CBSE - Quadratic Equations

(i) (x + 2)² = 2(x - 3)
⇒ x² + 2x + 1 = 2x - 6
⇒ x² + 7 = 0
It is of the form ax² + bx + c = 0.
Hence, the given equation is quadratic equation.

(ii) x² - 2x = (-2)(3 - x)
⇒ x² - 2x = -6 + 2x 
⇒ x² - 4x + 6 = 0
It is of the form ax² + bx + c = 0.
Hence, the given equation is quadratic equation.

(iii) (x - 2)(x + 1) = (x - 1)(x + 3)
⇒ x² - x - 2 = x² + 2x - 3
⇒ 3x - 1 =0
It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

(iv) (x - 3)(2x + 1) = x(x + 5)
⇒ 2x² - 5x - 3 = x² + 5x
⇒  x² - 10x - 3 = 0
It is of the form ax² + bx + c = 0.
Hence, the given equation is quadratic equation.

(v) (2x - 1)(x - 3) = (x + 5)(x - 1)
⇒ 2x² - 7x + 3 = x² + 4x - 5
⇒ x² - 11x + 8 = 0
It is of the form ax² + bx + c = 0.
Hence, the given equation is quadratic equation.

(vi) x² + 3x + 1 = (x - 2)²
⇒ x² + 3x + 1 = x² + 4 - 4x
⇒ 7x - 3 = 0
It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

(vii) (x + 2)³ = 2x(x² - 1)
⇒ x³ + 8 + x² + 12x = 2x³ - 2x
⇒ x³ + 14x - 6x² - 8 = 0
It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

(viii) x³ - 4x² - x + 1 = (x - 2)³
⇒  x³ - 4x² - x + 1 = x³ - 8 - 6x²  + 12x
⇒ 2x² - 13x + 9 = 0
It is of the form ax² + bx + c = 0.
Hence, the given equation is quadratic equation.

Let the breadth of the rectangular plot = x m
Hence, the length of the plot is (2x + 1) m.
Formula of area of rectangle = length × breadth = 528 m²
Putting the value of length and width, we get
(2x + 1) × x = 528
⇒ 2x² + x =528
⇒ 2x² + x - 528 = 0

Let the first integer number = x
Next consecutive positive integer will = x + 1
Product of both integers = x × (x +1) = 306
⇒ x² + x = 306
⇒ x² + x - 306 = 0

Let take Rohan's age = x years
Hence, his mother's age = x + 26
3 years from now
Rohan's age = x + 3
Age of Rohan's mother will = x + 26 + 3 = x + 29
The product of their ages 3 years from now will be 360 so that
(x + 3)(x + 29) = 360
⇒ x² + 29x + 3x + 87 = 360
⇒ x² + 32x + 87 - 360 = 0
⇒ x² + 32x - 273 = 0

Let the speed of train be x km/h.
Time taken to travel 480 km = 480/x km/h

In second condition, let the speed of train = (x - 8) km/h
It is also given that the train will take 3 hours to cover the same distance.
Therefore, time taken to travel 480 km = (480/x + 3) km/h

Speed × Time = Distance

(x - 8)(480/x + 3) = 480
⇒ 480 + 3x - 3840/x - 24 = 480
⇒ 3x - 3840/x = 24
⇒ 3x² - 8x - 1280 = 0

(i) x² – 3x – 10
= x² - 5x + 2x - 10
= x(x - 5) + 2(x - 5)
= (x - 5)(x + 2)
Roots of this equation are the values for which (x - 5)(x + 2) = 0

∴ x - 5 = 0 or x + 2 = 0

⇒ x = 5 or x = -2

(ii) 2x² + x – 6
= 2x² + 4x - 3x - 6
= 2x(x + 2) - 3(x + 2)
= (x + 2)(2x - 3)
Roots of this equation are the values for which (x + 2)(2x - 3) = 0

∴ x + 2 = 0 or 2x - 3 = 0

⇒ x = -2 or x = 3/2

(iii) √2 x² + 7x + 5√2
= √2 x² + 5x + 2x + 5√2 
= x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)
Roots of this equation are the values for which (√2x + 5)(x + √2) = 0

∴ √2x + 5 = 0 or x + √2 = 0

⇒ x = -5/√2 or x = -√2

(iv) 2x² – x + 1/8
= 1/8 (16x²  - 8x + 1)
= 1/8 (16x²  - 4x -4x + 1)
= 1/8 (4x(4x  - 1) -1(4x - 1))
= 1/8(4x - 1)²
Roots of this equation are the values for which (4x - 1)² = 0

