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Class 10 NCERT Solutions- Chapter 4 Quadratic Equations – Exercise 4.3
  • Last Updated : 03 Mar, 2021

Question 1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2x2– 7x + 3 = 0

Solution:

2x2 – 7x = – 3

Dividing by 2 on both sides, we get

x2\frac{7x}{2} = –\frac{3}{2}

x2 -2 × x ×\frac{7}{4} = –\frac{3}{2}



On adding (\frac{7}{4} )2 to both sides of equation, we get

(x)2 – 2×x×\frac{7}{4} +(\frac{7}{4} )2 = (\frac{7}{4} )2\frac{3}{2}

(x-\frac{7}{4} )2 = (\frac{49}{16} ) – (\frac{3}{2} ) (Using identity: a2 – 2ab + b2 = (a-b)2)

(x-\frac{7}{4} )2 =\frac{25}{16}

(x-\frac{7}{4} )2 = ±\frac{5}{4}

x =\frac{7}{4} ± \frac{5}{4}

x =\frac{7}{4} + \frac{5}{4} or x =\frac{7}{4} - \frac{5}{4}

x =\frac{12}{4} or x =\frac{2}{4}



x = 3 or x =\frac{1}{2}

(ii) 2x2+ x – 4 = 0

Solution:

2x2 + x = 4

Dividing both sides of the equation by 2, we get

x2 +\frac{x}{2} = 2

Now on adding(\frac{1}{4}) 2 to both sides of the equation, we get,

(x)2 + 2 × x ×\frac{1}{4} + (\frac{1}{4} )2 = 2 + (\frac{1}{4} )2

(x +\frac{1}{4} )2 =\frac{33}{16} (Using identity: a2 + 2ab + b2 = (a+b)2)

x +\frac{1}{4} = ±√(\frac{33}{16})

x =- \frac{1}{4} ± \frac{√33}{4}

x =\frac{-1± √33}{4}

Hence, x =\frac{-1+ √33}{4} or x =\frac{-1- √33}{4}

(iii) 4x2 + 4√3x + 3 = 0

Solution:

4x2 + 4√3x = -3

Dividing both sides of the equation by 4, we get

x2 + √3x = –\frac{3}{4}

Now on adding (\frac{√3}{2} )2 to both sides of the equation, we get,

(x)2 + 2 × x ×\frac{√3}{2} + (\frac{√3}{2} )2 = –\frac{3}{4} + (\frac{√3}{2} )2

(x +\frac{√3}{2} )2 = 0 (Using identity: a2 + 2ab + b2 = (a+b)2)

Hence, x = –\frac{√3}{2} or x = –\frac{√3}{2}

(iv) 2x2+ x + 4 = 0

Solution:

2x2 + x = -4

Dividing both sides of the equation by 2, we get

x2 +\frac{x}{2} = -2

Now on adding (\frac{1}{4} )2 to both sides of the equation, we get,

(x)2 + 2 × x ×\frac{1}{4} + (\frac{1}{4} )2 = – 2 + (\frac{1}{4} )2

(x +\frac{1}{4} )2 =\frac{-31}{16} (Using identity: a2 + 2ab + b2 = (a+b)2)

As we know, the square of numbers cannot be negative.

Hence, there is no real root for the given equation, 2x2 + x + 4 = 0.

Question 2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

(i) 2x2– 7x + 3 = 0

Solution:

On comparing the equation with ax2 + bx + c = 0, we get,

a = 2, b = -7 and c = 3

Then roots of the quadratic equation =\frac{-b± √(b^2-4ac)}{2a}

x =\frac{-(-7)± √((-7)^2-4(2)(3))}{2(2)}

x =\frac{7± √(49-24)}{4}

x =\frac{7± √25}{4}

x =\frac{7± 5}{4}

x =\frac{7+5}{4} or x =\frac{7-5}{4}

x =\frac{12}{4} or\frac{2}{4}

x = 3 or\frac{1}{2}

(ii) 2x2+ x – 4 = 0

Solution:

On comparing the equation with ax2 + bx + c = 0, we get,

a = 2, b = 1 and c = -4

Then roots of the quadratic equation =\frac{-b± √(b^2-4ac)}{2a}

x =\frac{-1± √(1^2-4(2)(-4))}{2(2)}

x =\frac{-1± √(1+32)}{4}

x =\frac{-1± √33}{4}

x =\frac{-1+√33}{4} or x =\frac{1-√33}{4}

(iii) 4x2 + 4√3x + 3 = 0

Solution:

On comparing the equation with ax2 + bx + c = 0, we get,

a = 4, b = 4√3 and c = 3

Then roots of the quadratic equation =\frac{-b± √(b^2-4ac)}{2a}

x =\frac{-(4√3)± √((4√3)^2-4(4)(3))}{2(4)}

x =\frac{-(4√3)± √(48-48)}{8}

x =\frac{-4√3}{8}

x =\frac{-√3}{2} or x =\frac{-√3}{2}

(iv) 2x2+ x + 4 = 0

Solution:

On comparing the equation with ax2 + bx + c = 0, we get,

a = 2, b = 1 and c = 4

Then roots of the quadratic equation =\frac{-b± √(b^2-4ac)}{2a}

x =\frac{-1± √(1^2-4(2)(4))}{2(2)}

x =\frac{-1± √(1-32)}{4}

x =\frac{-1± √(-31)}{4}

As we know, the square of a number can never be negative.

Hence, there is no real solution for the given equation.

Question 3. Find the roots of the following equations:

(i) x –\frac{1}{x} = 3, x ≠ 0

Solution:

After rearranging, we get

x2 – 3x -1 = 0

On comparing the equation with ax2 + bx + c = 0, we get,

a = 1, b = -3 and c = -1

Then roots of the quadratic equation =\frac{-b± √(b^2-4ac)}{2a}

x =\frac{-(-3)± √((-3)^2-4(1)(-1))}{2(1)}

x =\frac{3± √(9+4)}{2}

x =\frac{3±√(13)}{2}

x =\frac{3+√(13)}{2} or x =\frac{3-√(13)}{2}

(ii)\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30} , x ≠ -4,7

Solution:

\frac{(x-7-x-4)}{(x+4)(x-7)} = \frac{11}{30}

\frac{(-11)}{(x+4)(x-7)} = \frac{11}{30}

After rearranging,

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

x2 – 3x – 28 = 30

x2 – 3x + 2 = 0

On comparing the equation with ax2 + bx + c = 0, we get,

a = 1, b = -3 and c = 2

Then roots of the quadratic equation =\frac{-b± √(b^2-4ac)}{2a}

x =\frac{-(-3)± √((-3)^2-4(1)(2))}{2(1)}

x =\frac{3± √(9-8)}{2}

x =\frac{3±√1}{2}

x =\frac{3±1}{2}

x =\frac{4}{2} or x =\frac{2}{2}

x = 2 or x = 1

Question 4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is\frac{1}{3} . Find his present age.

Solution:

Let’s take,

Present age of Rehman is x years.

Three years ago, Rehman’s age was (x – 3) years.

Five years after, his age will be (x + 5) years.

According to the given condition,

\frac{1}{x-3} + \frac{1}{x-5} = \frac{1}{3}

\frac{(x+5+x-3)}{(x-3)(x+5)} = \frac{1}{3}

\frac{(2x+2)}{(x-3)(x+5)} = \frac{1}{3}

3(2x + 2) = (x-3)(x+5)

6x + 6 = x2 + 2x – 15

x2 – 4x – 21 = 0

x2 – 7x + 3x – 21 = 0 (by factorizing)

x(x – 7) + 3(x – 7) = 0

(x – 7)(x + 3) = 0

x = 7, -3

As, age cannot be negative.

Therefore, Rehman’s present age is 7 years.

Question 5. 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.

Solution:

Let’s take,

The marks of Shefali in Maths be x.

Then, the marks in English will be 30 – x.

According to the given condition,

(x + 2)(30 – x – 3) = 210

(x + 2)(27 – x) = 210

-x2 + 25x + 54 = 210

Multiply the equation by (-1),

x2 – 25x + 156 = 0

x2 – 12x – 13x + 156 = 0

x(x – 12) -13(x – 12) = 0

(x – 12)(x – 13) = 0

x = 12, 13

Hence, if the marks in Maths are 12, then marks in English will be 30 – 12 = 18

and, the marks in Maths are 13, then marks in English will be 30 – 13 = 17.

Question 6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Solution:

Let’s take,

Breadth = x

Length = x+30

Diagonal = x+60

Diagonal = √(Length2 + Breadth2)

According to the given condition,

√((x+30)2 + (x)2) = x+60

Squaring both sides,

x2 + (x + 30)2 = (x + 60)2

x2 + x2 + 900 + 60x = x2 + 3600 + 120x

x2 – 60x – 2700 = 0

x2 – 90x + 30x – 2700 = 0

x(x – 90) + 30(x -90) = 0

(x – 90)(x + 30) = 0

x = 90, -30

As, side of the field cannot be negative.

