Class 10 Maths Chapter 4: Quadratic Equations Notes & Formulas

Exam Weightage & Blueprint

Total: 4-6 Marks

This chapter falls under Unit II: Algebra (20 marks total). As per the latest syllabus: solve quadratic equations by factorisation and quadratic formula, determine nature of roots using discriminant, and solve real-life situational problems.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	High	Nature of Roots (Discriminant)
Short Answer	2 or 3	Medium	Solving by Factorisation/Formula
Long Answer	4 or 5	Medium	Word Problems (Speed/Age/Area)

CBSE 2026–27 Syllabus Status: Quadratic Equations

Unit II: Algebra • 5–6 Marks

This chapter is strictly structured according to the CBSE Rationalised Mathematics Syllabus. Focus your board preparation only on prescribed topics and avoid studying deleted sections:

In 2026–27 Board Syllabus:

Deleted / Not in Board Exam:

⏰ Last 24-Hour Checklist

Standard Form: $ax^2 + bx + c = 0, a \neq 0$.
Discriminant: $D = b^2 - 4ac$.
Quadratic Formula: $x = $\frac{-b ± $\sqrt{D}$}{2a}$$.

Nature of Roots: $D > 0, D = 0, D < 0$.
Speed Formula: Time = Distance / Speed.
Dimension Check: Length cannot be negative.

📐 Concepts & Solving Methods

Standard Form: A quadratic equation in variable $x$ is of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$.

1. Method of Factorisation (Splitting Middle Term)

Split the middle term $bx$ such that the product of the two parts equals $ac$.

Standard Integer Example: $2x^2 - 5x + 3 = 0$.
Split $-5x$ into $-2x$ and $-3x$ because $(-2)(-3) = 6 = (2)(3)$.
$2x(x - 1) - 3(x - 1) = 0 \implies (2x - 3)(x - 1) = 0 \implies x = \frac{3}{2}, 1$.

⭐ Radical / Square-Root Factoring (CBSE Favorite):
Solve $4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0$:
Product $ac = 4\sqrt{3} \times (-2\sqrt{3}) = -24$; Sum $b = 5$.
Numbers are $+8$ and $-3$:
$$4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0$$ $$4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = 0 \quad [since 3 = \sqrt{3} \times \sqrt{3}]$$ $$(4x - \sqrt{3})(\sqrt{3}x + 2) = 0 \implies x = \frac{\sqrt{3}}{4}, \; x = -\frac{2}{\sqrt{3}}$$

2. Quadratic Formula (Sridharacharya Formula)

For any quadratic equation $ax^2 + bx + c = 0$ ($a \neq 0$), roots are given by:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{D}}{2a} $$

💡 Equations Reducible to Quadratic Form:
CBSE frequently sets 3–4 mark questions with algebraic fractions: $$\frac{1}{x+a} - \frac{1}{x+b} = c \quad (x \neq -a, -b)$$ Always cross-multiply to eliminate denominators, bring to standard form $Ax^2 + Bx + C = 0$, and state non-permissible values $(x \neq -a, -b)$.

🧮 Nature of Roots (Discriminant)

The Discriminant is $D = b^2 - 4ac$. It determines the nature of roots without solving.

Value of D ($b^2 - 4ac$)	Nature of Roots	Roots
D > 0	Two Distinct Real Roots	$\frac{-b \pm \sqrt{D}}{2a}$
D = 0	Two Equal Real Roots (Coincident)	$-\frac{b}{2a}, -\frac{b}{2a}$
D < 0	No Real Roots (Imaginary)	Does not exist in $\mathbb{R}$

⚠️ Critical Board Exam Conditions on Parameter $k$:
1. "Real Roots": Condition is $D \ge 0$ (both distinct & equal roots are real). Do NOT write just $D > 0$.
2. "Equal Roots": Condition is $D = 0$.
3. Quadratic Condition $a \neq 0$: In $kx(x - 2) + 6 = 0 \implies kx^2 - 2kx + 6 = 0$, $D = (-2k)^2 - 4(k)(6) = 4k(k - 6) = 0 \implies k = 0$ or $k = 6$. But if $k = 0$, coefficient of $x^2$ becomes 0 (no longer quadratic). Hence, $k = 0$ is rejected, answer is $k = 6$.

⚡ High-Yield Conceptual Rules (Board MCQs & A-R):

Signs of Roots from Coefficients:
• If $c < 0$, roots are always of opposite signs regardless of $b$ (since product $= c/a < 0$).
• If $b = 0$, roots are equal in magnitude and opposite in sign: $x = \pm \sqrt{-c/a}$.
• If $c = 0$, one root is always $0$: $x(ax+b)=0 \implies x=0$ or $x=-b/a$.
Domain Restrictions in Rational Forms: In algebraic equations like $\frac{1}{x-1} - \frac{1}{x+5} = \frac{6}{7}$, always state domain restrictions ($x \neq 1, -5$) and reject any extraneous root that makes a denominator zero.

