Book Name |
: Ganit Prakash |

Subject |
: Mathematics |

Class |
: 10 (Madhyamik) |

Publisher |
: Prof Nabanita Chatterjee |

Chapter Name |
: Quadratic Equations With One Variable (1st Chapter) |

Table of Contents

Toggle**Application 1**

**I see whether the following equations can be written in the form of ax ^{2} + bx + c = 0, Where a, b, c are real numbers and a ≠ 0.**

**(iv) (x – 2) ^{2} = x^{3} – 4x + 4**

**Solution**

(x – 2)^{2} = x^{3} – 4x + 4

or, x^{2} – 4x + 4 = x^{3} – 4x + 4

or, x^{2} – x^{3} = 0

This is not a quadratic equation, as it is not in the form ax^{2} + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

**(v) (x – 3) ^{3} = 2x(x^{2 }– 1)**

**Solution**

x^{3} – 9x^{2} + 27x + 27 = 2x^{3} – 2x

or, x^{3} – 2x^{3} – 9x^{2} + 27x + 2x + 27 = 0

or, -x^{3} – 9x^{2} + 29x + 27 = 0

This is not a quadratic equation, as it is not in the form ax^{2} + 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, x^{2} + 2x – 24 = 0

Here, co-efficient 0f x^{2 }= 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) **

**x ^{2} – 7x + 2 **

**Solution**

This is a quadratic equation.

**Reason:**

As it is in the form ax^{2} + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

**Question (ii)**

**7x ^{3} – x(x+2)**

**Solution**

7x^{3} – x^{2} – 2x

This is not a quadratic equation.

**Reason:**

As it is not in the form ax^{2} + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

**Question (iii)**

**2x(x+5) + 1**

**Solution**

2x^{2} + 10x + 1

This is a quadratic equation.

**Reason:**

As it is in the form ax^{2} + 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 ax^{2} + 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 ax ^{2} + 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, x^{2} – x + 1 = 6x

or, x^{2} – x – 6x + 1 = 0

or, x^{2} – 7x + 1 = 0

This is a quadratic equation. As it is in the form ax^{2} + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

**(ii) x + 3/x = x ^{2}**

**Solution**

x2+3x=x2

or, x^{2} + 3 = x^{3}

or, x^{3} – x^{2} – 3 = 0

This is not a quadratic equation. As it is not in the form ax^{2} + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

**(iii) x ^{2} – 6√x + 2 = 0**

**Solution**

This is not a quadratic equation. As it is not in the form ax^{2} + bx + c = 0 [where a, b, c are real numbers and a ≠ 0]

**(iv) (x – 2) ^{2} = x^{2} – 4x + 4**

**Solution**

This is not a quadratic equation.

**Question 3**

**Let us determine the power of the variable for which the equation x ^{6} – x^{3} – 2 = 0 will become a quadratic equation.**

**Solution**

x^{6} – x^{3} – 2 = 0

or, (x^{3})^{2} – x^{3} – 2 = 0

This is a quadratic equation with respect to x^{3} .

**Question 4 (i) **

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

**Solution**

(a – 2)x^{2} + 3x + 5 = 0

If the value of a = 2, it will be 0x^{2} + 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 ax ^{2} + 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, 3x^{2} – 4 + x = 0

or, 3x^{2} + x – 4 = 0

This is in the form of ax^{2} + bx + c = 0 and the coefficient of x is 1.

**Question 4 (iii) **

**Let us express 3x ^{2} + 7x + 23 = (x + 4)(x + 3) + 2 in the form of the quadratic equation ax^{2} + bx + c = 0 (a ≠ 0).**

**Solution**

3x^{2} + 7x + 23 = (x + 4)(x + 3) + 2

or, 3x^{2} + 7x + 23 = x^{2 }+ 7x + 12 + 2

or, 3x^{2} = 7x – 7x + 23 – 14 – x^{2} = 0

or, 2x^{2} + 0.x + 9 = 0

**Question 4 (iv) **

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

**Solution**

(x + 2)^{3} = x(x^{2} – 1)

or, x^{3} + 6x^{2} + 12x + 8 = x^{3} – x

or, x^{3} – x^{3} + 6x^{2} + 12x + x + 8 = 0

or, 6x^{2} + 13x + 8 = 0

∴ co-efficient of x^{2} 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

x^{2} = 42 – x

or, x^{2} + 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, 4x^{2} – 1 = 143

or, 4x^{2} – 1 – 143 = 0

or, 4x^{2} – 144 = 0

or, 4(x^{2} – 36) = 0

or, (x^{2} – 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

x^{2} + (x + 1)^{2} = 313

or, x^{2} + x^{2} + 2x + 1 = 313

or, 2x^{2} + 2x – 312 = 0

∴ x^{2} + 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,

x^{2} + (x + 3)^{2} = 15^{2}

or, x^{2} + x^{2} + 6x + 9 = 225

or, 2x^{2} + 6x – 216 = 0

or, 2(x^{2} + 3x – 108) = 0

or, x^{2} + 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, -x^{2} + 320 – 4x = 0

