Quadratic Equations

Board: CBSE Class: 10

Chapter Overview

A quadratic equation in x is of the form ax² + bx + c = 0, where a != 0. Solving methods: factorization (splitting the middle term), completing the square, and the quadratic formula. The discriminant D = b² - 4ac determines the nature of roots: D > 0 (two real distinct roots), D = 0 (two equal real roots), D < 0 (no real roots). Word problems involving quadratic equations cover areas, time-speed-distance, and number problems.

Topics Covered

Standard Form
Factorization Method
Completing the Square
Quadratic Formula
Discriminant
Nature of Roots
Word Problems

Key Formulas

x = (-b +/- √b² - 4ac) / 2a

D = b² - 4ac

D > 0: real and distinct

D = 0: real and equal

D < 0: no real roots

Real-World Applications

Applications: Projectile motion, optimization problems, area calculations, engineering design.

Study Tips

Tip: Master the quadratic formula

Tip: Practice completing the square method

Tip: Check your answers by substitution

NCERT Exercise Solutions

Exercise 4.1

Check whether the following are quadratic equations:
(i) (x + 1)² = 2(x - 3)
(ii) x² - 2x = (-2)(3 - x)
(iii) (x - 2)(x + 1) = (x - 1)(x + 3)
(iv) (x - 3)(2x + 1) = x(x + 5)
(v) (2x - 1)(x - 3) = (x + 5)(x - 1)
(vi) x² + 3x + 1 = (x - 2)²
(vii) (x + 2)³ = 2x(x² - 1)
(viii) x³ - 4x² - x + 1 = (x - 2)³

Show Solution

Expand each equation and check if it is of the form ax² + bx + c = 0 with a != 0.

(i) (x + 1)² = 2(x - 3)
=> x² + 2x + 1 = 2x - 6
=> x² + 2x + 1 - 2x + 6 = 0
=> x² + 7 = 0
Here a = 1, b = 0, c = 7. Since a != 0, it is a quadratic equation.

(ii) x² - 2x = (-2)(3 - x)
=> x² - 2x = -6 + 2x
=> x² - 2x - 2x + 6 = 0
=> x² - 4x + 6 = 0
Here a = 1, b = -4, c = 6. Since a != 0, it is a quadratic equation.

(iii) (x - 2)(x + 1) = (x - 1)(x + 3)
=> x² + x - 2x - 2 = x² + 3x - x - 3
=> x² - x - 2 = x² + 2x - 3
=> x² - x - 2 - x² - 2x + 3 = 0
=> -3x + 1 = 0
=> 3x - 1 = 0
Here a = 0 (no x² term). Since a = 0, it is NOT a quadratic equation.

(iv) (x - 3)(2x + 1) = x(x + 5)
=> 2x² + x - 6x - 3 = x² + 5x
=> 2x² - 5x - 3 = x² + 5x
=> 2x² - 5x - 3 - x² - 5x = 0
=> x² - 10x - 3 = 0
Here a = 1, b = -10, c = -3. Since a != 0, it is a quadratic equation.

(v) (2x - 1)(x - 3) = (x + 5)(x - 1)
=> 2x² - 6x - x + 3 = x² - x + 5x - 5
=> 2x² - 7x + 3 = x² + 4x - 5
=> 2x² - 7x + 3 - x² - 4x + 5 = 0
=> x² - 11x + 8 = 0
Here a = 1, b = -11, c = 8. Since a != 0, it is a quadratic equation.

(vi) x² + 3x + 1 = (x - 2)²
=> x² + 3x + 1 = x² - 4x + 4
=> x² + 3x + 1 - x² + 4x - 4 = 0
=> 7x - 3 = 0
Here a = 0. Since a = 0, it is NOT a quadratic equation.

(vii) (x + 2)³ = 2x(x² - 1)
=> x³ + 6x² + 12x + 8 = 2x³ - 2x
=> x³ + 6x² + 12x + 8 - 2x³ + 2x = 0
=> -x³ + 6x² + 14x + 8 = 0
=> x³ - 6x² - 14x - 8 = 0
Here degree = 3. It is NOT a quadratic equation (it is cubic).

