## NCERT10th MathematicsExercise 4.3Topic: 4 Quadratic Equations

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## EXERCISE 4.3

**Q1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:**

**(i) 2x**

^{2}– 7x + 3 = 0 (ii) 2x^{2}+ x – 4 = 0**(iii) 4x**

^{2}+ 4√3x + 3 = 0 (iv) 2x^{2}+ x + 4 = 0### Explanation:

1) The quadratic equation is of the form ax

^{2}+ bx + c = 0, where a ≠ 0.ax^{2}+ bx + c = 0

x^{2}+ (b/a)x + (c/a) = 0

x^{2}+ (b/a)x = - (c/a)

x^{2}+ (b/a)x = - (c/a) ------------------ equation 1.

2) In the quadratic equation, for getting the perfect square,

last term = [(middle term)^{2}]/4 ------------------ equation 2.

3) From equation 1 and equation 2, last term = [(b/a)

^{2}]/4.4) Adding [(b/a)

^{2}]/4 to both sides of equation 1, we get,x^{2}+ (b/a)x + [(b/a)^{2}]/4 = [(b/a)^{2}]/4 - (c/a)

x^{2}+ (b/a)x + (b/2a)^{2}= (b/2a)^{2}- (c/a)

x^{2}+ (b/a)x + (b/2a)^{2}= (b)^{2}/4a^{2}- (c/a)

(x + b/2a)^{2}= (b^{2 }- 4ac)/4a^{2}

(x + b/2a) = ±√[(b^{2 }- 4ac)/4a^{2}]

(x + b/2a) = ±x = - b/2a ±√[(b^{2 }- 4ac)]/2a√[(b^{2 }- 4ac)]/2a

x = [- b ±√(b^{2 }- 4ac)]/2a ------------------ equation 3.

5) This is the method of the square.

### Solution:

**(i) 2x**

^{2}– 7x + 3 = 0

1) The given equation is 2x

^{2}– 7x + 3 = 0 ------------------ equation 1.2) Divide equation 1 by 2 we get,

2x^{2}– 7x + 3 = 0

x^{2}– (7/2)x + (3/2) = 0

x^{2}– (7/2)x = - (3/2) ------------------ equation 2.

3) From equation 1,

last term = [(middle term)^{2}]/4

last term = [(7/2)^{2}]/4

last term = 49/164) Add 49/16 to both sides of equation 2, and we get,

x^{2}- (7/2)x + 49/16 = 49/16 - 3/2

x^{2}- (7/2)x + 49/16 = 49/16 - 24/16

x^{2}- (7/2)x + 49/16 = (49 - 24)/16

(x - 7/4)^{2}= 25/16

(x - 7/4 = ±√(25/16)

(x - 7/4) = ± 5/4

x = 7/4 ± 5/4

x = (7 ± 5)/4

x = (7 + 5)/4, or x = (7 - 5)/4

x = 12/4, or x = 2/4

x = 3, or x = 1.

5) The roots of the quadratic equation 2x

^{2}– 7x + 3 = 0 are 3 and 1.**(ii) 2x**

^{2}+ x – 4 = 01) The given equation is 2x

^{2}+ x - 4 = 0 ------------------ equation 1.2) Divide equation 1 by 2 we get,

2x^{2}+ x - 4 = 0

x^{2}+ (1/2)x - (4/2) = 0

x^{2}+ (1/2)x = 2 ------------------ equation 2.

3) From equation 1,

last term = [(middle term)^{2}]/4

last term = [(1/2)^{2}]/4

last term = 1/164) Add 1/16 to both sides of equation 2, and we get,

x^{2}+ (1/2)x + 1/16 = 1/16 + 2

x^{2}+ (1/2)x + 1/16 = 1/16 + 32/16

x^{2}+ (1/2)x + 1/16 = (1 + 32)/16

(x + 1/4)^{2}= 33/16

(x + 1/4) = ±√(33/16)

(x + 1/4) = ±√33/4

x = - (1/4 ±√33/4)

x = (- 1 ±√33)/4

x = (- 1 +√33)/4, or x = (- 1 -√33)/4.

