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    Exercise 10.4.2

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6th class > > Exercise 10.4.2

Exercise 10.4.2

1. Find the roots of the following quadratic equations by factorisation.

(i) x2– 3x – 10 = 0

⇒ x2– 5x + 2x – 10 = 0
⇒ x(x-5)+2(x-5)=0
⇒ (x-5)(x+2)=0
⇒ x-5=0 and x+2=0
⇒ x=,.

(ii) 2x2 + x - 6 = 0

⇒ 2x2 + 4x - 3x - 6 = 0
⇒ 2x (x + 2) - 3 (x + 2) = 0
⇒ 2x - 3 = 0 and x + 2 = 0
⇒ x = /, .

(iii) √2x2 + 7x + 5√2 = 0

⇒ √2x2 + 5x + 2x + 5√2 = 0
⇒ √2x2 + 2x + 5x + 5√2 = 0
⇒ (√2x + 5) (x + √2) = 0
⇒ √2x + 5 = 0 or x + √2 = 0
⇒ x = - /√2, - √2.

(iv) 2x2 - x + 1/ 8 = 0

Multiplying both sides of the equation by 8:
⇒ 2(8) x2 - 8(x) + (8)(1/ 8) = (0)8
⇒ 16x2 - 8x + 1 = 0
⇒ 16x2 - 4x - 4x + 1 = 0
⇒ 4x (4x - 1) -1 (4x - 1) = 0
⇒ (4x - 1)2 = 0 4x - 1 = 0
x = /, /.

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

⇒ 100x2 - 20x + 1 = 0
⇒ 100x2 - 10x - 10x + 1 = 0
⇒ 10x(10x - 1) -1(10 x - 1) = 0
⇒ (10x - 1)2 = 0 10x - 1 = 0
x = /,/.

2. Solve the problems given in Example 1.

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(i) Let the number of marbles that John had be x.

The number of marbles Jivanti had will be = 45 - x

Both of them lost 5 marbles each:

Marbles with John = x - 5

Marbles with Jivanti = 45 - x - 5 = 40 - x

Product of current number of marbles = 124

Let us use this condition to form a quadratic equation.

(x - 5) (40 - x) = 124

(i) (x - 5) (40 - x) = 124

⇒ 40x - x2 - 200 + 5x = 124

⇒ −x2 + 45x - 200 -124 = 0

⇒ −x2 + 45x - 324 = 0

On multiplying both sides of the equation with negative sign, we get

⇒ x2 - 45x + 324 = 0

⇒ x2 - 36x - 9x + 324 = 0

⇒ x(x - 36) - 9 (x - 36) = 0

⇒ (x - 36) (x - 9) = 0

⇒ x = 36 and x = 9

John and Jivanti started with 36 and 9 marbles.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day.

(ii) Let the number of toys produced in a day be x.

Cost of each toy = (55 - x) rupees

Total cost of production = cost of each toy × Total number of toys

⇒ (55 - x) (x) = 750

⇒ 55x - x2 = 750

⇒ 55x - x2 - 750 = 0

⇒ x2 - 25x - 30x + 750 = 0

⇒ (x - 25) (x - 30) = 0

x - 25 = 0 and x - 30 = 0

x = 25 and x = 30

The number of toys produced on that day is 25 or 30.

3.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.

The product of the two numbers is given as 182. This means in terms of x this can be expressed as, x(27 - x) = 182

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

x(27 - x) = 182

27x - x2 = 182

27x - x2 - 182 = 0

x2 - 27 x + 182 = 0 [Rearranging the terms and multiplying both sides by negative sign]

x2 - 14x - 13x + 182 = 0

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

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

x = ,.

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

The consecutive Numbers be x,x+1.
Sum of their Squares x2+x+2 = 365.
⇒ x2+x2+2x+1 = 365 ⇒ 2x2+2x-364=0.
⇒ x2+x-182=0 ⇒ x2+14x-13x-182=0
⇒ x(x-14)-13x-182=0 ⇒ =0
x = ,, x cant be a negative Integers then x = ,Two consective integers = ,.

5. 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:

Use Pythagorean theorem states:

base2 + altitute2 = hypotenuse2

Substitute the given values into the theorem: x2+x−72 = 132.

Simplify and solve for x: x2+x−72 = 169

= 2x2-14x+49 = 169

Subtract 169 from both sides to set the equation to zero: 2x2-14x-120 = 0.

Divide the entire equation by 2 to simplify:x2-7 - 60 = 0.

Now, solve this quadratic equation using the quadratic formula:−b±b2−4ac2a

where a=1,b=-7,c=-60.

x =

This gives us two solutions: x = ,

Therefore, the base of the triangle is 12 cm, and the altitude is : x−7 = 12−7 =  cm.

So, the other two sides of the right triangle are: Base=12 cm Altitude=5 cm.

6. 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)=90

= 3x+2x2=90

= 2x2+3x-90=0

= 2x2+15x-12x-90=0

= 2x(x-6)+15(x-6)=0

= (2x+15)(x-6)=0

x=,/,

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