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NCERT Class 10 Mathematics — Chapter 4

Quadratic Equations

Standard form, factorization by splitting middle term, derivation of the quadratic formula, discriminant nature of roots, and multi-step real-world word problems.

Quick Key Takeaways:
Standard Form: ax2+bx+c=0ax^2 + bx + c = 0, where a,b,c∈Ra, b, c \in \mathbb{R} and a≠0a \neq 0.
Quadratic Formula (Sridharacharya Formula): x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Discriminant Continuum (D=b2−4acD = b^2 - 4ac): (1) D>0  ⟹  D > 0 \implies Two distinct real roots; (2) D=0  ⟹  D = 0 \implies Two equal real roots (x=−b2ax = -\frac{b}{2a}); (3) D<0  ⟹  D < 0 \implies No real roots.
Word Problem Constraints: Discard extraneous roots when variables represent physical quantities like speed (>0> 0), time (>0> 0), length (>0> 0), or age.
Graph & Coordinate VisualizerVisualize quadratic parabolas, vertex points, and real zeroes on the Cartesian grid
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Jump to Concept:1. 1. Quadratic Equation Architecture, Discriminant Laws & Root Properties2. 2. Nature of Roots ($D = b^2 - 4ac$) & Condition for Equal Roots3. 3. Important Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. CBSE Case-Study Modeling: Two Water Taps Filling a Swimming Pool5. 5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric
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1. Quadratic Equation Architecture, Discriminant Laws & Root Properties

Axioms & Foundational Theory

Comprehensive mathematical foundations, definitions, formulas, and visual conceptual model for Quadratic Equations.

• The Quadratic Formula Derivation (Completing the Square)
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Starting Equation: ax2+bx+c=0ax^2 + bx + c = 0 (a≠0a \neq 0).
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Step 1: Divide throughout by aa: x2+bax+ca=0  ⟹  x2+bax=−cax^2 + \frac{b}{a}x + \frac{c}{a} = 0 \implies x^2 + \frac{b}{a}x = -\frac{c}{a}
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Step 2: Add (b2a)2=b24a2\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} to both sides to complete the square:
x2+2(b2a)x+(b2a)2=b24a2−cax^2 + 2\left(\frac{b}{2a}\right)x + \left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a}
(x+b2a)2=b2−4ac4a2\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}
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Step 3: Taking square root on both sides:
x+b2a=±b2−4ac2a  ⟹  x=−b±b2−4ac2ax + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a} \implies \mathbf{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}
📊 Quadratic Equations: Discriminant & Nature of RootsVisual Model
DISCRIMINANT: D = b² - 4ac (for ax² + bx + c = 0)D > 0Two Real & DistinctRootsx = (-b ± √D) / 2aD = 0Two Real & EqualRoots (Coincident)x = -b / 2aD < 0No Real Roots(Imaginary / Complex)√Negative is not real

Visual schematic mapping the discriminant continuum ($D > 0, D = 0, D < 0$) to physical root behaviors.

2

2. Nature of Roots ($D = b^2 - 4ac$) & Condition for Equal Roots

Theorem Proofs & Derivations

Step-by-step rigorous geometric and algebraic theorem proofs and formula derivations for Quadratic Equations.

• The Three Discriminant Regimes
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Case 1 (D=b2−4ac>0D = b^2 - 4ac > 0): Two distinct real roots: x1=−b+D2a,x2=−b−D2ax_1 = \frac{-b + \sqrt{D}}{2a}, x_2 = \frac{-b - \sqrt{D}}{2a}.
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Case 2 (D=b2−4ac=0D = b^2 - 4ac = 0): Two equal real roots (coincident): x1=x2=−b2ax_1 = x_2 = -\frac{b}{2a}. Perfect square trinomial.
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Case 3 (D=b2−4ac<0D = b^2 - 4ac < 0): Negative under square root   ⟹  \implies No real roots (roots are complex conjugates).
3

3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

Topper Step Solutions

Standard CBSE board exam numericals and riders with complete step-by-step mathematical working and justifications.

