# Class 10 NCERT Solutions- Chapter 4 Quadratic Equations – Exercise 4.1

### Question 1. Check whether the following are quadratic equations :

### (i) (x + 1)^{2} = 2(x – 3)

**Solution:**

Here,

LHS = (x + 1)

^{2}= x

^{2}+ 2x + 1 (Using identity(a+b))^{2}= a^{2}+ 2ab + b^{2}and, RHS = 2(x–3)

= 2x – 6

As, LHS = RHS

x

^{2}+ 2x + 1 = 2x – 6x

^{2}+ 7 = 0 ………….(I)As, the eqn. (I) is in the form of

ax^{2}+ bx + c = 0 where (a≠0).Hence, the equation is

equation because highest power of x is 2.QUADRATIC

### (ii) x^{2} – 2x = (–2) (3 – x)

**Solution:**

Here,

LHS = x

^{2}– 2xand, RHS = (–2) (3 – x)

= 2x–6

As, LHS = RHS

x

^{2}– 2x = 2x – 6x

^{2}– 4x + 6 = 0 ………….(I)As, the eqn. (I) is in the form of

ax^{2}+ bx + c = 0 where (a≠0).Hence, the equation is

equation because highest power of x is 2.QUADRATIC

### (iii) (x – 2)(x + 1) = (x – 1)(x + 3)

**Solution:**

Here,

LHS = (x – 2)(x + 1)

= x

^{2}+ (–2+1)x + (–2)(1) (Using identity(x+a) (x+b) = x)^{2}+ (a+b)x + ab= x

^{2}– x – 2and, RHS = (x – 1)(x + 3)

= x

^{2}+ (–1+3)x + (–1)(3) (Using identity(x+a) (x+b) = x)^{2}+ (a+b)x + ab= x

^{2}+ 2x – 3As, LHS = RHS

x

^{2}– x – 2 = x^{2}+ 2x – 33x – 1 = 0 ………….(I)

As, the eqn. (I) is not in the form of

ax^{2}+ bx + c = 0 because (a=0).Hence, the equation is

equation because highest power of x is 1.NOT QUADRATIC

### (iv) (x – 3)(2x +1) = x(x + 5)

**Solution:**

Here,

LHS = (x – 3)(2x +1)

= 2x

^{2}+ x +(–3)(2x) + (–3)(1)= 2x

^{2}– 5x – 3and, RHS = x(x + 5)

= x

^{2}+ 5xAs, LHS = RHS

2x

^{2}– 5x – 3 = x^{2}+ 5xx

^{2}– 10x – 3 = 0 ………….(I)As, the eqn. (I) is in the form of

ax^{2}+ bx + c = 0 where (a≠0).Hence, the equation is

equation because highest power of x is 2.QUADRATIC

### (v) (2x – 1)(x – 3) = (x + 5)(x – 1)

**Solution:**

Here,

LHS = (2x – 1)(x – 3)

= 2x

^{2}+ (2x)(–3) +(–1)(x) + (–1)(–3)= 2x

^{2}– 7x + 3and, RHS = (x + 5)(x – 1)

= x

^{2}+ 5(x) + (–1)(x) + (5)(–1) (Using identity(x+a) (x+b) = x)^{2}+ (a+b)x + ab= x

^{2}+ 4(x) – 5As, LHS = RHS

2x

^{2}– 7x + 3 = x^{2}+ 4(x) – 5x

^{2}– 11x + 8 = 0 ………….(I)As, the eqn. (I) is in the form of

ax.^{2}+ bx + c = 0 where (a≠0)Hence, the equation is

equation because highest power of x is 2.QUADRATIC

### (vi) x^{2} + 3x + 1 = (x – 2)^{2}

**Solution:**

Here,

LHS = x

^{2}+ 3x + 1and, RHS = (x – 2)

^{2}= x

^{2}– 4x + 4 (Using identity(a–b)–^{2}= a^{2}2ab + b)^{2}As, LHS = RHS

x

^{2}+ 3x + 1 = x^{2}– 4x + 47x – 3 = 0 ………….(I)

As, the eqn. (I) is not in the form of

ax.^{2}+ bx + c = 0 because (a=0)Hence, the equation is

equation because highest power of x is 1.NOT QUADRATIC

### (vii) (x + 2)^{3} = 2x (x^{2} – 1)

**Solution:**

Here,

LHS = (x + 2)