∴ (4x - 1) = 0 or (4x - 1) = 0

⇒ x = 1/4 or x = 1/4

(v) 100x² – 20x + 1
= 100x² – 10x - 10x + 1
= 10x(10x - 1) -1(10x - 1)
= (10x - 1)²
Roots of this equation are the values for which (10x - 1)² = 0
∴ (10x - 1) = 0 or (10x - 1) = 0
⇒ x = 1/10 or x = 1/10

Let the number of John's marbles be x.
Therefore, number of Jivanti's marble = 45 - x
After losing 5 marbles,
Number of John's marbles = x - 5
Number of Jivanti's marbles = 45 - x - 5 = 40 - x
It is given that the product of their marbles is 124.

∴ (x - 5)(40 - x) = 124

⇒ x² – 45x + 324 = 0
⇒ x² – 36x - 9x + 324 = 0
⇒ x(x - 36) -9(x - 36) = 0
⇒ (x - 36)(x - 9) = 0
Either x - 36 = 0 or x - 9 = 0
⇒ x = 36 or x = 9
If the number of John's marbles = 36,
Then, number of Jivanti's marbles = 45 - 36 = 9
If number of John's marbles = 9,
Then, number of Jivanti's marbles = 45 - 9 = 36

Let the number of toys produced be x.
∴ Cost of production of each toy = Rs (55 - x)
It is given that, total production of the toys = Rs 750

∴ x(55 - x) = 750

⇒ x² – 55x + 750 = 0

⇒ x² – 25x - 30x + 750 = 0

⇒ x(x - 25) -30(x - 25) = 0

⇒ (x - 25)(x - 30) = 0

Either, x -25 = 0 or x - 30 = 0
⇒ x = 25 or x = 30
Hence, the number of toys will be either 25 or 30.

Let the first number be x and the second number is 27 - x.
Therefore, their product = x (27 - x)
It is given that the product of these numbers is 182.
Therefore, x(27 - x) = 182
⇒ x² – 27x - 182 = 0
⇒ x² – 13x - 14x + 182 = 0
⇒ x(x - 13) -14(x - 13) = 0
⇒ (x - 13)(x -14) = 0
Either x = -13 = 0 or x - 14 = 0
⇒ x = 13 or x = 14
If first number = 13, then
Other number = 27 - 13 = 14
If first number = 14, then
Other number = 27 - 14 = 13
Therefore, the numbers are 13 and 14.

Let the consecutive positive integers be x and x + 1.
Therefore, x² + (x + 1)² = 365
⇒ x² + x² + 1 + 2x = 365
⇒ 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
Either x + 14 = 0 or x - 13 = 0,

⇒ x = - 14 or x = 13
Since the integers are positive, x can only be 13.
∴ x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers will be 13 and 14.

Let the base of the right triangle be x cm.
Its altitude = (x - 7) cm
From Pythagoras theorem, we have
Base² + Altitude² = Hypotenuse²
∴ x² + (x - 7)² = 132
⇒ x² + x² + 49 - 14x = 169
⇒ 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
Either x - 12 = 0 or x + 5 = 0,
⇒ x = 12 or x = - 5
Since sides are positive, x can only be 12.
Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 - 7) cm = 5 cm.

Let the number of articles produced be x.
Therefore, cost of production of each article = Rs (2x + 3)
It is given that the total production is Rs 90.∴ x(2x + 3) = 0
⇒ 2x² + 3x - 90 = 0
⇒ 2x² + 15x -12x - 90 = 0
⇒ x(2x + 15) -6(2x + 15) = 0
⇒ (2x + 15)(x - 6) = 0
Either 2x + 15 = 0 or x - 6 = 0

⇒ x = -15/2 or x = 6

As the number of articles produced can only be a positive integer, therefore, x can only be 6.
Hence, number of articles produced = 6
Cost of each article = 2 × 6 + 3 = Rs 15.

