger pipe takes x−5 hours. In 1 hour: smaller fills 1/x, larger fills 1/(x−5). Together: 1/x + 1/(x−5) = 1/6. Multiply: 6(x−5+x) = x(x−5) → 6(2x−5) = x²−5x → 12x−30 = x²−5x → x²−17x+30 = 0. (x−15)(x−2) = 0 → x = 15 (x=2 rejected, x−5 = −3 not possible). Smaller pipe = 15 h, larger pipe = 10 h. Check: 1/15 + 1/10 = 2/30+3/30 = 5/30 = 1/6 ✓.

AP Exam Focus

Topic	Marks	Frequency
Quadratic formula / factorization	4	Very Common
Nature of roots (discriminant)	3	Very Common
Word problems (numbers/age)	5	Common
Word problems (speed-distance)	5	Common
Completing the square	3-4	Moderate
Sum and product of roots	3	Moderate

Self-Test Questions

Solve: 3x² − 10x + 8 = 0. (Answer: x = 2, x = 4/3)
Find the value of k for which x² + kx + 25 = 0 has equal roots. (Answer: k = ±10)
The sum of the squares of two consecutive even numbers is 244. Find them. (Answer: 10 and 12)
A ball is thrown upwards. Its height h(t) = −5t² + 40t. When will it hit the ground? (Answer: h=0 → −5t(t−8)=0 → t=8 seconds)
Solve by completing the square: x² + 8x − 9 = 0. (Answer: (x+4)²=25 → x=1 or x=−9)
Find the discriminant of 2x² + x − 3 = 0 and state the nature of roots. (Answer: Δ=1+24=25 > 0, perfect square → two distinct rational roots)
A rectangular garden has area 288 m². Its length is 4 m more than its breadth. Find dimensions. (Answer: b(b+4)=288 → b²+4b−288=0 → b=16, l=20) (Answer: 16 m × 20 m)

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Standard form

ax² + bx + c = 0, a ≠ 0

MUST bring all terms to one side before applying any method.

Quadratic formula

x = [−b ± √(b² − 4ac)] / 2a

Works for ANY quadratic. The ± gives two roots. Memorise this formula — it is the most important formula in Class 10 algebra.

Discriminant

Δ = b² − 4ac

Δ > 0: two distinct real roots. Δ = 0: two equal real roots. Δ < 0: no real roots. Δ is a perfect square: rational roots.

Equal roots condition

Δ = 0 → b² = 4ac

Used to find k when 'equal roots' is the condition. Solve b² − 4ac = 0 for k.

Sum of roots

α + β = −b/a

Same formula as for polynomial zeroes.

Product of roots

αβ = c/a

No sign change.

Factorisation shortcut

Find p, q such that p + q = b and p × q = ac; split bx as px + qx

Then factor by grouping. Works when roots are rational.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Not moving ALL terms to one side before applying the formula

✓ ax² + bx + c = 0 MUST have zero on the RHS. If the problem gives 3x² = 5x − 2, rewrite as 3x² − 5x + 2 = 0 FIRST.

WATCH OUT

✗ Forgetting the ± in the quadratic formula (writing only + or only −)

✓ The ± gives TWO roots. Always write BOTH: x₁ = (−b + √Δ)/2a and x₂ = (−b − √Δ)/2a.

WATCH OUT

✗ Forgetting to reject the negative/unrealistic root in word problems

✓ In physical problems (speed, age, length, area), negative values are almost always invalid. Write: 'Since [quantity] cannot be negative, x = [negative value] is rejected. Therefore x = [positive value].' AP board awards a mark for this step.

WATCH OUT

✗ In 'find k for equal roots', confusing the formula: using b² + 4ac = 0 instead of b² − 4ac = 0

✓ Discriminant = b² − 4ac (MINUS). Equal roots: b² − 4ac = 0, so b² = 4ac. The subtraction is the definition.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Quadratic Equations?