• 3-Mark Standard Board Question: Find the values of kk for which the quadratic equation (k−12)x2+2(k−12)x+2=0(k - 12)x^2 + 2(k - 12)x + 2 = 0 has equal roots.
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Step 1 (Coefficients):
a=(k−12),b=2(k−12),c=2a = (k - 12), b = 2(k - 12), c = 2. Condition: a≠0  ⟹  k≠12a \neq 0 \implies k \neq 12.
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Step 2 (Equal Roots Condition): D=b2−4ac=0D = b^2 - 4ac = 0.
[2(k−12)]2−4(k−12)(2)=0[2(k - 12)]^2 - 4(k - 12)(2) = 0
4(k−12)2−8(k−12)=04(k - 12)^2 - 8(k - 12) = 0
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Step 3 (Factoring):
4(k−12)[(k−12)−2]=04(k - 12)[(k - 12) - 2] = 0
4(k−12)(k−14)=04(k - 12)(k - 14) = 0
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Step 4 (Evaluating Roots):
k=12k = 12 or k=14k = 14.
Since k=12k = 12 makes the coefficient of x2x^2 zero (equation ceases to be quadratic), we must reject k=12k = 12.
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Final Boxed Answer: k=14\mathbf{k = 14}
• 5-Mark Heavyweight Board Problem / Rider: An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
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Step 1 (Variables Setup):
Let average speed of passenger train =x= x km/h.
Then average speed of express train =(x+11)= (x + 11) km/h.
Total distance =132= 132 km.
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Step 2 (Time Equations):
Time taken by passenger train t1=132xt_1 = \frac{132}{x} hours.
Time taken by express train t2=132x+11t_2 = \frac{132}{x + 11} hours.
Given: t1−t2=1t_1 - t_2 = 1.
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Step 3 (Algebraic Formulation):
132x−132x+11=1\frac{132}{x} - \frac{132}{x + 11} = 1
132[(x+11)−xx(x+11)]=1132\left[\frac{(x + 11) - x}{x(x + 11)}\right] = 1
132(11x2+11x)=1  ⟹  x2+11x−1452=0132\left(\frac{11}{x^2 + 11x}\right) = 1 \implies x^2 + 11x - 1452 = 0
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Step 4 (Solving Quadratic Equation):
Product =−1452= -1452, Sum =11= 11. Factors: +44+44 and −33-33.
x2+44x−33x−1452=0x^2 + 44x - 33x - 1452 = 0
x(x+44)−33(x+44)=0  ⟹  (x+44)(x−33)=0x(x + 44) - 33(x + 44) = 0 \implies (x + 44)(x - 33) = 0
x=33orx=−44x = 33 \quad \text{or} \quad x = -44
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Step 5 (Feasibility Check):
Speed cannot be negative, so we reject x=−44x = -44.
Passenger train speed x=33 km/hx = \mathbf{33\text{ km/h}}.
Express train speed x+11=33+11=44 km/hx + 11 = 33 + 11 = \mathbf{44\text{ km/h}}.
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Final Boxed Answer: Passenger Train Speed=33 km/h,Express Train Speed=44 km/h\mathbf{\text{Passenger Train Speed} = 33\text{ km/h}, \quad \text{Express Train Speed} = 44\text{ km/h}}
4

4. CBSE Case-Study Modeling: Two Water Taps Filling a Swimming Pool

Case Study Mastery (4 Marks)

Real-world mathematical modeling scenario with structured multi-part questions and full step solutions.

• Practical Application Context: Two Water Taps Filling a Swimming Pool
Two water taps together can fill a tank in 9389\frac{3}{8} hours (758\frac{75}{8} hours). The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately.
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Q1: Set up the quadratic equation for the time taken by the smaller tap. →\rightarrow Let smaller tap take xx hours. Larger tap takes (x−10)(x - 10) hours. Combined work rate: 1x+1x−10=875  ⟹  2x−10x(x−10)=875  ⟹  75(2x−10)=8(x2−10x)  ⟹  8x2−230x+750=0  ⟹  4x2−115x+375=0\frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75} \implies \frac{2x - 10}{x(x - 10)} = \frac{8}{75} \implies 75(2x - 10) = 8(x^2 - 10x) \implies \mathbf{8x^2 - 230x + 750 = 0} \implies \mathbf{4x^2 - 115x + 375 = 0}.
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Q2: Solve for xx and determine the individual filling times. →\rightarrow Factors of 4×375=15004 \times 375 = 1500 summing to −115-115 are −100-100 and −15-15. 4x2−100x−15x+375=0  ⟹  4x(x−25)−15(x−25)=0  ⟹  (x−25)(4x−15)=0  ⟹  x=254x^2 - 100x - 15x + 375 = 0 \implies 4x(x - 25) - 15(x - 25) = 0 \implies (x - 25)(4x - 15) = 0 \implies x = 25 or x=154=3.75x = \frac{15}{4} = 3.75. If x=3.75x = 3.75, larger tap takes 3.75−10=−6.253.75 - 10 = -6.25 hours (impossible). Hence, x=25 hoursx = \mathbf{25\text{ hours}}. Larger tap takes 25−10=15 hours25 - 10 = \mathbf{15\text{ hours}}.
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Q3: What is the combined fraction of tank filled in 1 hour? →\rightarrow 125+115=3+575=875\frac{1}{25} + \frac{1}{15} = \frac{3 + 5}{75} = \mathbf{\frac{8}{75}}.
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5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Important Solved Board Questions