5) The roots of the quadratic equation 2x

^{2}+ x - 4 = 0 are(- 1 +√33)/4 and (- 1 -√33)/4.

**(iii) 4x**

^{2}+ 4√3x + 3 = 01) The given equation is 4x

^{2}+ 4√3x + 3 = 0 ------------------ equation 1.2) Divide equation 1 by 4 we get,

4x^{2}+ 4√3x + 3 = 0

x^{2}+ √3x + (3/4) = 0

x^{2}+ √3x = - (3/4) ------------------ equation 2.

3) From equation 1,

last term = [(middle term)^{2}]/4

last term = [(√3)^{2}]/4

last term = 3/44) Add 3/4 to both sides of equation 2, and we get,

x^{2}+ √3x + (3/4) = (3/4) - (3/4)

x^{2}+ √3x + (3/4) = 0

(x + 3/2)^{2}= 0

(x + 3/2) = ± 0

x = - (3/2 ± 0)

x = - 3/2 ± 0

x = - 3/2, or x = - 3/2.

5) The roots of the quadratic equation 4x

^{2}+ 4√3x + 3 = 0 are -3/2 and -3/2.**(iv) 2x**

^{2}+ x + 4 = 01) The given equation is 2x

^{2}+ x + 4 = 0 ------------------ equation 1.2) Divide equation 1 by 2 we get,

2x^{2}+ x + 4 = 0

x^{2}+ (1/2)x + (4/2) = 0

x^{2}+ (1/2)x = - 2 ------------------ equation 2.

3) From equation 1,

last term = [(middle term)^{2}]/4

last term = [(1/2)^{2}]/4

last term = 1/164) Add 1/16 to both sides of equation 2, and we get,

x^{2}+ (1/2)x + 1/16 = 1/16 - 2

x^{2}+ (1/2)x + 1/16 = 1/16 - 32/16

x^{2}+ (1/2)x + 1/16 = (1 - 32)/16

(x + 1/4)^{2}= - 33/16

5) As the square of any real number can't be negative, so 2x

^{2}+ x + 4 = 0 has noreal roots.

**Q2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic**

**formula.**

**(i) 2x**

^{2}– 7x + 3 = 0 (ii) 2x^{2}+ x – 4 = 0**(iii) 4x**

^{2}+ 4√3x + 3 = 0 (iv) 2x^{2}+ x + 4 = 0### Explanation:

1) The quadratic equation is of the form ax

^{2}+ bx + c = 0, where a ≠ 0.ax^{2}+ bx + c = 0

x^{2}+ (b/a)x + (c/a) = 0

x^{2}+ (b/a)x = - (c/a)

x^{2}+ (b/a)x = - (c/a) ------------------ equation 1.

2) In the quadratic equation, for getting the perfect square,

last term = [(middle term)^{2}]/4 ------------------ equation 2.

3) From equation 1 and equation 2, last term = [(b/a)

^{2}]/4.4) Adding [(b/a)

^{2}]/4 to both sides of equation 1, we get,x^{2}+ (b/a)x + [(b/a)^{2}]/4 = [(b/a)^{2}]/4 - (c/a)

x^{2}+ (b/a)x + (b/2a)^{2}= (b/2a)^{2}- (c/a)

x^{2}+ (b/a)x + (b/2a)^{2}= (b)^{2}/4a^{2}- (c/a)

(x + b/2a)^{2}= (b^{2 }- 4ac)/4a^{2}

(x + b/2a) = ±√[(b^{2 }- 4ac)/4a^{2}]

(x + b/2a) = ±x = - b/2a ±√[(b^{2 }- 4ac)]/2a√[(b^{2 }- 4ac)]/2a

x = [- b ±√(b^{2 }- 4ac)]/2a ------------------ equation 3.

5) This is the method of the square.

