Chapter 4 : Quadratic Equations Class 10 Maths NCERT Solutions

Chapter 4: Quadratic Equations — Exercise 4.1

Quadratic equations are the secret behind the curves of a basketball throw and the arches of beautiful bridges. Mastering this exercise helps you identify these powerful equations and turn real-life situations into solvable math!

Q1. Check whether the following are quadratic equations:

i) (x + 1)^2 = 2(x – 3)

The Logic: A quadratic equation must look like ax^2 + bx + c = 0, where the highest power (degree) is 2 and ‘a’ is not zero. We just need to simplify both sides and check the highest power.
The Steps:
- Expand (x + 1)^2 using the formula (a + b)^2 = a^2 + 2ab + b^2.
- x^2 + 2x + 1 = 2x – 6
- Bring everything to one side: x^2 + 2x – 2x + 1 + 6 = 0
- x^2 + 7 = 0
The Result: Yes, it is a quadratic equation because the highest power of x is 2.

ii) x^2 – 2x = (-2)(3 – x)

The Logic: Simplify the right side by multiplying -2 with both terms inside the bracket.
The Steps:
- x^2 – 2x = -6 + 2x
- x^2 – 2x – 2x + 6 = 0
- x^2 – 4x + 6 = 0
The Result: Yes, it is a quadratic equation.

iii) (x – 2)(x + 1) = (x – 1)(x + 3)

Human Tip: Sometimes equations look quadratic because of the x terms, but when you multiply them out, the x^2 terms might cancel each other!
The Steps:
- Left side: x^2 + x – 2x – 2 = x^2 – x – 2
- Right side: x^2 + 3x – x – 3 = x^2 + 2x – 3
- Equating them: x^2 – x – 2 = x^2 + 2x – 3
- Subtract x^2 from both sides: -x – 2 = 2x – 3
- -3x + 1 = 0
The Result: No, it is not a quadratic equation because the x^2 term is gone (the degree is 1).

Q2. Represent the following situations in the form of quadratic equations:

i) The area of a rectangular plot is 528 m^2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

The Logic: We don’t need to solve it yet, just “translate” the English into Math. Let the smaller dimension be ‘x’.
The Steps:
- Let the breadth = x metres.
- The length is “one more than twice its breadth,” so Length = 2x + 1.
- Area of rectangle = Length x Breadth.
- 528 = (2x + 1) * x
- 528 = 2x^2 + x
The Result: 2x^2 + x – 528 = 0

ii) The product of two consecutive positive integers is 306. We need to find the integers.

Pro-Tip: “Consecutive” just means one after the other, like 5 and 6. If one is ‘x’, the next one is ‘x + 1’.
The Steps:
- Let the first integer = x.
- The second integer = x + 1.
- Their product is 306: x * (x + 1) = 306
- x^2 + x = 306
The Result: x^2 + x – 306 = 0

iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

The Logic: Always set ‘x’ as the present age of the person you want to find. Then, move “forward in time” by adding 3 years to both.
The Steps:
- Let Rohan’s present age = x.
- Mother’s present age = x + 26.
- 3 years from now: Rohan = x + 3; Mother = (x + 26) + 3 = x + 29.
- Product of ages = 360: (x + 3)(x + 29) = 360
- x^2 + 29x + 3x + 87 = 360
- x^2 + 32x + 87 – 360 = 0
The Result: x^2 + 32x – 273 = 0

iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

The Logic: Use the formula: Time = Distance / Speed. The “Time Gap” between the slow trip and the fast trip is 3 hours.
The Steps:
- Let the uniform speed of the train = x km/h.
- Time taken for 480 km = 480 / x.
- New speed = x – 8. New time = 480 / (x – 8).
- According to the question: (Time with slower speed) – (Time with normal speed) = 3
- [ 480 / (x – 8) ] – [ 480 / x ] = 3
- 480 [ (x – (x – 8)) / x(x – 8) ] = 3
- 480 [ 8 / (x^2 – 8x) ] = 3
- 3840 = 3(x^2 – 8x)
- 1280 = x^2 – 8x (dividing by 3)
The Result: x^2 – 8x – 1280 = 0

Summary Table

Concept	What to check?	Key Formula
Quadratic Check	Is the highest power 2?	ax^2 + bx + c = 0
Consecutive Nos	Numbers in a row	x and x + 1
Age Problems	Add years for “future”, Subtract for “past”	(x + n)
Speed/Time	The relationship between speed and time	Time = Distance / Speed

Chapter 4: Quadratic Equations — Exercise 4.2

Ready to solve the mystery of the “x”? This exercise focuses on Factorisation, a skill that’s like solving a puzzle where you find the two hidden numbers that make the whole equation click!