(viii) x³ - 4x² - x + 1 = (x - 2)³
=> x³ - 4x² - x + 1 = x³ - 6x² + 12x - 8
=> x³ - 4x² - x + 1 - x³ + 6x² - 12x + 8 = 0
=> 2x² - 13x + 9 = 0
Here a = 2, b = -13, c = 9. Since a != 0, it is a quadratic equation.
Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length is one more than twice its breadth. Find length and breadth.
(ii) The product of two consecutive positive integers is 306. Find the integers.
(iii) Rohan's mother is 26 years older than him. The product of their ages 3 years from now will be 360. Find Rohan's present age.
(iv) A train travels 480 km at uniform speed. If speed had been 8 km/h less, it would have taken 3 hours more. Find the speed.

Show Solution

Represent each situation using a variable and form a quadratic equation.

(i) Let breadth = b metres. Then length = 2b + 1 metres.
Area = length x breadth = (2b + 1)(b) = 528
=> 2b² + b = 528
=> 2b² + b - 528 = 0
This is the required quadratic equation.

(ii) Let the two consecutive positive integers be x and x + 1.
Product = x(x + 1) = 306
=> x² + x = 306
=> x² + x - 306 = 0
This is the required quadratic equation.

(iii) Let Rohan's present age = x years.
Mother's present age = x + 26 years.
3 years from now:
Rohan's age = x + 3
Mother's age = x + 29
Product = (x + 3)(x + 29) = 360
=> x² + 29x + 3x + 87 = 360
=> x² + 32x + 87 - 360 = 0
=> x² + 32x - 273 = 0
This is the required quadratic equation.

(iv) Let the speed of the train = x km/h.
Time at original speed = 480/x hours.
Time at reduced speed (x - 8) km/h = 480/(x - 8) hours.
Difference = 3 hours:
480/(x - 8) - 480/x = 3
=> (480x - 480(x - 8)) / (x(x - 8)) = 3
=> (480x - 480x + 3840) / (x(x - 8)) = 3
=> 3840 / (x(x - 8)) = 3
=> 3840 = 3x(x - 8)
=> 3x² - 24x - 3840 = 0
=> x² - 8x - 1280 = 0
This is the required quadratic equation.

Exercise 4.2

Find the roots of the following quadratic equations by factorization:
(i) x² - 3x - 10 = 0
(ii) 2x² + x - 6 = 0
(iii) √(2) x² + 7x + 5 √(2) = 0
(iv) 2x² - x + 1/8 = 0
(v) 100x² - 20x + 1 = 0

Show Solution

(i) x² - 3x - 10 = 0
=> x² - 5x + 2x - 10 = 0
=> x(x - 5) + 2(x - 5) = 0
=> (x - 5)(x + 2) = 0
=> x - 5 = 0 or x + 2 = 0
=> x = 5 or x = -2

(ii) 2x² + x - 6 = 0
=> 2x² + 4x - 3x - 6 = 0
=> 2x(x + 2) - 3(x + 2) = 0
=> (x + 2)(2x - 3) = 0
=> x + 2 = 0 or 2x - 3 = 0
=> x = -2 or x = 3/2

(iii) √(2) x² + 7x + 5 √(2) = 0
=> √(2) x² + 5x + 2x + 5 √(2) = 0
=> x(√(2) x + 5) + √(2)(√(2) x + 5) = 0
=> (√(2) x + 5)(x + √(2)) = 0
=> √(2) x + 5 = 0 or x + √(2) = 0
=> x = -5/√(2) or x = -√(2)

(iv) 2x² - x + 1/8 = 0
Multiply by 8: 16x² - 8x + 1 = 0
=> 16x² - 4x - 4x + 1 = 0
=> 4x(4x - 1) - 1(4x - 1) = 0
=> (4x - 1)(4x - 1) = 0
=> 4x - 1 = 0
=> x = 1/4 (repeated root)

(v) 100x² - 20x + 1 = 0
=> 100x² - 10x - 10x + 1 = 0
=> 10x(10x - 1) - 1(10x - 1) = 0
=> (10x - 1)(10x - 1) = 0
=> 10x - 1 = 0
=> x = 1/10 (repeated root)
Find the dimensions of a rectangle whose length is 4 m more than its breadth and the area is 192 m².