### Solution:

**(i) 2x**

^{2}– 7x + 3 = 01) The given equation is 2x

^{2}- 7x + 3 = 0 ------------------ equation 1.2) We know that the quadratic equation ax

^{2}+ bx + c = 0, where a ≠ 0 can be solvedusing the formula:

x = [- b ±√(b^{2 }- 4ac)]/2a ------------------ equation 2.

3) Equate the coefficient of equation 2x

^{2}- 7x + 3 = 0 with ax^{2}+ bx + c = 0, we have,a = 2, b = - 7, c = 3.

4) First we will find:

b^{2 }- 4ac = (- 7)^{2 }- 4(2)(3)

b^{2 }- 4ac = 49^{ }- 24

b^{2 }- 4ac = 25 ------------------ equation 3.

5) As b

^{2 }- 4ac = 25, it has real roots, so from equation 2 and equation 3, wehave,

x = [- b ±6) So, x = (7 + 5)/4 or x = (7 - 5)/4.√(b^{2 }- 4ac)]/2a

x = [- (- 7) ±√25]/2(2)

x = (7 ± 5)/4

x = (12)/4 or x = (2)/4

x = 3 or x = 1/2

7) So, x = 3 or x = 1/2.

**(ii) 2x**

^{2}+ x – 4 = 01) The given equation is 2x

^{2}+ x - 4 = 0 ------------------ equation 1.2) We know that the quadratic equation ax

^{2}+ bx + c = 0, where a ≠ 0 can be solvedusing the formula:

x = [- b ±√(b^{2 }- 4ac)]/2a ------------------ equation 2.

3) Equate the coefficient of equation 2x

^{2}+ x - 4 = 0 with ax^{2}+ bx + c = 0, we have,a = 2, b = 1, c = - 4.

4) First we will find:

b^{2 }- 4ac = (1)^{2 }- 4(2)(- 4)

b^{2 }- 4ac = 1^{ }+ 32

b^{2 }- 4ac = 33 ------------------ equation 3.

5) As b

^{2 }- 4ac = 33, it has real roots, so from equation 2 and equation 3, wehave,

x = [- b ±6) So, x = (- 1 +√(b^{2 }- 4ac)]/2a

x = [- (1) ±√33]/2(2)

x = (- 1 ±√33)/4

**√**33)/4 or x = (- 1 -

**√**33)/4.

**(iii) 4x**

^{2}+ 4√3x + 3 = 01) The given equation is 4x

^{2}+ 4**√**3x + 3 = 0 ------------------ equation 1.2) We know that the quadratic equation ax

^{2}+ bx + c = 0, where a ≠ 0 can be solvedusing the formula:

x = [- b ±√(b^{2 }- 4ac)]/2a ------------------ equation 2.

3) Equate the coefficient of equation 4x

^{2}+ 4**√**3x + 3 = 0 with ax^{2}+ bx + c = 0, soa = 4, b = 4√3, c = 3.

4) First we will find:

b^{2 }- 4ac = (4√3)^{2 }- 4(4)(3)

b^{2 }- 4ac = 48^{ }- 48

b^{2 }- 4ac = 0 ------------------ equation 3.

5) As, b

^{2 }- 4ac ≥ 0, it has real roots, so from equation 2 and equation 3, wehave,

x = [- b ±6) So, x = (- 1)/4 or x = (- 1)/4.√(b^{2 }- 4ac)]/2a

x = [- (1) ±√0]/2(4)

x = (- 1 ± 0)/4

**(iv) 2x**

^{2}+ x + 4 = 01) The given equation is 2x

^{2}+ x + 4 = 0 ------------------ equation 1.2) We know that the quadratic equation ax

^{2}+ bx + c = 0, where a ≠ 0 can be solvedusing the formula:

x = [- b ±√(b^{2 }- 4ac)]/2a ------------------ equation 2.

3) Equate the coefficient of equation 2x

^{2}+ x + 4 = 0 with ax^{2}+ bx + c = 0, soa = 2, b = 1, c = 4.

