Chapter – 1 : Quadratic Equations With One Variable | Chapter Solution Class 10

Book Name	: Ganit Prakash
Subject	: Mathematics
Class	: 10 (Madhyamik)
Publisher	: Prof Nabanita Chatterjee
Chapter Name	: Quadratic Equations With One Variable (1st Chapter)

Application 1

I see whether the following equations can be written in the form of ax² + bx + c = 0, Where a, b, c are real numbers and a ≠ 0.

(iv) (x – 2)² = x³ – 4x + 4

Solution

(x – 2)² = x³ – 4x + 4

or, x² – 4x + 4 = x³ – 4x + 4

or, x² – x³ = 0

This is not a quadratic equation, as it is not in the form ax² + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

(v) (x – 3)³ = 2x(x² – 1)

Solution

x³ – 9x² + 27x + 27 = 2x³ – 2x

or, x³ – 2x³ – 9x² + 27x + 2x + 27 = 0

or, -x³ – 9x² + 29x + 27 = 0

This is not a quadratic equation, as it is not in the form ax² + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

Application 4

The length of a rectangular area is 2 m more than its breadth and its area is 24 sqm. I construct a quadratic equation in one variable. [Let me do it myself]

Solution

Let the breadth of the rectangle = x m

the length of the rectangle = (x + 2) m

According to the given problem,

(x + 2)x = 24

or, x² + 2x – 24 = 0

Here, co-efficient 0f x² = 1, co-efficient 0f x = 2 and co-efficient 0f x^o = -24

Let us workout – 1.1

Question 1

I write the quadratic polynomials from the following polynomials by understanding them.

Question (i)

x² – 7x + 2 

Solution

This is a quadratic equation.

Reason:

As it is in the form ax² + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

Question (ii)

7x³ – x(x+2)

Solution

7x³ – x² – 2x

This is not a quadratic equation.

Reason:

As it is not in the form ax² + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

Question (iii)

2x(x+5) + 1

Solution

2x² + 10x + 1

This is a quadratic equation.

Reason:

As it is in the form ax² + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

Question (iv)

2x – 1

Solution

2x – 1

This is not a quadratic equation.

Reason:

As it is not in the form ax² + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

Question 2

Which of the following equations can be written in the form of ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0? Let us write it.

(i) x – 1 + 1/x = 6, (x ≠ 0)

Solution

x2-x+1x=6

or, x² – x + 1 = 6x

or, x² – x – 6x + 1 = 0

or, x² – 7x + 1 = 0

This is a quadratic equation. As it is in the form ax² + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

(ii) x + 3/x = x²

Solution

x2+3x=x2

or, x² + 3 = x³

or, x³ – x² – 3 = 0

This is not a quadratic equation. As it is not in the form ax² + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

(iii) x² – 6√x + 2 = 0

Solution

This is not a quadratic equation. As it is not in the form ax² + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

(iv) (x – 2)² = x² – 4x + 4

Solution

This is not a quadratic equation.

Question 3

Let us determine the power of the variable for which the equation x⁶ – x³ – 2 = 0 will become a quadratic equation.

Solution

x⁶ – x³ – 2 = 0

or, (x³)² – x³ – 2 = 0

This is a quadratic equation with respect to x³ .

Question 4 (i) 

Let us determine the value of ‘a’ for which the equation (a – 2)x² + 3x + 5 = 0 will not be a quadratic equation.

Solution

(a – 2)x² + 3x + 5 = 0

If the value of a = 2, it will be 0x² + 3x + 5 = 0

or, 3x + 5 = 0

So, if the value of a = 2, it will not be a quadratic equation.

Question 4 (ii) 

If x/(4 – x) = 1/3x, (x ≠ 0, x ≠ 4) be expressed in the form of ax² + bx + c = 0 (a ≠ 0), then let us determine the coefficient of x.

Solution

x4-x=13x x4-x-13x=0 3×2-(4-x)(4-x)3x=0

or, 3x² – 4 + x = 0

or, 3x² + x – 4 = 0

This is in the form of ax² + bx + c = 0 and the coefficient of x is 1.

Question 4 (iii) 

Let us express 3x² + 7x + 23 = (x + 4)(x + 3) + 2 in the form of the quadratic equation ax² + bx + c = 0 (a ≠ 0).