Show Solution

Let breadth = x metres.
Then length = (x + 4) metres.
Area = length x breadth = x(x + 4) = 192
=> x² + 4x = 192
=> x² + 4x - 192 = 0
=> x² + 16x - 12x - 192 = 0
=> x(x + 16) - 12(x + 16) = 0
=> (x + 16)(x - 12) = 0
=> x + 16 = 0 or x - 12 = 0
=> x = -16 or x = 12
Since breadth cannot be negative, x = 12.
Therefore, breadth = 12 m, length = 16 m.

Verification: Area = 12 x 16 = 192 m². Correct.

Exercise 4.3

Find the roots of the following quadratic equations by completing the square:
(i) 2x² - 7x + 3 = 0
(ii) 2x² + x - 4 = 0
(iii) 4x² + 4 √(3) x + 3 = 0
(iv) 2x² + x + 4 = 0

Show Solution

(i) 2x² - 7x + 3 = 0
Divide by 2: x² - (7/2)x + 3/2 = 0
=> x² - (7/2)x = -3/2
Add (7/4)² = 49/16 to both sides:
x² - (7/2)x + 49/16 = -3/2 + 49/16
=> (x - 7/4)² = (-24 + 49)/16 = 25/16
=> x - 7/4 = +/- 5/4
=> x = 7/4 +/- 5/4
=> x = 3 or x = 1/2

(ii) 2x² + x - 4 = 0
Divide by 2: x² + (1/2)x - 2 = 0
=> x² + (1/2)x = 2
Add (1/4)² = 1/16 to both sides:
x² + (1/2)x + 1/16 = 2 + 1/16
=> (x + 1/4)² = 33/16
=> x + 1/4 = +/- √(33)/4
=> x = (-1 +/- √(33))/4

(iii) 4x² + 4 √(3) x + 3 = 0
Divide by 4: x² + √(3)x + 3/4 = 0
=> x² + √(3)x = -3/4
Add (√(3)/2)² = 3/4 to both sides:
x² + √(3)x + 3/4 = -3/4 + 3/4 = 0
=> (x + √(3)/2)² = 0
=> x + √(3)/2 = 0
=> x = -√(3)/2 (repeated root)

(iv) 2x² + x + 4 = 0
Divide by 2: x² + (1/2)x + 2 = 0
=> x² + (1/2)x = -2
Add (1/4)² = 1/16:
x² + (1/2)x + 1/16 = -2 + 1/16
=> (x + 1/4)² = -31/16
Since RHS is negative, no real roots exist.
Find the roots of the following quadratic equations using the quadratic formula:
(i) x² - 3x - 10 = 0
(ii) 2x² + x - 6 = 0
(iii) √(2) x² + 7x + 5 √(2) = 0
(iv) 2x² - x + 1/8 = 0
(v) 100x² - 20x + 1 = 0

Show Solution

Quadratic formula: x = (-b +/- √(b² - 4ac)) / 2a

(i) x² - 3x - 10 = 0; a = 1, b = -3, c = -10
D = b² - 4ac = (-3)² - 4(1)(-10) = 9 + 40 = 49
x = (3 +/- √(49)) / 2 = (3 +/- 7) / 2
x = 5 or x = -2

(ii) 2x² + x - 6 = 0; a = 2, b = 1, c = -6
D = 1 - 4(2)(-6) = 1 + 48 = 49
x = (-1 +/- 7) / 4
x = 6/4 = 3/2 or x = -8/4 = -2

(iii) √(2) x² + 7x + 5 √(2) = 0; a = √(2), b = 7, c = 5 √(2)
D = 49 - 4 √(2) x 5 √(2) = 49 - 40 = 9
x = (-7 +/- 3) / (2 √(2))
x = -4/(2 √(2)) = -2/√(2) = -√(2) or x = -10/(2 √(2)) = -5/√(2)

(iv) 2x² - x + 1/8 = 0; a = 2, b = -1, c = 1/8
D = 1 - 4(2)(1/8) = 1 - 1 = 0
x = (1 +/- 0) / 4 = 1/4 (repeated root)

(v) 100x² - 20x + 1 = 0; a = 100, b = -20, c = 1
D = 400 - 400 = 0
x = (20 +/- 0) / 200 = 1/10 (repeated root)
Find the nature of roots of the following quadratic equations. If real roots exist, find them:
(i) 2x² - 3x + 5 = 0
(ii) 3x² - 4 √(3) x + 4 = 0
(iii) 2x² - 6x + 3 = 0

Show Solution

Discriminant D = b² - 4ac determines nature of roots:
D > 0: two distinct real roots
D = 0: two equal real roots
D < 0: no real roots

(i) 2x² - 3x + 5 = 0; a = 2, b = -3, c = 5
D = 9 - 4(2)(5) = 9 - 40 = -31
Since D < 0, no real roots exist.