7 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

7 questions~5 min worth ~10 marks in Andhra Pradesh (BIEAP) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Standard form: ax² + bx + c = 0. ALWAYS bring all terms to one side first.
•Factorisation: find p, q with pq = ac and p+q = b; split middle term and factor by grouping.
•Completing the square: move constant to RHS; add (b/2a)² to both sides; write LHS as perfect square; take ±√.
•Quadratic formula: x = [−b ± √(b²−4ac)] / 2a. The ± gives two roots.
•Discriminant Δ = b² − 4ac: Δ>0 (two real), Δ=0 (equal/repeated), Δ<0 (no real), Δ=perfect square (rational roots).
•Equal roots: Δ = 0 → b² = 4ac. Solve for the unknown k.
•Sum of roots = −b/a. Product of roots = c/a.
•Word problems: always REJECT negative roots for physical quantities (speed, age, length, area). Write the rejection explicitly.
•For 'find k for real roots': need Δ ≥ 0. For 'find k for equal roots': need Δ = 0. For 'find k for no real roots': need Δ < 0.
•Verify your answer by substituting BOTH roots back into the original equation.

Andhra Pradesh (BIEAP) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 8–10 marks

Question type	Marks each	Typical count	What it tests
Very Short Answer (VSA)	1	2	Definition, formula recall
Short Answer (SA)	2	2	Direct application
Long Answer (LA)	4	1	Multi-step problem with full working

Prep strategy

AP SSC board rewards clearly shown working — never skip steps
Memorise all formulas: AP board MCQs test formula recall directly
Practise 4-mark questions with full working under timed conditions
For word problems: identify variables, write equation, solve, state answer with units

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Projectile motion in physics

The height of a thrown ball h(t) = −½gt² + v₀t + h₀ is a quadratic. Setting h = 0 finds when it lands.

Area and perimeter problems

Rectangles with a given area lead to quadratic equations when one dimension is expressed in terms of the other.

Profit maximisation in business

Revenue = price × quantity. When price depends on quantity, Revenue becomes quadratic. The maximum is at the vertex.

Lens formula in optics

1/f = 1/v − 1/u can be rearranged to a quadratic when two of the three quantities involve the same unknown.

Signal timing in circuits

RC charging/discharging circuits lead to quadratic equations when finding the time to reach a specific voltage level.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Factorisation check: compute Δ first. If Δ is a perfect square, factorisation will work cleanly. If not, use the formula.
Word problems: write 'Let x = ...' with units. Set up the equation explicitly. Solve. Reject invalid roots WITH a written reason. Verify in the original word condition.
For 'find k' problems: write the discriminant expression fully. Set it to 0 (equal roots) or ≥ 0 (real roots). Solve the resulting quadratic or inequality.
Completing the square in exam: move constant, add (b/2)² inside (for a=1) or (b/2a)² for general a. Write each step — examiners award step marks.
Time management: factorisation takes ~1 minute; formula takes ~2 minutes. Use factorisation when ac is small and factors are obvious.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

The vertex of the parabola y = ax² + bx + c is at x = −b/2a, y = c − b²/4a. The vertex gives the maximum or minimum value — important in optimization problems.
Vieta's formulas for quadratics: α + β = −b/a and αβ = c/a. Can be used to find sums of powers: α² + β² = (α+β)² − 2αβ; α³ + β³ = (α+β)³ − 3αβ(α+β).
The quadratic formula can be derived by completing the square on ax² + bx + c = 0 — a valuable proof exercise for EAMCET preparation.
In competitive maths: if α is a root of ax² + bx + c = 0, then α is also a root of cx² + bx + a = 0 (reciprocal equation). This is used in EAMCET problems.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

AP Board SSC6-10 marks guaranteed — quadratic formula, discriminant, and word problems in every paper

EAMCETVery high — quadratic equations appear in algebra, and roots are needed for polynomial questions

JEE FoundationVery high — discriminant conditions, nature of roots, and quadratic inequalities

IMO FoundationQuadratic Diophantine equations (integer solutions) are common olympiad problems

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

The square root term √(b²−4ac) determines whether the formula gives real or complex numbers. If Δ < 0, the square root is imaginary, so no real roots. If Δ = 0, both roots are the same (−b/2a). If Δ > 0, two different real values.

Yes! The formula ALWAYS works. Factorisation is faster when roots are nice integers or simple fractions, but the formula handles irrational and messy roots equally well.

It converts the standard form ax² + bx + c = 0 into (x + p)² = q, which can be solved by square roots. The quadratic formula is actually derived by completing the square on ax² + bx + c = 0.

If the problem involves PRODUCTS of unknowns, SQUARES, or quantities like area (l × w), time × rate, then you need a quadratic. If the relationship is purely additive, a linear equation suffices.

Say 'no REAL solution'. The equation still has two COMPLEX (imaginary) solutions. For AP SSC level, 'no real roots' is the correct answer.

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