(i) 2x² – 7x + 3 = 0
⇒ 2x² – 7x = - 3
On dividing both sides of the equation by 2, we get
⇒ x² – 7x/2  = -3/2
⇒ x² – 2 × x ×  7/4 = -3/2
On adding (7/4)² to both sides of equation, we get
⇒ (x)² - 2 × x × 7/4 + (7/4)² = (7/4)² - 3/2

⇒ (x - 7/4)² = 49/16 - 3/2
⇒ (x - 7/4)² = 25/16
⇒ (x - 7/4) = ± 5/4
⇒ x = 7/4 ± 5/4
⇒ x = 7/4 + 5/4 or x = 7/4 - 5/4
⇒ x = 12/4 or x = 2/4
⇒ x = 3 or 1/2

(ii) 2x² + x – 4 = 0
⇒ 2x² + x = 4
On dividing both sides of the equation, we get
⇒ x² + x/2 = 2
On adding (1/4)² to both sides of the equation, we get
⇒ (x)² + 2 × x × 1/4 + (1/4)² = 2 + (1/4)²
⇒ (x + 1/4)² = 33/16
⇒ x + 1/4 = ± √33/4
⇒ x = ± √33/4 - 1/4
⇒ x = ± √33-1/4
⇒ x = √33-1/4 or x = -√33-1/4

(iii) 4x² + 4√3x + 3 = 0
⇒ (2x)² + 2 × 2x × √3 + (√3)² = 0
⇒ (2x + √3)² = 0
⇒ (2x + √3) = 0 and (2x + √3) = 0
⇒ x = -√3/2 or x = -√3/2

(iv) 2x² + x + 4 = 0
⇒ 2x² + x = -4
On dividing both sides of the equation, we get
⇒ x² + 1/2x = 2
⇒ x² + 2 × x × 1/4 = -2
On adding (1/4)² to both sides of the equation, we get
⇒ (x)² + 2 × x × 1/4 + (1/4)² = (1/4)² - 2
⇒ (x + 1/4)² = 1/16 - 2
⇒ (x + 1/4)² = -31/16
However, the square of number cannot be negative.
Therefore, there is no real root for the given equation.

(i) 2x² – 7x + 3 = 0
On comparing this equation with ax² + bx + c = 0, we get
a = 2, b = -7 and c = 3
By using quadratic formula, we get
x = -b±√b² - 4ac/2a
⇒ x = 7±√49 - 24/4
⇒ x = 7±√25/4
⇒ x = 7±5/4
⇒ x = 7+5/4 or x = 7-5/4
⇒ x = 12/4 or 2/4
∴  x = 3 or 1/2

(ii) 2x² + x - 4 = 0
On comparing this equation with ax² + bx + c = 0, we get
a = 2, b = 1 and c = -4
By using quadratic formula, we get
x = -b±√b² - 4ac/2a
⇒x = -1±√1+32/4
⇒x = -1±√33/4

∴ x = -1+√33/4 or x = -1-√33/4

(iii) 4x² + 4√3x + 3 = 0
On comparing this equation with ax² + bx + c = 0, we get
a = 4, b = 4√3 and c = 3
By using quadratic formula, we get
x = -b±√b² - 4ac/2a
⇒ x = -4√3±√48-48/8
⇒ x = -4√3±0/8
∴ x = -√3/2 or x = -√3/2

(iv) 2x² + x + 4 = 0
On comparing this equation with ax² + bx + c = 0, we get
a = 2, b = 1 and c = 4
By using quadratic formula, we get
x = -b±√b² - 4ac/2a
⇒ x = -1±√1-32/4
⇒ x = -1±√-31/4
The square of a number can never be negative.
∴There is no real solution of this equation.

(i) x-1/x = 3
⇒ x² - 3x -1 = 0
On comparing this equation with ax² + bx + c = 0, we get
a = 1, b = -3 and c = -1
By using quadratic formula, we get
x = -b±√b² - 4ac/2a
⇒ x = 3±√9+4/2
⇒ x = 3±√13/2
∴ x = 3+√13/2 or x = 3-√13/2

(ii) 1/x+4 - 1/x-7 = 11/30
⇒ x-7-x-4/(x+4)(x-7) = 11/30
⇒ -11/(x+4)(x-7) = 11/30
⇒ (x+4)(x-7) = -30
⇒ x² - 3x - 28 = 30
⇒ x² - 3x + 2 = 0
⇒ x² - 2x - x + 2 = 0
⇒ x(x - 2) - 1(x - 2) = 0
⇒ (x - 2)(x - 1) = 0
⇒ x = 1 or 2