(ii) 3x² - 4 √(3) x + 4 = 0; a = 3, b = -4 √(3), c = 4
D = 48 - 4(3)(4) = 48 - 48 = 0
Since D = 0, two equal real roots exist.
x = (4 √(3) +/- 0) / 6 = (2 √(3))/3 (repeated root)

(iii) 2x² - 6x + 3 = 0; a = 2, b = -6, c = 3
D = 36 - 4(2)(3) = 36 - 24 = 12
Since D > 0, two distinct real roots exist.
x = (6 +/- √(12)) / 4 = (6 +/- 2 √(3)) / 4 = (3 +/- √(3)) / 2
Find two numbers whose sum is 27 and product is 182.

Show Solution

Let first number = x.
Then second number = 27 - x (since sum is 27).
Product: x(27 - x) = 182
=> 27x - x² = 182
=> -x² + 27x - 182 = 0
=> x² - 27x + 182 = 0
=> x² - 13x - 14x + 182 = 0
=> x(x - 13) - 14(x - 13) = 0
=> (x - 13)(x - 14) = 0
=> x = 13 or x = 14

If x = 13, the numbers are 13 and 14.
If x = 14, the numbers are 14 and 13.
Hence the two numbers are 13 and 14.
Find two consecutive positive integers, the sum of whose squares is 365.

Show Solution

Let the two consecutive positive integers be x and x + 1.
Sum of squares: x² + (x + 1)² = 365
=> x² + x² + 2x + 1 = 365
=> 2x² + 2x + 1 - 365 = 0
=> 2x² + 2x - 364 = 0
=> x² + x - 182 = 0
=> x² + 14x - 13x - 182 = 0
=> x(x + 14) - 13(x + 14) = 0
=> (x + 14)(x - 13) = 0
=> x = -14 or x = 13
Since x is positive, x = 13.
The two consecutive integers are 13 and 14.

Verification: 13² + 14² = 169 + 196 = 365. Correct.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Show Solution

Let base = x cm.
Then altitude = (x - 7) cm.
Hypotenuse = 13 cm.
By Pythagoras theorem:
x² + (x - 7)² = 13²
=> x² + x² - 14x + 49 = 169
=> 2x² - 14x + 49 - 169 = 0
=> 2x² - 14x - 120 = 0
=> x² - 7x - 60 = 0
=> x² - 12x + 5x - 60 = 0
=> x(x - 12) + 5(x - 12) = 0
=> (x - 12)(x + 5) = 0
=> x = 12 or x = -5
Since length cannot be negative, x = 12.
Base = 12 cm, Altitude = 5 cm.

Verification: 12² + 5² = 144 + 25 = 169 = 13². Correct.
A cottage industry produces a certain number of pottery articles in a day. On a particular day, the cost of production of each article (in rupees) was 3 more than twice the number of articles produced that day. If the total cost of production was 90 rupees, find the number of articles produced and the cost of each article.

Show Solution

Let number of articles produced = x.
Cost per article = 2x + 3 rupees.
Total cost = x(2x + 3) = 90
=> 2x² + 3x = 90
=> 2x² + 3x - 90 = 0
=> 2x² - 12x + 15x - 90 = 0
=> 2x(x - 6) + 15(x - 6) = 0
=> (x - 6)(2x + 15) = 0
=> x = 6 or x = -15/2
Since number of articles cannot be negative, x = 6.
Cost per article = 2(6) + 3 = 15 rupees.

Verification: 6 articles x 15 rupees = 90 rupees. Correct.
Find the roots of the equation x² + 3x - 10 = 0 by completing the square method.

