Class 10 Quadratic Equations Notes

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)

⏰ Last 24-Hour Checklist

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

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 $-3$:
$$4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0$$ $$4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = 0 \quad [\text{since } 3 = \sqrt{3} \times \sqrt{3}]$$ $$(4x - \sqrt{3})(\sqrt{3}x + 2) = 0 \implies x = \frac{\sqrt{3}}{4}, \; x = -\frac{2}{\sqrt{3}}$$

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 $k = 6$.

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, $D = 0 \Rightarrow b^2 - 4ac = 0$.
$k^2 - 4(2)(3) = 0 \Rightarrow k^2 = 24 \Rightarrow k = \pm 2\sqrt{6}$.

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).

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

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 $\left(x = -\frac{b}{2a}\right)$
✅ $D < 0$ → 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: $\frac{\text{Dist}}{\text{lower speed}} - \frac{\text{Dist}}{\text{higher speed}} = \text{Time saved}$
✅ Tap/Work formula: $\frac{1}{\text{time}_1} + \frac{1}{\text{time}_2} = \frac{1}{\text{total time}}$
✅ 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.

Chapter 4: Quadratic Equations

Overview