∴ x^{2 }+ 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(x^{2} + 5x)

or, 1500 = 2(x^{2} + 5x)

or, 2(x^{2} + 5x) = 1500

or, x^{2} + 5x = 750

or, x^{2} + 5x – 750 = 0

∴ x^{2} + 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

= x^{2}/100

According to the problem,

x+x^{2}/00=336

or, 100x + x^{2} = 33600

or, x^{2} + 100x – 33600 = 0

∴ x^{2} + 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(x^{2} – 4)

or, 5(x^{2} – 4) = 21x

or, 5x^{2} – 20 – 21x = 0

∴ 5x^{2} – 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 = x^{2} + 3x

or, x^{2} + 3x = 4x + 6

or, x^{2} + 3x – 4x – 6 = 0

or, x^{2} – x – 6 = 0

∴ x^{2} – 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, x^{2} + 6x + 12 = 11x+6

or, x^{2} – 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 + 4x^{2} – 1800 = 450

or, 4x^{2} + 170x – 450 = 0

or, 2x^{2} + 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 x ^{2} + kx + 3 = 0 [Let me do it myself].**

**Solution**

As one root of equation x^{2} + 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 – a^{2}x)(bx – 1) + (2b – b^{2}x)(ax – 1) = 0

or, 2abx – a^{2}bx^{2} – 2a + a^{2}x +2abx – ab^{2}x^{2} – 2b + b^{2}x = 0

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

or, – (a^{2}b + ab^{2})x + (a^{2} + b^{2} + 4ab)x – (2a + 2b) = 0

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

or, ab (a + b)x^{2} – (a + b)^{2}x – 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, (x^{2} + 6x + 9 + x^{2} – 6x + 9) × 2 = 5 × (x^{2} – 9)

or, 4x^{2} + 36 = 5x^{2} – 45

or, 5x^{2} – 4x^{2} – 45 – 36 = 0

or, x^{2} – 9^{2} = 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)**

**x ^{2} + 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)**

**8x ^{2} + 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)**

**x ^{2} – √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 7x ^{2} + kx – 3 = 0**

**Solution**

7x^{2} + kx – 3 = 0

As 2/3 is one root of the equation 7x^{2} + 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 x ^{2} + 3ax + k = 0.**

**Solution**

x^{2} + 3ax + k = 0

As (-a) is the root of the equation

∴ x^{2} + 3ax + k = 0

(-a)^{2} + 3a(-a) + k = 0

a^{2} – 3a^{2} + k = 0

-2a^{2} + k = 0

∴ k = 2a^{2}

**Question 3**

**If 2/3 and -3 are the two roots of the quadratic equation ax ^{2} + 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) **

**3y ^{2 }– 20 = 160 – 2y^{2}**

**Solution**

3y^{2 }– 20 = 160 – 2y^{2}

or, 3y^{2}+ 2y^{2} = 160 + 20

or, 5y^{2} = 180

or, 5y^{2} – 180 = 0

or, 5(y^{2} – 36) = 0

or, y^{2} – 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, 4x^{2} + 4x + 1 + x^{2} + 2x + 1 – 6x – 47 = 0

or, 5x^{2} – 45 = 0

or, 5(x^{2}-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, x^{2} – 7x – 9x+ 63 – 195 = 0

or, x^{2} – 16x – 132 = 0

or, x^{2} – 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, (3x^{2} – 24)/x = x/3

or, (3x^{2}-24) × 3 = x^{2}

or, 9x^{2} – 72 – x^{2} = 0

or, 8x^{2} – 72 = 0

or, 8(x^{2}-9) = 0

or, (x^{2} – 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, (x^{2} + 9)/3x = 15/x

or, x^{2} + 9 = 45

or, x^{2} -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, (10x^{2} – 1)/x = 3

or, 10x^{2} – 1 = 3x

or, 10x^{2} – 3x – 1 =0

or, 10x^{2} – 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