Q1. Find the roots of the following quadratic equations by factorisation:

i) x^2 – 3x – 10 = 0

The Logic: We use the “Splitting the Middle Term” method. We need two numbers that multiply to -10 and add up to -3.
Human Tip: Think of the factors of 10. (5 and 2). Since we need -3, we use -5 and +2.
The Steps:
- x^2 – 5x + 2x – 10 = 0
- x(x – 5) + 2(x – 5) = 0
- (x – 5)(x + 2) = 0
- So, x – 5 = 0 or x + 2 = 0
The Result: Roots are 5 and -2.

ii) 2x^2 + x – 6 = 0

The Logic: Multiply the first and last numbers (2 x -6 = -12). We need two numbers that multiply to -12 and add up to +1.
The Steps:
- Numbers are +4 and -3.
- 2x^2 + 4x – 3x – 6 = 0
- 2x(x + 2) – 3(x + 2) = 0
- (x + 2)(2x – 3) = 0
The Result: Roots are -2 and 3/2.

iii) root 2 x^2 + 7x + 5 root 2 = 0

Pro-Tip: Don’t let the roots scare you! Multiply (root 2) x (5 root 2) = 5 x 2 = 10. We need numbers that multiply to 10 and add to 7. (Hint: 5 and 2).
The Steps:
- root 2 x^2 + 2x + 5x + 5 root 2 = 0
- root 2 x(x + root 2) + 5(x + root 2) = 0
- (x + root 2)(root 2 x + 5) = 0
The Result: Roots are -root 2 and -5/root 2.

iv) 2x^2 – x + 1/8 = 0

Human Tip: Fractions make things messy. Let’s clear it by multiplying the whole equation by 8!
The Steps:
- 16x^2 – 8x + 1 = 0
- Multiply to 16, add to -8. (Numbers: -4, -4).
- 16x^2 – 4x – 4x + 1 = 0
- 4x(4x – 1) – 1(4x – 1) = 0
- (4x – 1)(4x – 1) = 0
The Result: Roots are 1/4 and 1/4.

Q3. Find two numbers whose sum is 27 and product is 182.

The Logic: Let the first number be ‘x’. Then the second is (27 – x). Their product is 182.
The Steps:
- x(27 – x) = 182
- 27x – x^2 = 182
- x^2 – 27x + 182 = 0
- Factors of 182 that add to 27 are 13 and 14.
- (x – 13)(x – 14) = 0
The Result: The numbers are 13 and 14.

Q4. Find two consecutive positive integers, sum of whose squares is 365.

The Logic: Consecutive means one after another, like x and (x + 1).
The Steps:
- x^2 + (x + 1)^2 = 365
- x^2 + x^2 + 2x + 1 = 365
- 2x^2 + 2x – 364 = 0
- Divide by 2: x^2 + x – 182 = 0
- (x + 14)(x – 13) = 0
- Since integers are positive, x = 13.
The Result: Integers are 13 and 14.

Q5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

The Logic: Use our old friend, the Pythagoras Theorem (a^2 + b^2 = c^2). Let base = x, then altitude = (x – 7).
The Steps:
- x^2 + (x – 7)^2 = 13^2
- x^2 + x^2 – 14x + 49 = 169
- 2x^2 – 14x – 120 = 0
- x^2 – 7x – 60 = 0
- (x – 12)(x + 5) = 0
- x = 12 (Base cannot be negative).
The Result: Base = 12 cm, Altitude = 5 cm.

Summary Table

Method	When to Use	Key Formula
Splitting Middle Term	Standard form ax^2 + bx + c	Find p, q such that p+q=b and pq=ac
Consecutive Integers	Numbers in a row	x and x + 1
Roots/Zeros	Solving the equation	Value of x that makes equation 0

Chapter 4: Quadratic Equations — Exercise 4.3

This exercise is the “Final Boss” of Quadratic Equations! It teaches you how to predict the future of an equation using the Nature of Roots, helping you decide if a problem is even solvable before you spend time on it.

Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

i) 2x^2 – 3x + 5 = 0

The Logic: Before solving, we check the Discriminant (D). Think of D as a magnifying glass that tells us if the roots are real, equal, or imaginary.
The Steps:
- Here, a = 2, b = -3, c = 5.
- Discriminant (D) = b^2 – 4ac
- D = (-3)^2 – 4(2)(5)
- D = 9 – 40 = -31
The Result: Since D is less than 0, the equation has no real roots.

ii) 3x^2 – 4 root 3 x + 4 = 0

The Logic: If D = 0, it means the equation is a “Perfect Square” and both roots will be exactly the same.
The Steps:
- a = 3, b = -4 root 3, c = 4.
- D = (-4 root 3)^2 – 4(3)(4)
- D = (16 x 3) – 48 = 48 – 48 = 0.
- Since D = 0, real and equal roots exist.
- Roots = -b / 2a = -(-4 root 3) / (2 x 3) = 4 root 3 / 6.
The Result: The roots are (2 root 3) / 3 and (2 root 3) / 3.

iii) 2x^2 – 6x + 3 = 0

The Logic: If D is a positive number, we have two distinct (different) real roots.
The Steps:
- a = 2, b = -6, c = 3.
- D = (-6)^2 – 4(2)(3) = 36 – 24 = 12.
- Since D > 0, roots exist.
- x = [-b + root D] / 2a and x = [-b – root D] / 2a
- x = [6 + root 12] / 4 and x = [6 – root 12] / 4
- root 12 = 2 root 3.
- x = [6 + 2 root 3] / 4 = (3 + root 3) / 2
The Result: The roots are (3 + root 3) / 2 and (3 – root 3) / 2.

Q2. Find the values of k for each of the following quadratic equations, so that they have two equal roots:

i) 2x^2 + kx + 3 = 0

The Logic: “Two equal roots” is a huge hint! It means we must set the Discriminant (D) to zero.
The Steps:
- a = 2, b = k, c = 3.
- For equal roots, b^2 – 4ac = 0.
- k^2 – 4(2)(3) = 0
- k^2 – 24 = 0 => k^2 = 24.
- k = root 24 = plus/minus 2 root 6.
The Result: k = 2 root 6 or k = -2 root 6.

ii) kx(x – 2) + 6 = 0

Human Tip: First, simplify the equation into the standard form ax^2 + bx + c = 0!
The Steps:
- kx^2 – 2kx + 6 = 0.
- a = k, b = -2k, c = 6.
- For equal roots, D = 0.
- (-2k)^2 – 4(k)(6) = 0
- 4k^2 – 24k = 0
- 4k(k – 6) = 0.
- So, k = 0 or k = 6. (But k cannot be 0 because the x^2 term would disappear).
The Result: k = 6.

Q3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m^2? If so, find its length and breadth.

The Logic: We turn the word problem into a quadratic equation. If the Discriminant comes out positive, it’s possible!
The Steps:
- Let breadth = x. Then length = 2x.
- Area = length x breadth = (2x)(x) = 2x^2.
- 2x^2 = 800 => x^2 = 400.
- x = root 400 = 20.
The Result: Yes, it is possible. Breadth = 20m and Length = 40m.

Q4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

The Logic: Let’s test the situation using the Discriminant.
The Steps:
- Let age of Friend A = x. Then Friend B = 20 – x.
- Four years ago: (x – 4)(20 – x – 4) = 48.
- (x – 4)(16 – x) = 48.
- 16x – x^2 – 64 + 4x = 48
- -x^2 + 20x – 112 = 0 => x^2 – 20x + 112 = 0.
- D = (-20)^2 – 4(1)(112) = 400 – 448 = -48.
The Result: Since D < 0, this situation is not possible.

Q5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m^2? If so, find its length and breadth.

The Steps:
- Perimeter = 2(l + b) = 80 => l + b = 40.
- Let l = x, then b = 40 – x.
- Area = x(40 – x) = 400.
- 40x – x^2 = 400 => x^2 – 40x + 400 = 0.
- D = (-40)^2 – 4(1)(400) = 1600 – 1600 = 0.
The Result: Yes, it is possible. Length = 20m and Breadth = 20m (It’s a square!).

Summary Table

Discriminant (D) Value	Nature of Roots	Exam Strategy
D > 0	2 Distinct Real Roots	Use Quadratic Formula to solve
D = 0	2 Equal Real Roots	Roots are simply -b/2a
D < 0	No Real Roots	Stop! The situation is not possible

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Chapter 4 : Quadratic Equations Class 10 Maths NCERT Solutions

Contents

Class 10 Maths NCERT Solutions - Chapter wise

Chapter 1: Real Numbers

Chapter 2: Polynomials

Chapter 3 : Pair of Linear Equations in Two Variables