问题1.通过分解找到以下二次方程的根：
(i)x 2 – 3x – 10 = 0
解决方案：
Here, LHS = x2– 3x – 10
= x2 – 5x + 2x – 10
= x(x – 5) + 2(x – 5)
= (x – 5)(x + 2)
The roots of this equation, x2 – 3x – 10 = 0 are the values of x for which
(x – 5)(x + 2) = 0
Hence, x – 5 = 0 or x + 2 = 0
⇒ x = 5 or x = -2
(ii)2x 2 + x – 6 = 0
解决方案：
Here, LHS = 2x2 + x – 6
= 2x2 + 4x – 3x – 6
= 2x(x + 2) – 3(x + 2)
= (2x– 3)(x + 2)
The roots of this equation, 2x2 + x – 6 = 0 are the values of x for which
(2x– 3)(x + 2) = 0
Hence, 2x– 3 = 0 or x + 2 = 0
⇒ x = 3/2 or x = –2
(ⅲ)√2×2 + 7×+5√2= 0
解决方案：
Here, LHS = √2x2 + 7x + 5√2
= √2x2 + 5x + 2x + 5√2
= x(√2x + 5) + √2(√2x + 5)
= (√2x + 5) (x +√2)
The roots of this equation, √2x2 + 7x + 5√2 = 0 are the values of x for which
(√2x + 5) (x +√2) = 0
Hence, √2x + 5 = 0 or x +√2 = 0
⇒ x = –5/√2 or x = –√2
(iv)2x 2 – x + 1/8 = 0
解决方案：
Here, LHS = 2x2 – x + 1/8
= 1/8(16x2 – 8x + 1)
= 1/8(16x2 – 4x -4x + 1)
= 1/8(4x(4x-1) -1 (4x-1))
= 1/8 (4x-1) (4x-1)
The roots of this equation, 2x2 – x + 1/8 = 0 are the values of x for which
1/8 (4x-1) (4x-1) = 0
(4x-1)2 = 0
Hence, 4x-1 = 0 or 4x-1 = 0
⇒ x = 1/4 or x = 1/4
(v)100x 2 – 20x +1 = 0
解决方案：
Here, LHS = 100x2 – 20x + 1
= 100x2 – 10x – 10x + 1
= 10x(10x – 1) – 1(10x – 1)
= (10x – 1) (10x – 1)
The roots of this equation, 100x2 – 20x + 1 = 0 are the values of x for which
(10x – 1) (10x – 1) = 0
(10x – 1)2 = 0
Hence, 10x – 1 = 0 or 10x – 1 = 0
⇒ x = 1/10 or x = 1/10
问题2：用数学方法解决以下情况：
(i)John和Jivanti共有45颗弹珠。他们俩每人损失了5个大理石，现在他们拥有的大理石数量为124。我们想知道他们必须从多少个大理石开始。
解决方案：
Let’s say,
The number of marbles John have = x.
So, number of marble Jivanti have = 45 – x
After losing 5 marbles each,
Number of marbles John have = x – 5
Number of marble Jivanti have = 45 – x – 5 = 40 – x
Here, According to the given condition
(x – 5)(40 – x) = 124
x2 – 45x + 324 = 0
x2 – 36x – 9x + 324 = 0
x(x – 36) -9(x – 36) = 0
(x – 36)(x – 9) = 0
Hence, x – 36 = 0 or x – 9 = 0
x = 36 or x = 9
Therefore,
If, John’s marbles = 36, then, Jivanti’s marbles = 45 – 36 = 9
And if John’s marbles = 9, then, Jivanti’s marbles = 45 – 9 = 36
(ii)手工业每天生产一定数量的玩具。发现每个玩具的生产成本(卢比)为55减去一天生产的玩具数量。在特定的一天，生产的总成本为₹750。我们想知道当天生产的玩具数量。
解决方案：
Let us say,
Number of toys produced in a day be x.
Therefore, cost of production of each toy = Rs(55 – x)
Given, total cost of production of the toys = Rs 750
So, x(55 – x) = 750
x2 – 55x + 750 = 0
x2 – 25x – 30x + 750 = 0
x(x – 25) -30(x – 25) = 0
(x – 25)(x – 30) = 0
Hence, x – 25 = 0 or x – 30 = 0
x = 25 or x = 30
Hence, the number of toys produced in a day, will be either 25 or 30.
问题3.找到两个数字之和为27，乘积为182。
解决方案：
Let’s say,
First number be x and the second number is 27 – x.
Therefore, the product of two numbers will be:
x(27 – x) = 182
x2 – 27x – 182 = 0
x2 – 13x – 14x + 182 = 0
x(x – 13) -14(x – 13) = 0
(x – 13)(x -14) = 0
Hence, x – 13 = 0 or x – 14= 0
x = 13 or x = 14
Hence, if first number = 13, then second number = 27 – 13 = 14
And if first number = 14, then second number = 27 – 14 = 13
Hence, the numbers are 13 and 14.
问题4.查找两个连续的正整数，其平方和为365。
解决方案：
Let’s say,
Two consecutive positive integers be x and x + 1.
Here, According to the given condition,
x2 + (x + 1)2 = 365
x2 + x2 + 1 + 2x = 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
Hence, x – 13 = 0 or x + 14= 0
x = 13 or x = – 14
As, here it is said positive integers, so x can be 13, only.
So,
x = 13
and, x + 1 = 13 + 1 = 14
Hence, two consecutive positive integers will be 13 and 14.
问题5.直角三角形的高度比其底边小7厘米。如果斜边为13厘米，则找到另外两个侧面。
解决方案：
Let’s say,
Base of the right triangle be x cm.
So, the altitude of right triangle = (x – 7) cm
Base2 + Altitude2 = Hypotenuse2 (Pythagoras theorem)
x2 + (x – 7)2 = 132
x2 + x2 + 49 – 14x = 169 (using identity (a-b)2 = a2 – 2ab + b2)
2x2 – 14x – 120 = 0
x2 – 7x – 60 = 0 (Dividing by 2)
x2 – 12x + 5x – 60 = 0
x(x – 12) + 5(x – 12) = 0
(x – 12)(x + 5) = 0
Hence, x – 12 = 0 or x + 5= 0
x = 12 or x = – 5
As, here side will be a positive integers, so x can be 12, only.
Therefore, the base of the given triangle is 12 cm and,
the altitude of this triangle will be (12 – 7) cm = 5 cm.
问题6：家庭手工业每天生产一定数量的陶器产品。在某一天观察到，每件商品的生产成本(以卢比计)比当日生产的商品数量多3倍。如果当天的总生产成本为₹90，请找到生产的商品数量以及每件商品的成本。
解决方案：
Let’s say,
Number of articles produced be x.
So, cost of production of each article = ₹ (2x + 3)
Here, According to the given condition
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
Hence, 2x +15 = 0 or x – 6= 0
x = –15/2 or x = 6
As the number of articles produced can only be a positive integer,
So, x = 6.
Hence, number of articles produced = 6
Cost of each article = 2 × 6 + 3 = ₹ 15.