4) First we will find:

b^{2 }- 4ac = (1)^{2 }- 4(2)(4)

b^{2 }- 4ac = 1^{ }- 32

b^{2 }- 4ac = - 31 ------------------ equation 3.

5) As b

^{2 }- 4ac = - 31 < 0, so 2x^{2}+ x + 4 = 0 has no real roots.**Q3. Find the roots of the following equations:**

**(i) x - (1/x) = 3, x ≠ 0 (ii) 1/(x + 4) - 1/(x - 7) = 11/30, x ≠ - 4, x ≠ 7**

### Explanation:

1) Convert your equation in to the quadratic form ax

^{2}+ bx + c = 0, where a ≠ 0.2) Solve this equation to get the values of the variable x.

### Solution:

**(i) x - (1/x) = 3, x ≠ 0**

1) The given equation is

x - (1/x) = 3

(x^{2}- 1)/x = 3

(x^{2}- 1) = 3x

x^{2}- 3x -1 = 0 ------------------ equation 1.

2) We know that the quadratic equation ax

^{2}+ bx + c = 0, where a ≠ 0 can be solvedusing the formula:

x = [- b ±√(b^{2 }- 4ac)]/2a ------------------ equation 2.

3) Equate the coefficient of equation x

^{2}- 3x -1 = 0 with ax^{2}+ bx + c = 0, soa = 1, b = - 3, c = 1.

4) First we will find:

b^{2 }- 4ac = (- 3)^{2 }- 4(1)(1)

b^{2 }- 4ac = 9^{ }- 4

b^{2 }- 4ac = 5 ------------------ equation 3.

5) As b

^{2 }- 4ac = 5 > 0, it has real roots, so from equation 2 and equation 3, wehave,

x = [- b ±6) So, x = (- 1)/4 or x = (- 1)/4.√(b^{2 }- 4ac)]/2a

x = [- (1) ±√0]/2(4)

x = (- 1 ± 0)/4

**(ii) 1/(x + 4) - 1/(x - 7) = 11/30, x ≠ - 4, x ≠ 7**

1) The given equation is

1/(x + 4) - 1/(x - 7) = 11/30

[1(x - 7) - 1(x + 4)]/(x + 4)(x - 7) = 11/30

(x - 7 - x - 4)/(x + 4)(x - 7) = 11/3030(x - 7 - x - 4) = 11 (x + 4)(x - 7)

30(- 11) = 11 (x + 4)(x - 7)

- 30 = (x + 4)(x - 7)

- 30 = x(x - 7) + 4 (x - 7)

- 30 = x^{2}- 7x + 4x - 28

- 30 = x^{2}- 3x - 28

x^{2}- 3x - 28 + 30 = 0

x^{2}- 3x + 2 = 0 ------------------ equation 1.

2) We know that the quadratic equation ax

^{2}+ bx + c = 0, where a ≠ 0 can be solvedusing the formula:

x = [- b ±√(b^{2 }- 4ac)]/2a ------------------ equation 2.

3) Equate the coefficient of equation x

^{2}- 3x + 2 = 0 with ax^{2}+ bx + c = 0, soa = 1, b = - 3, c = 2.

4) First we will find:

b^{2 }- 4ac = (- 3)^{2 }- 4(1)(2)

b^{2 }- 4ac = 9^{ }- 8

b^{2 }- 4ac = 1 ------------------ equation 3.

5) As b

^{2 }- 4ac = 1 > 0, it has real roots, so from equation 2 and equation 3, wehave,

x = [- b ±6) So, x = (4)/2 or x = (2)/2, i.e. x = 2, or x = 1.√(b^{2 }- 4ac)]/2a

x = [- (- 3) ±√1]/2(1)

x = (3 ± 1)/2

**Q4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3 Find his present age.**

1) Let Rehman's present age be x.

2) So, 3 years ago, his age was (x - 3).

3) 5 years later, his age will be (x + 3).

