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

 

Let us work out 1.5

Question 1

Let us write the nature of two roots of the following quadratic equations :

(i) 2x² + 7x + 3 = 0

(ii) 3x² – 2√6 x + 2 = 0

(iii) 2x² – 7x + 9 = 0

(iv) 5/3 x² – 2/3 x + 1 = 0

Solution

Question 2

By calculating, let us write the value(s) of k for which each of the following quadratic equations has real and equal roots —

(i) 49x² + kx + 1 = 0

Solution

(ii) 3x² – 5x + 2k = 0

Solution

(iii) 9x² – 24x + k = 0

Solution

(iv) 2x² + 3x + k = 0

Solution

(v) x² – 2(5 + 2k)x + 3(7 + 10k) = 0

Solution

(vi) (3k + 1)x² + 2(k + 1)x + k = 0

Solution

Question 3

Let us form the quadratic equations from two roots given below —

(i) 4, 2

Solution

(ii) -4, -3

Solution

(iii) -4, 3

Solution

(iv) 5, -3

Solution

Question 4

Find m for which the two roots of the quadratic equation 4x² + 4(3m – 1)x + (m + 7) = 0 are reciprocal to each other.

Solution

Question 5

If two roots of the quadratic equation (b – c)x² + (c – a)x + (a – b) = 0 are equal, then let us prove that, 2b = a + c.

Solution

Question 6

If two roots of the quadratic equation (a² + b²)x² – 2(ac + bd)x + (c² + d²) = 0 are equal, then let us prove that, a/b = c/d.

Solution

Question 7

Let us prove that, the quadratic equation 2(a² + b²)x² + 2(a + b)x + 1 = 0 has no real root, if a ≠ b.

Solution

Question 8

If two roots of the quadratic equation 5x² + 2x – 3 = 0 are α and β, then let us determine the value of:

(i) α² + β²

(ii) α³ + β³

(iii) 1/α + 1/β

(iv) α²/β + β²/α

Solution