^{3}= x

^{3}+ 2^{3}+ 3x(2)(x+2) (Using identity(x+y))^{3}= x^{3}+ y^{3}+ 3xy(x+y)= x

^{3}+ 8 + 6x^{2}+12xand, RHS = 2x (x

^{2}– 1)= 2x

^{3}– 2xAs, LHS = RHS

x

^{3}+ 8 + 6x^{2}+12x = 2x^{3}– 2xx

^{3}– 6x^{2}– 14x – 8 = 0 ………….(I)As, the eqn. (I) is not in the form of

ax.^{2}+ bx + c = 0 where (a≠0)Hence, the equation is

equation because highest power of x is 3.NOT QUADRATIC

### (viii) x^{3} – 4x^{2} – x + 1 = (x – 2)^{3}

**Solution:**

Here,

LHS = x

^{3}– 4x^{2}– x + 1and, RHS = (x – 2)

^{3}= x

^{3}– 2^{3}– 3x(2)(x–2) (Using identity(x–y)–^{3}= x^{3}y–^{3}3xy(x-y))= x

^{3}– 8 – 6x^{2}+12xAs, LHS = RHS

x

^{3}– 4x^{2}– x + 1 = x^{3}– 8 – 6x^{2}+12x2x

^{2}– 13x + 9 = 0 ………….(I)As, the eqn. (I) is in the form of

ax.^{2}+ bx + c = 0 where (a≠0)Hence, the equation is

equation because highest power of x is 2.QUADRATIC

### Question 2. 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 meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

**Solution:**

Let’s consider,

Breadth of the rectangular plot = b m

Then, length of the plot = (2b + 1) m.

As,

Area of rectangle = length × breadth528 m

^{2}= (2x + 1) × x2x

^{2}+ x =5282x

^{2}+ x – 528 = 0 ……………….(I)As, the eqn. (I) is in the form of

ax.^{2}+ bx + c = 0 where (a≠0)Hence, the equation is

equation because highest power of x is 2.QUADRATICTherefore, the length and breadth of plot, satisfies the quadratic equation,

2x, which is the required representation of the problem mathematically.^{2}+ x – 528 = 0

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

**Solution:**

Let’s consider,

The first integer number = x

Next consecutive positive integer will be = x + 1

Product of two consecutive integers = x × (x +1)

x

^{2}+ x = 306x

^{2}+ x – 306 = 0 ……………….(I)As, the eqn. (I) is in the form of

ax.^{2}+ bx + c = 0 where (a≠0)Hence, the equation is

equation because highest power of x is 2.QUADRATICTherefore, the two integers x and x+1, satisfies the quadratic equation,

x, which is the required representation of the problem mathematically.^{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.

**Solution:**

Let’s consider,

Age of Rohan’s = x years

So, Rohan’s mother’s age = x + 26

After 3 years,

Age of Rohan’s = x + 3

Age of Rohan’s mother will be = x + 26 + 3 = x + 29

The product of their ages after 3 years will be equal to 360, such that

(x + 3)(x + 29) = 360

x

^{2}+ 29x + 3x + 87 = 360x

^{2}+ 32x + 87 – 360 = 0x

^{2}+ 32x – 273 = 0 ……………….(I)As, the eqn. (I) is in the form of

ax.^{2}+ bx + c = 0 where (a≠0)Hence, the equation is

equation because highest power of x is 2.QUADRATICTherefore, the age of Rohan and his mother, satisfies the quadratic equation,

x, which is the required representation of the problem mathematically.^{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.

**Solution:**

Let’s consider,

The speed of train = x km/hr

Time taken to travel 480 km = (480/x) hr

Time = Distance / SpeedHere, According to the given condition,

The speed of train = (x – 8) km/hr

As, the train will take 3 hours more to cover the same distance.

Hence, Time taken to travel 480 km = 480/(x+3) km/hr

As,

Speed × Time = Distance(x – 8)(480/(x + 3) = 480

480 + 3x – 3840/x – 24 = 480

3x – 3840/x = 24

3x

^{2}– 8x – 1280 = 0 ……………….(I)As, the eqn. (I) is in the form of

ax.^{2}+ bx + c = 0 where (a≠0)Hence, the equation is

QUADRATICequation because highest power of x is 2.Therefore, the speed of the train, satisfies the quadratic equation,

3x, which is the required representation of the problem mathematically.^{2}– 8x – 1280 = 0

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