Quadratic Root Finder

Enter coefficients for $ax^2 + bx + c = 0$

Solved Examples (Board Marking Scheme)

Q1. Find the discriminant of $2x^2 - 4x + 3 = 0$ and find the nature of roots. (2 Marks)

Step 1: Identify Coefficients 0.5 Mark

Step 2: Calculate Discriminant 1 Mark

Step 3: Conclusion 0.5 Mark

Q2. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides. (4 Marks)

Step 1: Form Equation 1 Mark

Step 2: Simplify 1 Mark

Step 3: Solve 1.5 Marks

Step 4: Conclusion 0.5 Mark

Previous Year Questions (PYQs)

2023 (2 Marks): Find the value of $k$ for which the equation $2x^2 + kx + 3 = 0$ has two equal roots.
Ans: For equal roots, $D = 0 \Rightarrow b^2 - 4ac = 0$.
$k^2 - 4(2)(3) = 0 \Rightarrow k^2 = 24 \Rightarrow k = ± 2\sqrt{6}$.

2020 (3 Marks): Solve for x: $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$.
Ans: Split $7x$ into $2x + 5x$. Roots are $-\sqrt{2}, -\frac{5}{\sqrt{2}}$.

2019 (4 Marks): A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less. Find the speed.
Ans: Eq: $\frac{360}{x} - \frac{360}{x+5} = 1$. Solving gives $x = 40$ km/h ($x=-45$ rejected).

CBSE Section E • 4 Marks

Case Study: High-Speed Vande Bharat Rail Travel

Scenario: A modern semi-high-speed express train travels a distance of 360 km at a uniform speed. If the speed had been 5 km/h more, the train would have taken 1 hour less for the same journey.

Q1: Taking the original uniform speed as $x$ km/h, write the equation representing the time difference. 1 Mark

Q2: Convert the condition into a standard quadratic equation in $x$. 1 Mark

Q3: Find the original uniform speed of the train. (OR) Find the discriminant of the resulting quadratic and state whether its roots are rational or irrational. 2 Marks

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Formula Error: Forgetting the $ \pm $ in the quadratic formula or calculating $b^2 - 4ac$ incorrectly (sign errors).

Rejecting Roots: Forgetting to explain WHY a negative root is rejected (e.g., "Speed cannot be negative").

Square Root: writing $\sqrt{16} = \pm 4$. Note: $\sqrt{16}$ is 4. The equation $x^2=16$ gives $x= \pm 4$.

💡 Scoring Tips

Identify Form: Always bring equation to $ax^2+bx+c=0$ first.

Units: In word problems, don't forget units (km/h, m, years) in the final answer.

Check D: Calculate $D$ first. If $D$ is a perfect square, calculation is likely correct.

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. The quadratic equation $ax^2 + bx + c = 0$ has no real roots if:

2. The roots of the equation $x^2 - 3x - 10 = 0$ are:

3. Which of the following is NOT a quadratic equation?

4. For a quadratic equation to have equal roots, the discriminant must be:

5. The sum of roots of $3x^2 - 5x + 2 = 0$ is:

Self-Assessment Mock Test (10 Marks)

Q1 (1M): For what value of $k$ does the equation $kx^2 - 6x - 2 = 0$ have real roots?

Q2 (2M): Solve for $x$: $x^2 - (\sqrt{3}+1)x + \sqrt{3} = 0$.

Q3 (3M): Find the value of $k$ for which $kx(x - 2) + 6 = 0$ has two equal roots.

Q4 (4M): Solve for $x$: $\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}$, where $x \neq -4, 7$.

📝 More Solved Board Questions

Q3. Solve $6x^2 - x - 2 = 0$ by factorisation. 2 Marks

Sol. We need two numbers whose product = $6 \times (-2) = -12$ and sum = $-1$.

Those numbers are $-4$ and $3$: $(-4)(3) = -12$, $-4 + 3 = -1$

$6x^2 - 4x + 3x - 2 = 2x(3x-2) + 1(3x-2) = (2x+1)(3x-2) = 0$

$x = -\frac{1}{2}$ or $x = \frac{2}{3}$

Q4. Find $k$ for which $kx^2 + 2x + 1 = 0$ has real and equal roots. 2 Marks

Sol. For equal roots: $D = 0$

$D = (2)^2 - 4(k)(1) = 4 - 4k = 0$

$4k = 4 \Rightarrow$ $k = 1$

Q5. A train travels 480 km at a uniform speed. If speed were 8 km/h more, it would take 2 hours less. Find the speed. 4 Marks

Sol. Let speed = $x$ km/h. Time = $\frac{480}{x}$ hours.