Examiner step-marking allocations, common algebraic/arithmetic deduction traps, and presentation guidelines.

• Step-by-Step Marking Rubric & Presentation Guidelines
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1 Mark: Deriving correct quadratic equation from word problem text.
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1 Mark: Applying factorization or quadratic formula without sign errors.
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2 Marks: Computing algebraic roots with complete intermediate steps.
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1 Mark: Explicitly stating why the negative/fractional root is physically inadmissible and writing final boxed answer with units.
• Common Calculation Traps & Verification Checklist
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Trap 1: Forgetting to check a≠0a \neq 0 when kk appears in the x2x^2 coefficient (e.g. (k−12)x2(k-12)x^2).
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Trap 2: Failing to write reasons for discarding negative roots in real-world contexts.
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Trap 3: Misinterpreting time rate problems (work done in 1 hour is reciprocal of total time).
Authentic Board Question (3 Marks)Topic: Quadratic Equations Reducible Forms
Solve for xx: 1x+4−1x−7=1130\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}, where x≠−4,7x \neq -4, 7.

Official CBSE Step-by-Step Marking Breakdown:

Step 1: LCM & Simplification: Take LCM of denominators and simplify the numerator: (x−7)−(x+4)(x+4)(x−7)=−11x2−3x−28\frac{(x-7) - (x+4)}{(x+4)(x-7)} = \frac{-11}{x^2 - 3x - 28}.
1 Mark
Step 2: Cross-Multiplication & Standard Form: Cancel 1111 from both sides: −1x2−3x−28=130  ⟹  x2−3x+2=0\frac{-1}{x^2 - 3x - 28} = \frac{1}{30} \implies x^2 - 3x + 2 = 0.
1 Mark
Step 3: Factorisation & Roots: Factorise (x−1)(x−2)=0  ⟹  x=1(x-1)(x-2) = 0 \implies x = 1 or x=2x = 2.
1 Mark
Model Student Answer (Target: Full 3/3 Marks):
Given: 1x+4−1x−7=1130\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}
Taking LCM on LHS:
(x−7)−(x+4)(x+4)(x−7)=1130\frac{(x-7) - (x+4)}{(x+4)(x-7)} = \frac{11}{30}
−11x2−3x−28=1130\frac{-11}{x^2 - 3x - 28} = \frac{11}{30}
Dividing both sides by 1111:
−1x2−3x−28=130\frac{-1}{x^2 - 3x - 28} = \frac{1}{30}
Cross-multiplying:
−(30)=x2−3x−28-(30) = x^2 - 3x - 28
x2−3x+2=0x^2 - 3x + 2 = 0
Factorising by splitting the middle term:
x2−2x−x+2=0  ⟹  x(x−2)−1(x−2)=0x^2 - 2x - x + 2 = 0 \implies x(x-2) - 1(x-2) = 0
(x−1)(x−2)=0(x-1)(x-2) = 0
x=1x = 1 or x=2x = 2.
Examiner Mark Deduction Traps:
•Always check that final answers do not violate the initial domain condition (x≠−4,7x \neq -4, 7).
•Cancelling common factors across numerators before cross-multiplying prevents high-degree arithmetic errors.

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.
For ax2+bx+c=0ax^2 + bx + c = 0, the discriminant is D=b2−4acD = b^2 - 4ac. It determines the nature of roots: (1) If D>0D > 0, there are two distinct real roots: x=−b±D2ax = \frac{-b \pm \sqrt{D}}{2a}. (2) If D=0D = 0, there are two equal real roots: x=−b2ax = -\frac{b}{2a}. (3) If D<0D < 0, there are no real roots.

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