4) According to the problem,

[1/(x - 3)] + [1/(x + 5)] = 1/3

[1(x + 5) + 1(x - 3)]/[(x + 5)(x - 3)] = 1/3

(x + 5 + x - 3)/(x + 5)(x - 3) = 1/33(2x + 2) = (x + 5)(x - 3)

6x + 6 = x^{2}+ 5x - 3x - 15

x^{2}+ 2x - 15 - 6x - 6 = 0

x^{2}- 4x - 21 = 0 ------------------ equation 1.

5) Equate the coefficient of equation x

^{2}- 4x - 21 = 0 with ax^{2}+ bx + c = 0, soa = 1, b = - 4, c = - 21

6) First we will find:

b^{2 }- 4ac = (- 4)^{2 }- 4(1)(- 21)

b^{2 }- 4ac = 16^{ }+ 84

b^{2 }- 4ac = 100 ------------------ equation 2.

7) As b

^{2 }- 4ac = 100 > 0, it has real roots, so from equation 1 and equation 2, wehave,

x = [- b ±√(b^{2 }- 4ac)]/2a

x = [- (- 4) ±√100]/2(1)x = (4 ± 10)/2

8) So, x = (14)/2 or x = (- 6)/2, i.e. x = 7, or x = - 3.

9) As age can't be negative, x = 7, so Rehman's present age is 7 years.

**Q5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.**

1) Let Shefali's marks in Mathematics be x.

2) Her English marks will be (30 - x).

3) According to the problem,

(x + 2)(30 - x - 3) = 210

(x + 2)(27 - x) = 210

x(27 - x) + 2(27 - x) = 210

27x - x27x - 2x - x^{2}+ 54 - 2x = 210^{2}+ 54 = 210

25x - x^{2}= 210 - 54

25x - x^{2}= 156

x^{2}- 25x + 156 = 0 ------------------ equation 1.

4) Equate the coefficient of equation x

^{2}- 25x + 156 = 0 with ax^{2}+ bx + c = 0, soa = 1, b = - 25, c = 156

6) First we will find:

b^{2 }- 4ac = (- 25)^{2 }- 4(1)(156)

b^{2 }- 4ac = 625^{ }- 624

b^{2 }- 4ac = 1 ------------------ equation 2.

7) As b

^{2 }- 4ac = 1 > 0, it has real roots, so from equation 1 and equation 2, wehave,

x = [- b ±√(b^{2 }- 4ac)]/2a

x = [- (- 25) ±√1]/2(1)x = (25 ± 1)/2

8) So, x = (26)/2 or x = (24)/2, i.e. x = 13, or x = 12.

9) If x = 13, then

she got 13 marks in Mathematics and 30 - 13 = 17 marks in English.

10) If x = 12, then

she got 12 marks in Mathematics and 30 - 12 = 18 marks in English.

**Q6. The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.**

1) Let the shorter side of a rectangle be x m.

2) So, the longer side of a rectangle will be (x + 30) m

3) According to the problem,

x^{2}+ (x + 30)^{2}= (x + 60)^{2}

x^{2}= (x + 60)^{2}- (x + 30)^{2}use a^{2}- b^{2}= (a - b)(a + b)

x^{2}= [(x + 60) - (x + 30)] [(x + 60) + (x + 30)]

x^{2}= (x + 60 - x - 30) (x + 60 + x + 30)

xx^{2}= 30(2x + 90)^{2}= 60x + 2700

x^{2}- 60x - 2700 = 0 ------------------ equation 1.

4) Quadratic equation: x

^{2}- 60x - 2700 = 0here last term is 2700 and its sign is minus, factorise 1 x 2700 in such a way

that their difference will be 60.

1 x 2700 = (30) x (- 90) (30 - 90 = - 60).

x^{2}+ 30x - 90x - 2700 = 0x(x + 30) - 90(x + 30) = 0(x + 30)(x - 90) = 0

5) So, (x + 30) = 0, or (x - 90) = 0

6) So, x = - 30 or x = 90.