Solution

3x² + 7x + 23 = (x + 4)(x + 3) + 2

or, 3x² + 7x + 23 = x² + 7x + 12 + 2

or, 3x² = 7x – 7x + 23 – 14 – x² = 0

or, 2x² + 0.x + 9 = 0

Question 4 (iv) 

Let us express the equation (x + 2)³ = x(x² – 1) in the form of ax² + bx + c = 0 (a ≠ 0) and write the co-efficient of x², x and x^o.

Solution

(x + 2)³ = x(x² – 1)

or, x³ + 6x² + 12x + 8 = x³ – x

or, x³ – x³ + 6x² + 12x + x + 8 = 0

or, 6x² + 13x + 8 = 0

∴ co-efficient of x² is 6,

co-efficient of x is 13,

co-efficient of x^o is 8.

Question 5

Let us construct quadratic equations in one variable from the following statement.

Question 5 (i)

Divide 42 into two parts such that one part is equal to the square of the other part.

Solution

Let one part is x and the other part is 42 – x.

According to question

x² = 42 – x

or, x² +  x – 42 = 0

Question 5 (ii)

The product of two consecutive positive odd numbers is 143.

Solution

Let one positive odd number is 2x – 1 and the next odd number is 2x + 1.

∴ (2x – 1)(2x + 1) = 143

or, 4x² – 1 = 143

or, 4x² – 1 – 143 = 0

or, 4x² – 144 = 0

or, 4(x² – 36) = 0

or, (x² – 36) = 0

Question 5 (iii)

The sum of the squares of two consecutive numbers is 313.

Solution

let one number be x and the other number is x + 1.

According to Question

x² + (x + 1)² = 313

or, x² + x² + 2x + 1 = 313

or, 2x² + 2x – 312 = 0

∴ x² + x – 156 = 0

Question

Let us construct the quadratic equations in one variable from the following statements.

Question 6 (i)

The length of the diagonal of a rectangular area is 15 m and the length exceeds its breadth by 3 m.

Solution

Let the breadth and the length of the rectangle are ‘x’ m. and (x + 3) m respectively.

According to the given problem,

x² + (x + 3)² = 15²

or, x² + x² + 6x + 9 = 225

or, 2x² + 6x – 216 = 0

or, 2(x² + 3x – 108) = 0

or, x² + 3x – 108 = 0

Question 6 (ii)

One person bought some kg of sugar in Rs 80. If he would get 4 kg more sugar with that money, then the price of 1 kg sugar would be less by Rs 1.

Solution

Let the price of x kg sugar is Rs 80

∴ The price of 1 kg of sugar is 80/x

According to the given problem,

or, 80 – x + 320/x – 4 – 80 = 0

or, -x² + 320 – 4x = 0

∴ x² + 4x -320 = 0 is the required equation.

Question 6 (iii)

The distance between the two stations is 300 km. A train went to the second station from the first station with uniform velocity. If the velocity of the train is 5 km/hour more, then the time taken by the train to reach the second station would be lesser by 2 hours.

Solution

Let the speed of the train is ‘x’ km/hr

∴ To go 300 km the train takes 300/x hr

If the speed of the train is (x + 5) km/hr, the time required will 300/(x + 5) hr.

According to the problem,

or, 300x + 1500 – 300 x = 2(x² + 5x)

or, 1500 = 2(x² + 5x)

or, 2(x² + 5x) = 1500

or, x² + 5x = 750

or, x² + 5x – 750 = 0

∴ x² + 5x – 750 = 0  is the required equation.

Question 6 (iv)

A clock seller sold a clock by purchasing it at Rs. 336. The amount of his profit percentage is as much as the amount with which he bought the clock.

Solution

Let the cost price of the watch is Rs. x.

∴ Profit = x% of x

= x×x100

= x²/100

According to the problem,

x+x²/00=336

or, 100x + x² = 33600

or, x² + 100x – 33600 = 0

∴ x² + 100x – 33600 = 0 is the required equation.

Question 6 (v)

If the velocity of the stream is 2 km/hr, then the time taken by Ratanmajhi to cover the 21 km downstream and upstream is 10 hours.

Solution

Let the speed of the boat in still water is x km/hr.

∴ The speed of the boat with the current is (x + 2) km/hr and the speed of the boat against the current is (x-2) km/hr.

According to the problem,

or, 21x – 42 + 21x + 42 = 10 (x + 2)(x – 2)

or, 42x = 10(x² – 4)

or, 5(x² – 4) = 21x

or, 5x² – 20 – 21x = 0

∴ 5x² – 21x – 20 = 0 is the required equation.