Let the present age of Rehman be x years.
Three years ago, his age was (x - 3) years.
Five years hence, his age will be (x + 5) years.
It is given that the sum of the reciprocals of Rehman's ages 3 years ago and 5 years from now is 1/3. ∴ 1/x-3 + 1/x-5 = 1/3
x+5+x-3/(x-3)(x+5) = 1/3
2x+2/(x-3)(x+5) = 1/3
⇒ 3(2x + 2) = (x-3)(x+5)
⇒ 6x + 6 = x² + 2x - 15
⇒ x² - 4x - 21 = 0
⇒ x² - 7x + 3x - 21 = 0
⇒ x(x - 7) + 3(x - 7) = 0
⇒ (x - 7)(x + 3) = 0
⇒ x = 7, -3
However, age cannot be negative.
Therefore, Rehman's present age is 7 years

Let the marks in Maths be x.
Then, the marks in English will be 30 - x.
According to the question,
(x + 2)(30 - x - 3) = 210
(x + 2)(27 - x) = 210
⇒ -x² + 25x + 54 = 210
⇒ x² - 25x + 156 = 0
⇒ x² - 12x - 13x + 156 = 0
⇒ x(x - 12) -13(x - 12) = 0
⇒ (x - 12)(x - 13) = 0
⇒ x = 12, 13
If the marks in Maths are 12, then marks in English will be 30 - 12 = 18
If the marks in Maths are 13, then marks in English will be 30 - 13 = 17

Let the shorter side of the rectangle be x m.
Then, larger side of the rectangle = (x + 30) m

⇒ x² + (x + 30)² = (x + 60)²

⇒ x² + x² + 900 + 60x = x² + 3600 + 120x

⇒ x² - 60x - 2700 = 0

⇒ x² - 90x + 30x - 2700 = 0

⇒ x(x - 90) + 30(x -90)

⇒ (x - 90)(x + 30) = 0

⇒ x = 90, -30

However, side cannot be negative. Therefore, the length of the shorter side will be 90 m.
Hence, length of the larger side will be (90 + 30) m = 120 m.

Let the larger and smaller number be x and y respectively.
According to the question,
x² - y² = 180 and y² = 8x
⇒ x² - 8x = 180
⇒ x² - 8x - 180 = 0
⇒ x² - 18x + 10x - 180 = 0
⇒ x(x - 18) +10(x - 18) = 0
⇒ (x - 18)(x + 10) = 0
⇒ x = 18, -10

Let the larger and smaller number be x and y respectively.
According to the question,
x² - y² = 180 and y² = 8x
⇒ x² - 8x = 180
⇒ x² - 8x - 180 = 0
⇒ x² - 18x + 10x - 180 = 0
⇒ x(x - 18) +10(x - 18) = 0
⇒ (x - 18)(x + 10) = 0
⇒ x = 18, -10

However, the larger number cannot be negative as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.
Therefore, the larger number will be 18 only.
x = 18
∴ y² = 8x = 8 × 18 = 144
⇒ y = ±√44 = ±12
∴ Smaller number = ±12
Therefore, the numbers are 18 and 12 or 18 and - 12.

Let the speed of the train be x km/hr.
Time taken to cover 360 km = 360/x hr.According to the question,
⇒ (x + 5)(360-1/x) = 360
⇒ 360 - x + 1800-5/x = 360
⇒ x² + 5x + 10x - 1800 = 0
⇒ x(x + 45) -40(x + 45) = 0
⇒ (x + 45)(x - 40) = 0
⇒ x = 40, -45
However, speed cannot be negative.
Therefore, the speed of train is 40 km/h.

Let the time taken by the smaller pipe to fill the tank be x hr.
Time taken by the larger pipe = (x - 10) hr
Part of tank filled by smaller pipe in 1 hour = 1/x

Part of tank filled by larger pipe in 1 hour = 1/x - 10

It is given that the tank can be filled in  = 75/8 hours by both the pipes together. Therefore,
1/x + 1/x-10 = 8/75
x-10+x/x(x-10) = 8/75
⇒ 2x-10/x(x-10) = 8/75
⇒ 75(2x - 10) = 8x² - 80x
⇒ 150x - 750 = 8x² - 80x
⇒ 8x² - 230x +750 = 0
⇒ 8x² - 200x - 30x + 750 = 0
⇒ 8x(x - 25) -30(x - 25) = 0
⇒ (x - 25)(8x -30) = 0
⇒ x = 25, 30/8
Time taken by the smaller pipe cannot be 30/8 = 3.75 hours. As in this case, the time taken by the larger pipe will be negative, which is logically not possible.

Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 - 10 =15 hours respectively.

Let the average speed of passenger train be x km/h.
Average speed of express train = (x + 11) km/h
It is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.