7) As the length of any side can't be negative, ignore x = -30, so we have x = 90.

8) The length of the shorter side is 90 m and the longer side is 90 + 30 = 120 m.

**Q7. The difference of the squares of the two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.**

1) Let the larger number be x.

2) So, the square of the smaller number will be 8x

3) According to the problem,

x^{2}- 8x = 180

x^{2}- 8x - 180 = 0 ------------------ equation 1.

4) Quadratic equation: x

^{2}- 8x - 180 = 0here last term is 180 and its sign is minus, factorise 1 x 180 in such a way

that their difference will be 8.

1 x 180 = (10) x (- 18) (10 - 18 = - 8).

x^{2}+ 10x - 18x - 180 = 0x(x + 10) - 18(x + 10) = 0(x + 10)(x - 18) = 0

5) So, (x + 10) = 0, or (x - 18) = 0

6) So, x = - 10 or x = 18.

7) If x = - 10, the smaller number will be - 10 (8) = - 80, as a square of anynumber can't be negative, we can't take x = -10 as the lager number.

8) So, our larger number will be 18.

9) According to the problem, a square of a smaller number = 8x = 8(18).

square of a smaller number = 8(18)

square of a smaller number = 2 x 2 x 2 x 2 x 3 x 3, so,

smaller number = ± (2 x 2 x 3)

smaller number = ± 12.

10) The numbers are 12 and 18 or - 12 and 18.

**Q8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.**

1) Let the uniform speed of a train be x km/h.

2) To cover 360 km, the time will be (360/x) hrs.

3) If the speed is increased by 5 km/h, then time will be decreased by 1 hr, so

new speed = (x + 5) km/h, and new time = [(360/x) -1] hrs.

4) According to the condition,

(x + 5)[(360/x) -1] = 360

(x + 5)[(360 - x)/x] = 360

(x + 5)(360 - x) = 360x

x(360 - x) + 5(360 - x) = 360x

360x - x^{2}+ 1800 - 5x = 360x

- x^{2}+ 1800 - 5x = 0

x^{2}^{ }+ 5x - 1800 = 0 ---------- equation 1

4) Quadratic equation: x

^{2}+ 5x - 1800 = 0here last term is 1800 and its sign is minus, factorise 1 x 1800 in such a way

that their difference will be 5.

1 x 1800 = (45) x (- 40) (45 - 40 = 5).

x^{2}+ 45x - 40x - 1800 = 0x(x + 45) - 40(x + 45) = 0(x + 45)(x - 40) = 0

5) So, (x + 45) = 0, or (x - 40) = 0

6) So, x = - 45 or x = 40.

7) As x is the speed, it can't be negative, so we have x = 40.8) The speed of the train is 40 km/h.

**Q9.Two water taps together can fill a tank in**

**9 and 3/8 hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.**

1) Let the time taken by the smaller pipe to fill the tank be x hrs.

2) Time taken by the larger pipe will be (x - 10) hrs.

3) The part of a tank filled by a smaller pipe in 1 hr will be 1/x.

4) The part of a tank filled by a larger pipe in 1 hr will be 1/(x - 10).

5) Total time taken by both the pipes to fill the tank = 9 and 3/8 = (72+3)/8 = 75/8.

6) According to the problem, the part of a tank filled by both pipes in 1 hour is 8/75.

7) So, we have,

1/x + 1/(x - 10) = 8/75

[(x - 10) + x]/x(x - 10) = 8/7575(2x - 10) = 8x(x - 10)

150x - 750 = 8x^{2}- 80x

8x^{2}- 80x - 150x + 750 = 0

8x^{2}- 230x + 750 = 0 ---------- equation 1

8) Quadratic equation: 8x

^{2}- 230x + 750 = 0here last term is 750 and its sign is plus, factorise 8 x 750 in such a way

that their sum will be - 230.

8 x 750 = (- 200) x (- 30)

(- 200 - 30 = - 230).