Question 6 (vi)

The time taken to clean out a garden by Majid is 3 hours more than Mahim. Both of them together can complete the work in 2 hours.

Solution

Let Mahim alone takes x hrs. to finish the work & Majid alone take (x + 3) hrs to finish the work.

According to the problem,

or, 4x + 6 = x² + 3x

or, x² + 3x = 4x + 6

or, x² + 3x – 4x – 6 = 0

or, x² – x – 6 = 0

∴ x² – x – 6 = 0 is the required equation.

Question 6 (vii)

The unit digit of a two-digit number exceeds its ten’s digit by 6 and the product of two digits is less by 12 from the number.

Solution

Let in a two digit number, the digit in the tenth place is x and the digit in the unit place is (x+6).

∴ The number is 10x + (x+6).

According the given problem, x(x+6) + 12 = 10x + x + 6

or, x² + 6x + 12 = 11x+6

or, x² – 5x + 6 = 0 is the required equation.

Question 6 (viii)

There is a road of equal width around the outside of a rectangular playground having a length of 45 m. and breadth of 40 m and the area of the road is 450 sq.m.

Solution

Let the width of the path = ‘x’ m.

According to the problem,

(45+ 2x) × (40 + 2x) – 45 × 40 = 450

or, 1800 + 90x + 80x + 4x² – 1800 = 450

or, 4x² + 170x – 450 = 0

or, 2x² + 85x – 225 = 0 is the required equation.

Application 8

Let me write by calculating, for what value of k, 1 will be a root of the quadratic equation x² + kx + 3 = 0 [Let me do it myself].

Solution

As one root of equation x² + kx + 3 = 0 is 1.

∴ 12+ k. 1+3=0

or, 1+k+3=0

or k+ 4 = 0

∴ k = -4.

Application 13

I solve and write the two roots of the quadratic equation a/(ax – 1) + b/(bx – 1) = a + b [Let me do it myself]

Solution

or, (2a – a²x)(bx – 1) + (2b – b²x)(ax – 1) = 0

or, 2abx – a²bx² – 2a + a²x +2abx – ab²x² – 2b + b²x = 0

or, – a²bx² – ab²x² + 2abx + a²x + 2abx + b²x – (2a + 2b) = 0

or, – (a²b + ab²)x + (a² + b² + 4ab)x – (2a + 2b) = 0

or, (a²b + ab²)x – {(a + b)² + 2ab}x + 2(a + b) = 0

or, ab (a + b)x² – (a + b)²x – 2abx + 2(a + b) = 0

or, x(a + b){abx – (a+b)} – 2{abx – (a+b)}=0

or, {abx – (a+b)} {x(a+b) – 2} = 0

Either (a + b)x – 2 = 0

∴ x = 2/(a+b)

or, abx – (a + b) = 0

∴ x = (a+b)/ab

Application 15

I solve the quadratic equation (x + 3)/(x – 3) + (x – 3)/(x + 3) = 2½ (x ≠ -3, 3) [Let do it myself]

Solution

or, (x² + 6x + 9 + x² – 6x + 9) × 2 = 5 × (x² – 9)

or, 4x² + 36 = 5x² – 45

or, 5x² – 4x² – 45 – 36 = 0

or, x² – 9² = 0

or, (x + 9)(x – 9) = 0

Either, x + 9 = 0  ∴ x = – 9

or, x – 9 = 0  ∴ x = 9

Let us workout – 1.2

Question 

1. In each of the following cases, let us justify and write whether the given values are the roots of the given quadratic equation :

Question 1 (i)

x² + x + 1 = 0; 1, -1

Solution

When x = 1

(1)² + 1 + 1 = 3

When x = – 1

(-1)² – 1 + 1 = 1

∴ 1 and -1 are not the roots of the given equation.

Question 1 (ii)

8x² + 7x = 0, 0, – 2

Solution

When x = 0

∴ 8(0)² + 7(0) = 0

When x = – 2

∴ 8(-2)² + 7(-2) = 32 – 14 = 18

∴ 0 is the root while -2 is not the root of the given equation.

Question 1 (iii)

x + 1/x = 13/6

Solution

When x = 5/6

∴

When x = 4/3

∴

∴ 5/6 and 4/3 are not the roots of the given equation.

Question 1 (iv)

x² – √3 x – 6 = 0

Solution

When x = – √3

∴ (- √3)² – √3 (- √3) – 6

= 3 + 3 – 6

= 0

When x = 2√3

∴ (2√3)² – √3 (2√3) – 6

= 12 – 6 – 6

= 0

∴ (- √3) and (2√3) are the roots of the given equation.

