Solve the followings Questions.

1. Find the roots of the following Quadratic Equations by factorization.

(i) x²-3x-10 = 0

Answer:
x²-3x-10 = 0
x²-5x+2x-10 = 0
x(x-5)+2(x-5) = 0
(x-5)(x+2) = 0
⇒ x-5 = 0 or, x+2 = 0
x = 5 or x =-2
Hence the required roots are 5, -2.

(ii) 2x²+x-6=0

Answer:
2x²+x-6 = 0
2x²+4x-3x-6 = 0
2x(x+2)-3(x+2) = 0
(2x-3)(x+2) = 0
⇒ 2x-3 = 0 or, x+2 = 0
x = 3/2 or x = -2
Hence the required roots are 3/2 , -2.

(iii) √2×2+7x+5√2 = 0

Answer:
√2x²+7x+5√2 = 0
√2x²+2x+5x+ 5√2 = 0
√2x(x+√2)+5(x+√2) = 0
(x+√2)(√2x +5) = 0
⇒ x+√2 = 0 or, √2x +5 = 0
x = -√2 or x = –5/√2

Hence the required roots are -√2, –5/√2.

(iv) 2x²-x+1/8=0

Answer:
2x²-x+1/8=0
16x²-8x+1 = 0
(4x)²-2.4x.1+12 = 0
(4x-1)2= 0
⇒ 4x-1 = 0  or 4x-1 = 0
x = 1/4   or x = 1/4
Hence the required roots are 1/4 , 1/4

(v) 100x²-20x+1 = 0

Answer:

100x²-20x+1 = 0
(10x)²-2×10x.1+12= 0
(10x-1)² = 0
10x-1 = 0 or 10x-1 = 0
x = 1/10 or x = 1/10
Hence the required roots are 1/10  or 1/10
 

Solve the following problems given :

2. (i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each others, and product of the marbles they now have is 124. We would like to find how many marbles they had to start with?
Answer:
Let the number of marbles John had be x.
Then the number of marbles Jivanti had be 45-x
They lost 5 marbles to each others.
Hence now the number of marbles of John is x-5 and that of Jivanti is 45-x-5 = 40-x.
Then by given condition we have
(x-5)(40-x) = 124
40x-200-x²+5x = 124
x²-45x+324 = 0
x²-36x-9x+324 = 0
x(x-36)-9(x-36) = 0
(x-36)(x-9) = 0
x-36 = 0
x = 36
or x-9 = 0
x = 9
Therefore, they had 36 and 9 marvels respectively.

(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.
Answer:
Let the number of toys produced on that day is x.
Therefore, the cost of production of each toy on that day is (55-x)
So, the total cost of production on that day is = x(55-x)
Then using the given condition, we have,
x(55-x) = 750
55x-x² =750
x²-55x+750 = 0
x²-25x-30x+750 = 0
x(x-25)-30(x-25) = 0
(x-25)(x-30) = 0
x-25 = 0
x =25
or x-30 =0
x = 30
Therefore, the number of toys produced on that day is 25 or 30.

3. Find the two numbers whose sum is 27 and product is 182.

Answer:
Let one number is x. Then another number is 27-x.
Then by the given condition,
x(27-x) = 182
27x- x²-182 = 0
x²-27x+182 = 0
x²-14x-13x+182 = 0
x(x-14)-13(x-14) = 0
(x-14)(x-13) = 0
x-14 = 0
x =14
or, x-13 = 0
x =13
Hence the required two numbers are 13 and 14.

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

Answer:
Let the two consecutive positive integers be x and x+1
Then by the given condition,
x²+(x+1)² = 365
x²+ x²+2x+1 = 365
2x²+2x-364 = 0
x2+x-182 = 0
x²+14x-13x-182 = 0
x(x+14)-13(x+14) = 0
(x+14)(x-13) = 0
x = -14 or 13
Since the numbers are positive so x = -14 not possible.
So, the numbers are 13 and 13+1 = 14.

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.