8x^{2}- 200x - 30x + 750 = 08x(x - 25) - 30(x - 25) = 0(x - 25)(8x - 30) = 0

9) So, (x - 25) = 0, or (8x - 30) = 0

10) So, x = 25 or x = 30/8.

11) If we consider x = 30/8, the time taken by a larger pipe will be (30/8 - 10) = (30 - 80)/8 = - 50/8 which negative, so ignore x = 30/8.

12) So consider x = 25. i.e. time taken by smaller pipe to fill the tank will be 25 hrs

and the time taken by larger pipe to fill the tank will be 15 hrs.

**Q10. An express train takes 1 hour less than a passenger train to travel 132 km between**

**Mysore and Bangalore (without taking into consideration the time they stop at**

**intermediate stations). If the average speed of the express train is 11km/h more than that**

**of the passenger train, find the average speed of the two trains.**

1) Let the average speed of a passenger train be x km/h.

2) So the average speed of the express train will be (x + 11) km/h.

3) We know that time = distance/speed, so to cover 132 km,

a) for passenger train, time = 132/x hrs.

b) for express train, time = 132/(x + 11) hrs.

4) As the express train takes 1 hour less than the passenger train to cover 132 km,

so

[132/x] - [132/(x + 11)] = 1

132[1/x - 1/(x + 11)] = 1

132[(x + 11) - x ]/x(x + 11)] = 1

132(11)/x(x + 11) = 1

132(11) = x(x + 11)

1452 = x^{2}+ 11x

x^{2}+ 11x - 1452 = 0 ------------ equation 1

5) Quadratic equation: x

^{2}+ 11x - 1452 = 0here last term is 1452 and its sign is minus, factorise 1 x 1452 in such a way

that their difference will be 11.

1 x 1452 = (44) x (- 33) (44 - 33 = 11).

x^{2}+ 44x - 33x - 1452 = 0x(x + 44) - 33(x + 1452) = 0(x + 44)(x - 33) = 0

6) So, (x + 44) = 0, or (x - 33) = 0

7) So, x = - 44 or x = 33.

8) As the speed can't be negative, so ignore x = - 44, so the average speed ofpassenger train will be 33 km/h and the average speed of the express train will be 44 km/h.

**Q11. The sum of the areas of the two squares is 468**

**m**

^{2}**. If the difference of their perimeters is 24 m,**

**find the sides of the two squares.**

1) Let the side of the first square be x m.

2) So, its area will be x

^{2}.3) According to the problem, the area of the other square will be (468 - x

^{2}). the sideof this square will be√(468 - x^{2}).

4) The perimeter of the first square will be 4x and that of the other will be

4√(468 - x^{2})

5) As the difference between their perimeters is 24,

4x - 4√(468 - x^{2}) = 24

4[x -√(468 - x^{2})] = 24

x -√(468 - x^{2}) = 6

x - 6 =√(468 - x^{2}) squaring both side, we get

(x - 6)^{2}= (468 - x^{2})

x^{2}- 12x + 36 = 468 - x^{2}

x^{2 }+ x^{2}- 12x + 36 - 468 = 0

2x^{2}- 12x - 432 = 0

x^{2}- 6x - 216 = 0 ------------------ equation 1.

6) Quadratic equation: x

^{2}- 6x - 216 = 0here last term is 216 and its sign is minus, factorise 1 x 216 in such a way

that their difference will be - 6.

1 x 216 = (12) x (- 18) (12 - 18 = - 6).

x^{2}+ 12x - 18x - 216 = 0x(x + 12) - 18(x + 12) = 0(x + 12)(x - 18) = 0

5) So, (x + 12) = 0, or (x - 18) = 0

6) So, x = - 12 or x = 18.

7) As the length of any side can't be negative, ignore x = - 12, so we have x = 18.

8) The length of the side of the first square will be 18 m.

9) So the Area of the other square is

= 468 - x^{2}

= 468 - 1810) The sides of the squares are 12 m and 18 m.^{2}= 468 - 324

= 144. Therefore side of other square will be√144 = 12 m.

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