Show Solution

x² + 3x - 10 = 0
=> x² + 3x = 10
Add (3/2)² = 9/4 to both sides:
x² + 3x + 9/4 = 10 + 9/4
=> (x + 3/2)² = (40 + 9)/4 = 49/4
=> x + 3/2 = +/- 7/2
=> x = -3/2 +/- 7/2
=> x = 2 or x = -5
Find the roots of the equation 2x² - 2 √(2) x + 1 = 0 using the quadratic formula.

Show Solution

2x² - 2 √(2) x + 1 = 0; a = 2, b = -2 √(2), c = 1
D = (2 √(2))² - 4(2)(1) = 8 - 8 = 0
Since D = 0, roots are equal.
x = (2 √(2) +/- 0) / 4 = √(2)/2 (repeated root)

Exercise 4.4

Find the value of k for which the quadratic equation kx(x - 2) + 6 = 0 has equal roots.

Show Solution

kx(x - 2) + 6 = 0
=> kx² - 2kx + 6 = 0
Comparing with ax² + bx + c = 0:
a = k, b = -2k, c = 6

For equal roots, discriminant D = 0:
b² - 4ac = 0
=> (-2k)² - 4(k)(6) = 0
=> 4k² - 24k = 0
=> 4k(k - 6) = 0
=> k = 0 or k = 6

Since k = 0 makes the equation linear (not quadratic), we take k = 6.

Verification: For k = 6, equation is 6x² - 12x + 6 = 0
=> 6(x² - 2x + 1) = 0
=> 6(x - 1)² = 0
=> x = 1 (repeated root). Correct.
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.

Show Solution

Let age of one friend = x years.
Then age of other friend = (20 - x) years.
Four years ago:
Friend 1 age = x - 4
Friend 2 age = 20 - x - 4 = 16 - x
Product: (x - 4)(16 - x) = 48
=> 16x - x² - 64 + 4x = 48
=> -x² + 20x - 64 = 48
=> -x² + 20x - 112 = 0
=> x² - 20x + 112 = 0
Discriminant: D = 400 - 4(1)(112) = 400 - 448 = -48
Since D < 0, no real roots exist.
Therefore, this situation is NOT possible.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.

Show Solution

Let breadth = x metres.
Then length = 2x metres.
Area = length x breadth = 2x(x) = 2x² = 800
=> x² = 400
=> x = +/- 20
Since breadth cannot be negative, x = 20 m.
Breadth = 20 m, Length = 40 m.

Verification: Area = 40 x 20 = 800 m². Correct.
Yes, it is possible. The mango grove measures 40 m x 20 m.
Is the following situation possible? If so, determine the dimensions. The sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Show Solution

Let side of first square = x m.
Let side of second square = y m.
Sum of areas: x² + y² = 468
Difference of perimeters: 4x - 4y = 24
=> x - y = 6
=> x = y + 6

Substitute in first equation:
(y + 6)² + y² = 468
=> y² + 12y + 36 + y² = 468
=> 2y² + 12y + 36 - 468 = 0
=> 2y² + 12y - 432 = 0
=> y² + 6y - 216 = 0
=> y² + 18y - 12y - 216 = 0
=> y(y + 18) - 12(y + 18) = 0
=> (y + 18)(y - 12) = 0
=> y = -18 or y = 12
Since side cannot be negative, y = 12 m.
Then x = 12 + 6 = 18 m.

Yes, possible. Sides are 18 m and 12 m.
Verification: 18² + 12² = 324 + 144 = 468. Correct.
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. Find the speed of the train.

Show Solution

Let original speed = x km/h.
Original time = 480/x hours.
Reduced speed = (x - 8) km/h.
New time = 480/(x - 8) hours.
Difference: 480/(x - 8) - 480/x = 3
=> (480x - 480(x - 8)) / (x(x - 8)) = 3
=> (480x - 480x + 3840) / (x(x - 8)) = 3
=> 3840 / (x(x - 8)) = 3
=> 3840 = 3x(x - 8)
=> 3x² - 24x - 3840 = 0
=> x² - 8x - 1280 = 0
=> x² - 40x + 32x - 1280 = 0
=> x(x - 40) + 32(x - 40) = 0
=> (x - 40)(x + 32) = 0
=> x = 40 or x = -32
Since speed cannot be negative, x = 40 km/h.

Verification: At 40 km/h, time = 480/40 = 12 hours.
At 32 km/h, time = 480/32 = 15 hours.
Difference = 3 hours. Correct.