At new speed: $\frac{480}{x+8} = \frac{480}{x} - 2$

$480x - 480(x+8) = -2x(x+8)$

$-3840 = -2x^2 - 16x$

$x^2 + 8x - 1920 = 0$

Using formula: $x = \frac{-8 \pm \sqrt{64 + 7680}}{2} = \frac{-8 \pm 88}{2}$

$x = 40$ (taking positive value; $x = -48$ rejected as speed > 0)

Speed = 40 km/h

Q6. Solve for $x$: $\frac{1}{x-1} - \frac{1}{x+5} = \frac{6}{7}, \; (x \neq 1, -5)$. 3 Marks (Board Standard)

Sol. Taking LCM on LHS: $\frac{(x+5) - (x-1)}{(x-1)(x+5)} = \frac{6}{7}$

$\frac{6}{x^2 + 4x - 5} = \frac{6}{7} \implies \frac{1}{x^2 + 4x - 5} = \frac{1}{7}$

$x^2 + 4x - 5 = 7 \implies x^2 + 4x - 12 = 0$

Factorise: $(x + 6)(x - 2) = 0 \implies \mathbf{x = -6 or x = 2}$.

Q7. Two water taps together can fill a tank in $9\frac{3}{8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time taken by each tap. 5 Marks (Long Answer)

Sol. Let the smaller tap take $x$ hours. Then larger tap takes $(x - 10)$ hours.

Portion filled in 1 hour: Smaller tap $= \frac{1}{x}$, Larger tap $= \frac{1}{x - 10}$.

Together in 1 hour $= \frac{1}{9\frac{3}{8}} = \frac{1}{75/8} = \frac{8}{75}$.

$$\frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75} \implies \frac{x - 10 + x}{x(x - 10)} = \frac{8}{75}$$

$$\frac{2x - 10}{x^2 - 10x} = \frac{8}{75} \implies 75(2x - 10) = 8(x^2 - 10x)$$

$$150x - 750 = 8x^2 - 80x \implies 8x^2 - 230x + 750 = 0$$

Divide by 2: $4x^2 - 115x + 375 = 0$

Split $-115x$: $4x^2 - 100x - 15x + 375 = 0 \implies 4x(x - 25) - 15(x - 25) = 0$

$(4x - 15)(x - 25) = 0 \implies x = 25$ or $x = \frac{15}{4} = 3.75$.

If $x = 3.75$, larger tap time $= 3.75 - 10 = -6.25$ hours (impossible, time cannot be negative).

Therefore, Smaller tap $= 25$ hours, Larger tap $= 15$ hours.

🎯 Board Pattern (2018–2025): Word problems (Train Speed, Two Taps, Upstream/Downstream) appear every year as 4–5 mark questions. Always: (1) define the variable with units, (2) form the fractional equation, (3) reduce to $ax^2+bx+c=0$, (4) solve via splitting or quadratic formula, and (5) explicitly justify why one root is rejected!

📋 Board Revision Checklist

✅ Standard form: $ax^2 + bx + c = 0$, $a \neq 0$
✅ Factorisation: find two numbers with product $= ac$ and sum $= b$
✅ Radical splitting: express $3 = \sqrt{3} \times \sqrt{3}$ or $2 = \sqrt{2} \times \sqrt{2}$ when grouping
✅ Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
✅ Discriminant: $D = b^2 - 4ac$
✅ $D > 0$ → Two distinct real roots
✅ $D = 0$ → Two equal real roots $(x = -\frac{b}{2a})$
✅ $D < 0$ → No real roots
✅ For real roots: use condition $D \geq 0$ (not just $D > 0$)
✅ Condition $a \neq 0$: for $kx(x-2)+6=0$, reject $k=0$ so $k=6$
✅ Speed difference formula: $\frac{Dist}{lower speed} - \frac{Dist}{higher speed} = Time saved$
✅ Tap/Work formula: $\frac{1}{time_1} + \frac{1}{time_2} = \frac{1}{total time}$
✅ Always reject negative values for length/speed/age with written justification

💡 Exam Tip:
“Equal roots” → $D = 0$. “Real roots” → $D \geq 0$. “No real roots” → $D < 0$. Getting this distinction right is guaranteed 1 mark in MCQ/SA questions.

Frequently Asked Questions: Class 10 Maths Chapter 4

What is the weightage of Quadratic Equations in CBSE Class 10 Maths?

Quadratic Equations accounts for 5 to 7 marks in Unit 2: Algebra, often including 1 MCQ on discriminant/roots and a 3-mark or 5-mark word problem.

What is the discriminant formula and how does it determine the nature of roots?

The discriminant is D = b^2 - 4ac. If D > 0, there are two distinct real roots. If D = 0, there are two equal real roots. If D < 0, there are no real roots.

Is the Completing the Square method included in CBSE Class 10?

No, the Completing the Square method has been omitted from the rationalized CBSE syllabus. Students should solve equations using factorisation (splitting the middle term) or the quadratic formula x = (-b +- sqrt(b^2 - 4ac)) / (2a).

How do I avoid sign errors when using the quadratic formula?

Always rearrange the equation into standard form ax^2 + bx + c = 0 and explicitly write down values of a, b, and c with their signs before computing D = b^2 - 4ac.

Chapter 4 Study Modules & Practice Resources

Complete Notes & Formulas • NCERT Solutions • Previous Year Questions • Chapter Tests & Quizzes • Practice Worksheets

Chapter 4: Quadratic Equations

Overview