10类NCERT解决方案-第4章二次方程式-练习4.1

问题1.检查以下是否是二次方程式：

(i)(x +1) ² = 2(x – 3)

解决方案：

Here,

LHS = (x + 1)²

= x² + 2x + 1        (Using identity (a+b)² = a² + 2ab + b²)

and, RHS = 2(x–3)

= 2x – 6

As, LHS = RHS

x² + 2x + 1 = 2x – 6

x² + 7 = 0 ………….(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

(ii)x ² – 2x =(–2)(3 – x)

解决方案：

Here,

LHS = x² – 2x

and, RHS = (–2) (3 – x)

= 2x–6

As, LHS = RHS

x² – 2x = 2x – 6

x² – 4x + 6 = 0 ………….(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

(iii)(x – 2)(x + 1)=(x – 1)(x + 3)

解决方案：

Here,

LHS = (x – 2)(x + 1)

= x² + (–2+1)x + (–2)(1)    (Using identity (x+a) (x+b) = x² + (a+b)x + ab)

= x² – x – 2

and, RHS = (x – 1)(x + 3) 

= x² + (–1+3)x + (–1)(3)   (Using identity (x+a) (x+b) = x² + (a+b)x + ab)

= x² + 2x – 3

As, LHS = RHS

x² – x – 2 = x² + 2x – 3

3x – 1 = 0 ………….(I)

As, the eqn. (I) is not in the form of ax² + bx + c = 0 because (a=0).

Hence, the equation is NOT QUADRATIC equation because highest power of x is 1.

(iv)(x – 3)(2x +1)= x(x + 5)

解决方案：

Here,

LHS = (x – 3)(2x +1)

= 2x² + x +(–3)(2x) + (–3)(1)

= 2x² – 5x – 3

and, RHS = x(x + 5)

= x² + 5x

As, LHS = RHS

2x² – 5x – 3 = x² + 5x

x² – 10x – 3 = 0 ………….(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

(v)(2x – 1)(x – 3)=(x + 5)(x – 1)

解决方案：

Here,

LHS = (2x – 1)(x – 3)

= 2x² + (2x)(–3) +(–1)(x) + (–1)(–3)

= 2x² – 7x + 3

and, RHS = (x + 5)(x – 1) 

= x² + 5(x) + (–1)(x) + (5)(–1)     (Using identity (x+a) (x+b) = x² + (a+b)x + ab)

= x² + 4(x) – 5

As, LHS = RHS

2x² – 7x + 3 = x² + 4(x) – 5

x² – 11x + 8 = 0 ………….(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

(vi)x ² + 3x + 1 =(x – 2) ²

解决方案：

Here,

LHS = x² + 3x + 1

and, RHS = (x – 2)²

= x² – 4x + 4      (Using identity (a–b)² = a² – 2ab + b²)

As, LHS = RHS

x² + 3x + 1 = x² – 4x + 4

7x – 3 = 0 ………….(I)

As, the eqn. (I) is not in the form of ax² + bx + c = 0 because (a=0).

Hence, the equation is NOT QUADRATIC equation because highest power of x is 1.

(vii)(x + 2) ³ = 2x(x ² – 1)

解决方案：

Here,

LHS = (x + 2)³

= x³ + 2³ + 3x(2)(x+2)   (Using identity (x+y)³ = x³ + y³ + 3xy(x+y))

= x³ + 8 + 6x² +12x

and, RHS = 2x (x² – 1) 

= 2x³  – 2x      

As, LHS = RHS

x³ + 8 + 6x² +12x = 2x³  – 2x 

x³ – 6x² – 14x – 8 = 0 ………….(I)

As, the eqn. (I) is not in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is NOT QUADRATIC equation because highest power of x is 3.

(viii)x ³ – 4x ² – x + 1 =(x – 2) ³

解决方案：

Here,

LHS = x³ – 4x² – x + 1

and, RHS = (x – 2)³

= x³ – 2³ – 3x(2)(x–2)     (Using identity (x–y)³ = x³ – y³ – 3xy(x-y))

= x³ – 8 – 6x² +12x

As, LHS = RHS

x³ – 4x² – x + 1 = x³ – 8 – 6x² +12x

2x² – 13x + 9 = 0 ………….(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

问题2.用二次方程式表示以下情况：

(i)矩形图的面积为528 m ² 。地块的长度(以米为单位)是其宽度的两倍多。我们需要找到情节的长度和广度。

解决方案：

Let’s consider,

Breadth of the rectangular plot = b m

Then, length of the plot = (2b + 1) m.

As, Area of rectangle = length × breadth 

528 m² = (2x + 1) × x

2x² + x =528

2x² + x – 528 = 0 ……………….(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

Therefore, the length and breadth of plot, satisfies the quadratic equation, 2x² + x – 528 = 0, which is the required representation of the problem mathematically.

(ii)两个连续的正整数的乘积是306。我们需要找到这些整数。

解决方案：

Let’s consider,

The first integer number = x

Next consecutive positive integer will be = x + 1

Product of two consecutive integers = x × (x +1)

x² + x = 306

x² + x – 306 = 0 ……………….(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

Therefore, the two integers x and x+1, satisfies the quadratic equation, x² + x – 306 = 0, which is the required representation of the problem mathematically.

(iii)罗汉的母亲比他大26岁。从现在开始他们3年的年龄(以年为单位)的乘积将是360。我们想找到Rohan的当前年龄。

解决方案：

Let’s consider,

Age of Rohan’s = x  years

So, Rohan’s mother’s age = x + 26

After 3 years,

Age of Rohan’s = x + 3

Age of Rohan’s mother will be = x + 26 + 3 = x + 29

The product of their ages after 3 years will be equal to 360, such that

(x + 3)(x + 29) = 360

x² + 29x + 3x + 87 = 360

x² + 32x + 87 – 360 = 0

x² + 32x – 273 = 0 ……………….(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

Therefore, the age of Rohan and his mother, satisfies the quadratic equation, x² + 32x – 273 = 0, which is the required representation of the problem mathematically.

(iv)火车以匀速行驶480公里。如果速度降低了8 km / h，则覆盖相同距离要多花3个小时。我们需要找到火车的速度。

解决方案：

Let’s consider,

The speed of train = x  km/hr

Time taken to travel 480 km = (480/x) hr

Time = Distance / Speed

Here, According to the given condition,

The speed of train = (x – 8) km/hr

As, the train will take 3 hours more to cover the same distance.

Hence, Time taken to travel 480 km = 480/(x+3) km/hr

As,

Speed × Time = Distance

(x – 8)(480/(x + 3) = 480

480 + 3x – 3840/x – 24 = 480

3x – 3840/x = 24

3x² – 8x – 1280 = 0 ……………….(I)

As, the eqn. (I) is in the form of ax² + bx + c = 0 where (a≠0).

Hence, the equation is QUADRATIC equation because highest power of x is 2.

Therefore, the speed of the train, satisfies the quadratic equation, 3x² – 8x – 1280 = 0, which is the required representation of the problem mathematically.