Question 2 (i)

Let us calculate and write the value of k for which 2/3 will be a root the quadratic equation 7x² + kx – 3 = 0

Solution

7x² + kx – 3 = 0

As 2/3 is one root of the equation 7x² + kx – 3 = 0

∴ 7(2/3)² + k (2/3) – 3 = 0

or, 7 (4/9) + 2k/3 – 3 = 0

or, 28/9 + 2k/3 – 3 = 0

or, 2k/3 = 3 – 28/9

or, 2k/3 = -1/9

∴ k = – 1/6

Question 2 (ii)

Let us calculate and write the value of k for which -a will be a root of the quadratic equation x² + 3ax + k = 0.

Solution

x² + 3ax + k = 0

As (-a) is the root of the equation

∴ x² + 3ax + k = 0

(-a)² + 3a(-a) + k = 0

a² – 3a² + k = 0

-2a² + k = 0

∴ k = 2a²

Question 3

If 2/3 and -3 are the two roots of the quadratic equation ax² + 7x + b = 0, then let me calculate the values of a and b.

Solution

When x = 2/3

∴ a(2/3)² + 7(2/3) + b = 0

or, 4a/9 + 14/3 + b = 0

or, (4a + 42 + 9b)/9 = 0

or, 4a + 9b = -42 —– (i)

When x = -3

a(-3)² + 7 (-3) + b = 0

9a-21 + b = 0

9a + b = 21 —– (ii)

Now, solving 4a+ 9b = 42 and 9a + b = 21,

we get, a = 3 & b = – 6

Question

Let us solve

Question 4 (i) 

3y² – 20 = 160 – 2y²

Solution

3y² – 20 = 160 – 2y²

or, 3y²+ 2y² = 160 + 20

or, 5y² = 180

or, 5y² – 180 = 0

or, 5(y² – 36) = 0

or, y² – 36 = 0

or, (y + 6) (y – 6) = 0

∴ When y + 6 = 0 ⇒ y = – 6

and when y 6 = 0 ⇒ y = 6

Question 4 (ii)

(2x+1)² + (x + 1)² = 6x + 47

Solution

(2x + 1)² + (x + 1)² = 6x + 47

or, 4x² + 4x + 1 + x² + 2x + 1 – 6x – 47 = 0

or, 5x² – 45 = 0

or, 5(x²-9) = 0

(x)²– (3)² = 0

(x + 3) (x-3)= 0

∴ Either x + 3 = 0 ⇒ x = -3

Or, x – 3=0 ⇒ x = 3

Question 4 (iii)

(x-7) (x-9) = 195

Solution

(x-7) (x-9) = 195

or, x² – 7x – 9x+ 63 – 195 = 0

or, x² – 16x – 132 = 0

or,  x² – 22x + 6x – 132 = 0

or, x(x-22)+6 (x-22) = 0

=> (x – 22) (x+6) = 0

Either x – 22 = 0 ⇒ x = 22

or, x + 6 = 0 ⇒ x = – 6

Question 4 (iv)

3x – 24/x = x/3, x ≠ 0

Solution

3x – 24/x = x/3

or, (3x² – 24)/x = x/3

or, (3x²-24) × 3 = x²

or, 9x² – 72 – x² = 0

or, 8x² – 72 = 0

or, 8(x²-9) = 0

or, (x² – 9) = 0

or, (x)² – (3)² = 0

(x + 3) (x-3)= 0

Either x + 3 = 0 ⇒ x = -3

Or, x – 3 = 0 ⇒ x = 3

Question 4 (v)

x/3 + 3/x = 15/x, x ≠ 0

Solution

x/3 + 3/x = 15/x

or, (x² + 9)/3x = 15/x

or, x² + 9 = 45

or, x² -36 = 0

or, (x + 6) (x-6)= 0

∴ x = 6 and – 6

Question 4 (vi)

10x – 1/x = 3, x ≠ 0

Solution

10x – 1/x = 3

or, (10x² – 1)/x = 3

or, 10x² – 1 = 3x

or, 10x² – 3x – 1 =0

or, 10x² – 5x + 2x – 1 = 0

or, 5x(2x – 1) + 1(2x – 1)= 0

or, (2x – 1) (5x + 1) = 0

Either, 2x – 1 = 0 ⇒ x = 1/2

or, 5x + 1 = 0 ⇒ x = -1/5