Hence, the length of the shorter side will be 90 m.

and, the length of the larger side will be (90 + 30) m = 120 m.

Question 7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Solution:

Let’s take,

The larger number = x

and, smaller number = y

According to the given condition,

x2 – y2 = 180 and y2 = 8x (It means x has to be positive, because it is obtained by squaring a number)

x2 – 8x = 180

x2 – 8x – 180 = 0

x2 – 18x + 10x – 180 = 0

x(x – 18) +10(x – 18) = 0

(x – 18)(x + 10) = 0

x = 18, -10

As x cannot be negative,

Hence, the larger number will be 18.

x = 18

So, As y2 = 8x

= 8 × 18

= 144

y = ±√144 = ±12

So, Smaller number = ±12

Hence, the numbers are 18 and 12 or 18 and -12.

Question 8. 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.

Solution:

Let’s take

The speed of the train = x km/hr.

As, Speed =\frac{Distance}{Time}

Time taken to cover 360 km =\frac{360}{x} hr.

As per the question given,

(x + 5)(\frac{360}{x} – 1) = 360

(x + 5)(\frac{360 - x}{x} ) = 360

(x + 5)(360 – x) = 360x

x2 + 5x -1800 = 0

x2 + 45x – 40x + 1800 = 0

x(x + 45) -40(x + 45) = 0

(x + 45)(x – 40) = 0

x = 40, -45

As we know, the value of speed cannot be negative.

Hence, the speed of train is 40 km/h.

Question 9. Two water taps together can fill a tank in 9\frac{3}{8} hours. The tap of 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.

Solution:

Let’s take

The time taken by the smaller pipe to fill the tank = x hr.

Time taken by the larger pipe = (x – 10) hr

Part of tank filled by smaller pipe in 1 hour =\frac{1}{x}

Part of tank filled by larger pipe in 1 hour =\frac{1}{x-10}

According to the given condition,

9\frac{3}{8} hrs taken to fill with both the pipe.

So,

\frac{1}{x} + \frac{1}{x-10} = \frac{8}{75}

\frac{x-10+x}{x(x-10)} = \frac{8}{75}

\frac{2x-10}{x(x-10)} = \frac{8}{75}

75(2x – 10) = 8x2 – 80x

150x – 750 = 8x2 – 80x

8x2 – 230x +750 = 0

8x2 – 200x – 30x + 750 = 0

8x(x – 25) -30(x – 25) = 0

(x – 25)(8x -30) = 0

x = 25,\frac{30}{8}

Time taken by the smaller pipe cannot be\frac{30}{8} hours, as the time taken by the larger pipe will become negative.

Hence, time taken by the smaller pipe = 25hours

and, by the larger pipe =15 hours

Question 10. 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.

Solution:

Let’s take

The average speed of passenger train = x km/h.

Average speed of express train = (x + 11) km/h

According to the given condition,

\frac{132}{x} + \frac{132}{x+11} = 1

\frac{132(x+11-x)}{x(x+11)} = 1

\frac{132 × 11}{x(x+11)} = 1

132 × 11 = x(x + 11)

x2 + 11x – 1452 = 0

x2 + 44x -33x -1452 = 0

x(x + 44) -33(x + 44) = 0

(x + 44)(x – 33) = 0

x = – 44, 33

As we know, Speed cannot be negative.

Hence, the speed of the passenger train will be 33 km/h

and, the speed of the express train will be 33 + 11 = 44 km/h.

Question 11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Solution:

Let the sides of the two squares be x and y meter.

Perimeter = 4x and 4y respectively

Area = x2 and y2 respectively.

According to the given condition,

4x – 4y = 24

x – y = 6

x = y + 6 ………………….(I)

and,

x2 + y2 = 468

(6 + y)2 + y2 = 468 (From (I))

36 + y2 + 12y + y2 = 468

2y2 + 12y + 432 = 0

y2 + 6y – 216 = 0

y2 + 18y – 12y – 216 = 0

y(y +18) -12(y + 18) = 0

(y + 18)(y – 12) = 0

y = -18, 12

As we know, the side of a square cannot be negative.

Hence, the sides of the squares are 12 m

and, (12 + 6) m = 18 m.

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