Answer:
Let base of the right triangle is x cm. Then the altitude is x-7 cm.
Then by the property of right triangle and using the given property we have
x²+(x-7)² = (13)²
x²+ x²-14x+49 = 169
2x²-14x-120 = 0
x²-7x-60 = 0
x²-12x+5x-60 = 0
x(x-12)+5(x-12) = 0
(x-12)(x+5) = 0
x-12 = 0
x =12
or, x+5 = 0
x = -5
Since length can’t be negative, so x = -5 is not possible.
Hence the base is 12 cm and the altitude is (12-7)cm = 5 cm

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

Answer:
Let the number of articles produced on that day is x. then the price on that day is 2x+3
Then using the given condition we have,
x(2x+3) = 90
2x²+3x-90 = 0
2x²+15x-12x-90 = 0
x(2x+15)-6(2x+15)=0
(2x+15)(x-6) = 0
2x+15 = 0
x = – 15/2
or, x-6 = 0
x = 6
Since Articles can’t be negative then x = – 15/2 is not possible.
Therefore, the number of articles produced on that day = 6
Cost of each article = 2×6+3 = 15

CHAPTER NAME	OLD NCERT	NEW NCERT
Real Numbers	EXERCISE 1.1
	EXERCISE 1.2	1.1	CLICK HERE
	EXERCISE 1.3	1.2	CLICK HERE
	EXERCISE 1.4
Polynomials	EXERCISE 2.1	2.1	CLICK HERE
	EXERCISE 2.2	2.2	CLICK HERE
	EXERCISE 2.3
	EXERCISE 2.4
Pair of Linear Equations in Two Variables	EXERCISE 3.1
	EXERCISE 3.2	3.1	CLICK HERE
	EXERCISE 3.3	3.2	CLICK HERE
	EXERCISE 3.4	3.3	CLICK HERE
	EXERCISE 3.5
	EXERCISE 3.6
	EXERCISE 3.7
Quadratic Equations	EXERCISE 4.1	4.1	CLICK HERE
	EXERCISE 4.2	4.2	CLICK HERE
	EXERCISE 4.3
	EXERCISE 4.4	4.3	CLICK HERE
Arithmetic Progressions	EXERCISE 5.1	5.1	CLICK HERE
	EXERCISE 5.2	5.2	CLICK HERE
	EXERCISE 5.3	5.3	CLICK HERE
	EXERCISE 5.4	5.4 (Optional)	CLICK HERE
Triangles	EXERCISE 6.1	6.1	CLICK HERE
	EXERCISE 6.2	6.2	CLICK HERE
	EXERCISE 6.3	6.3	CLICK HERE
	EXERCISE 6.4
	EXERCISE 6.5
	EXERCISE 6.6
Coordinate Geometry	EXERCISE 7.1	7.1	CLICK HERE
	EXERCISE 7.2	7.2	CLICK HERE
	EXERCISE 7.3
	EXERCISE 7.4
Introduction to Trigonometry	EXERCISE 8.1	8.1	CLICK HERE
	EXERCISE 8.2	8.2	CLICK HERE
	EXERCISE 8.3
	EXERCISE 8.4	8.3	CLICK HERE
Some Applications of Trigonometry	EXERCISE 9.1	9.1	CLICK HERE
Circles	EXERCISE 10.1	10.1	CLICK HERE
	EXERCISE 10.2	10.2	CLICK HERE
Construction
Areas Related to Circles	EXERCISE 12.1
	EXERCISE 12.2	11.1	CLICK HERE
	EXERCISE 12.3
Surface Areas and Volumes	EXERCISE 13.1	12.1	CLICK HERE
	EXERCISE 13.2	12.2	CLICK HERE
	EXERCISE 13.3
	EXERCISE 13.4
	EXERCISE 13.5
Statistics	EXERCISE 14.1	13.1	CLICK HERE
	EXERCISE 14.2	13.2	CLICK HERE
	EXERCISE 14.3	13.3	CLICK HERE
	EXERCISE 14.4
Probability	EXERCISE 15.1	14.1	CLICK HERE
	EXERCISE 15.2