Exercise 4.2

Find the roots of the following quadratic equations by factorisation:

(i)

(i) x2 - 3x -10 = 0

Solution

x2- 3x - 10 = 0

x2 - x + x - 10 = 0

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

(x - 5) (x + 2) = 0

x - 5 = and x + 2 = 0

x = and x =

Therefore, roots are : - 2, 5

(ii)

(ii) 2x2 + x - 6 = 0

Solution

2x2 + 4x - 3x - 6 = 0

2x (x + ) - (x + ) = 0

(2x - 3) (x + 2) = 0

2x - 3 = and x + 2 = 0

2x = and x =

x = and x =

Therefore, roots are: 32, -2

(iii)

(iii) 2 x2 + 7x + 52 = 0

Solution

2 x2 + x + x + 52 = 0

(2x + 5) (x + 2) = 0

2x + 5 = or x + 2 =

2x = or x = - 2

x = −52 or x = - 2

Therefore, roots are: −52, - 2

(iv)

(iv) 2x2 - x + 18 = 0

Solution

2x2 - x + 18 = 0

Multiplying both sides of the equation by 8:

2(8) x2 - (x) + (8)(18) = (0)8

x2 - 8x + = 0

16x2- x - x + 1 = 0

4x (4x - 1) (4x - 1) = 0

(4x - 1) (4x - 1) =

4x−12 = 0

4x - = 0

x = and x =

Roots are: 14, 14

(v)

(v) 100x2 - 20x + 1 = 0

Solution

100x2 - 20x + 1 = 0

100x2 - x - x + 1 = 0

10x(10x - 1) (10 x - 1) = 0

(10x - )(10x - ) = 0

10x−12 = 0

10x - = 0

x = and x =

Roots are: 110, 110.

2. Find two numbers whose sum is 27 and product is 182.

Solution

Let us find the 2 numbers whose sum is 27 and product is 182.

For this, let us consider one of the numbers to be x.

Then the other number will be 27 - .

The product of the two numbers is given as 182. This means in terms of x this can be expressed as, x( - x) =

This can be written in the form of the following quadratic equation:

x(27 - x) = 182

x - x2 = 182

27x - x2 - = 0

x2 - x + = 0 [Note:Rearranging the terms and multiplying both sides by negative sign]

x2 - x - x + 182 = 0

x (x - 14) - (x - 14) = 0

(x -13) (x - 14) = 0

x - 13 = and x - 14 = 0

x = and x =

Therefore, the required numbers are 13, 14. In other words, the two numbers whose sum is 27 and product is 182 are 13 and 14.

3. Find two consecutive positive integers, sum of whose squares is 365.

Solution

Let the first integer be x.

The next consecutive positive integer will be x + 1.

According to the given question, the sum of squares of x and x + 1 is . i.e.,

x2 + x+12 = 365

x2 + (x2 + + 1) = 365 [ Note : a+b2 = a2 + 2ab + b2]

x2 + 2x + 1 = 365

2x2 + 2x + 1 - = 0

2x2 + 2x - = 0

2(x2 + x - ) = 0

x2 + x - = 0

x2 + x - 13x - 182 = 0

x (x + 14) - (x + 14) = 0

(x - 13) (x + 14) =

x - 13 = and x + 14 = 0

Therefore x = and x = .

4. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution

In a right triangle, altitude is one of the sides.

Let the base be x cm.

The altitude will be (x - 7) cm.

We can now apply the Pythagoras theorem to the given right triangle.

Pythagoras theorem: Hypotenuse2 = side12 + side22

132 = x2 + x−72

= x2 +x2 - x +

169 = x2 - 14x + 49

2x2 - 14x + 49 - = 0

2x2 - 14x - = 0

2x2−14x−1202 = 0

x2 - x - = 0

x2 - x + x - 60 = 0

x(x - 12) + (x - 12) = 0

(x + 5) (x - ) = 0

x - 12 = 0 and x + 5 = 0

x = and x =

We know that the value of the base cannot be .

Therefore, Base = 12 cm, Altitude = 12 - 7 = 5 cm.

5. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.

Solution

Let there be x articles.

Then the production cost of each article = 3 + 2x.

Total production cost = x (3 + 2x) =

x + 2x2 = 90

2x2 + 3x - = 0

2x2 + 15x - x - 90 = 0

2x (x-6) + (x-6) = 0

(2x + 15) (x-6) =

x = ,

Number cannot be .

So, number of articles = 6 and the cost of each article = ₹ 15.

Quadratic Equations

Glossary

Exercise 4.2

(i)

(ii)

(iii)

(iv)

(v)

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Quadratic Equations

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Glossary

Exercise 4.2

(i)

(ii)

(iii)

(iv)

(v)