⇒ 132 × 11 = x(x + 11)

⇒ x² + 11x - 1452 = 0

⇒ x² +  44x -33x -1452 = 0

⇒ x(x + 44) -33(x + 44) = 0

⇒ (x + 44)(x - 33) = 0

⇒ x = - 44, 33

Speed cannot be negative.

Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.

Let the sides of the two squares be x m and y m. Therefore, their perimeter will be 4x and 4yrespectively and their areas will be x² and y² respectively.
It is given that
4x - 4y = 24
x - y = 6
x = y + 6
Also, x² + y² = 468
⇒ (6 + y²) + y² = 468
⇒ 36 + y² + 12y + y² = 468
⇒ 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, 12
However, side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.

(i) Consider the equation
x² - 3x + 5 = 0
Comparing it with ax² + bx + c = 0, we get
a = 2, b = -3 and c = 5
Discriminant = b² - 4ac
= ( - 3)² - 4 (2) (5) = 9 - 40
= - 31
As b² - 4ac < 0,
Therefore, no real root is possible for the given equation.

(ii) 3x² - 4√3x + 4 = 0
Comparing it with ax² + bx + c = 0, we get
a = 3, b = -4√3 and c = 4
Discriminant = b² - 4ac
= (-4√3)² - 4(3)(4)
= 48 - 48 = 0
As b² - 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be -b/2a and -b/2a.-b/2a = -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3
Therefore, the roots are 2/√3 and 2/√3.

(iii) 2x² - 6x + 3 = 0

Comparing this equation with ax² + bx + c = 0, we get
a = 2, b = -6, c = 3
Discriminant = b² - 4ac

= (-6)² - 4 (2) (3)
= 36 - 24 = 12
As b² - 4ac > 0,
Therefore, distinct real roots exist for this equation:

(i) 2x² + kx + 3 = 0
Comparing equation with ax² + bx + c = 0, we get
a = 2, b = k and c = 3
Discriminant = b² - 4ac

= (k)² - 4(2) (3)
= k² - 24
For equal roots,
Discriminant = 0
k² - 24 = 0
k² = 24
k = ±√24 = ±2√6

(ii) kx(x - 2) + 6 = 0
or kx² - 2kx + 6 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = k, b = - 2k and c = 6
Discriminant = b² - 4ac
= ( - 2k)2 - 4 (k) (6)
= 4k2 - 24k
For equal roots,
b2 - 4ac = 0
4k2 - 24k = 0
4k (k - 6) = 0
Either 4k = 0
or k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms 'x²' and 'x'.
Therefore, if this equation has two equal roots, k should be 6 only.

Let the breadth of mango grove be l.
Length of mango grove will be 2l.
Area of mango grove = (2l) (l)= 2l²
2l² = 800
l² = 800/2 = 400
l² - 400 =0
Comparing this equation with al² + bl + c = 0, we get
a = 1, b = 0, c = 400
Discriminant = b² - 4ac

= (0)² - 4 × (1) × ( - 400) = 1600
Here, b² - 4ac > 0
Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.
l = ±20
However, length cannot be negative.
Therefore, breadth of mango grove = 20 m
Length of mango grove = 2 × 20 = 40 m

Let the age of one friend be x years.
then the age of the other friend will be (20 - x) years.
4 years ago,

Age of 1st friend = (x - 4) years
Age of 2nd friend = (20 - x - 4) = (16 - x) years
A/q we get that,
(x - 4) (16 - x) = 48

16x - x² - 64 + 4x = 48
 - x² + 20x - 112 = 0
x² - 20x + 112 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = 1, b = -20 and c = 112
Discriminant = b² - 4ac
= (-20)² - 4 × 112
= 400 - 448 = -48
b² - 4ac < 0
Therefore, there will be no real solution possible for the equations. Such type of condition doesn't exist.

Let the length and breadth of the park be l and b.
Perimeter = 2 (l + b) = 80
l + b = 40
Or, b = 40 - l
Area = l×b = l(40 - l) = 40l - l²40l -  l² = 400
l² -  40l + 400 = 0
Comparing this equation with al² + bl + c = 0, we get
a = 1, b = -40, c = 400
Discriminant = b² - 4ac
(-40)² - 4 × 400
= 1600 - 1600 = 0
b² - 4ac = 0
Therefore, this equation has equal real roots. And hence, this situation is possible.
Root of this equation,l = -b/2a
l = (40)/2(1) = 40/2 = 20
Therefore, length of park, l = 20 m
And breadth of park, b = 40 - l = 40 - 20 = 20 m.