10类NCERT解决方案-第4章二次方程式-练习4.3

问题1.通过完成平方的方法，找到以下二次方程式的根(如果存在)：

(i)2x ² – 7x + 3 = 0

解决方案：

2x² – 7x = – 3

Dividing by 2 on both sides, we get

x² –$\frac{7x}{2}$= –$\frac{3}{2}$

x² -2 × x ×$\frac{7}{4}$= –$\frac{3}{2}$

On adding ($\frac{7}{4}$)² to both sides of equation, we get

(x)² – 2×x×$\frac{7}{4}$+($\frac{7}{4}$)² = ($\frac{7}{4}$)²–$\frac{3}{2}$

(x-$\frac{7}{4}$)² = ($\frac{49}{16}$) – ($\frac{3}{2}$) (Using identity: a² – 2ab + b² = (a-b)²)

(x-$\frac{7}{4}$)² =$\frac{25}{16}$

(x-$\frac{7}{4}$)² = ±$\frac{5}{4}$

x =$\frac{7}{4} ± \frac{5}{4}$

x =$\frac{7}{4} + \frac{5}{4}$or x =$\frac{7}{4} - \frac{5}{4}$

x =$\frac{12}{4}$or x =$\frac{2}{4}$

x = 3 or x =$\frac{1}{2}$

(ii)2x ² + x – 4 = 0

解决方案：

2x² + x = 4

Dividing both sides of the equation by 2, we get

x² +$\frac{x}{2}$= 2

Now on adding$(\frac{1}{4})$² to both sides of the equation, we get,

(x)² + 2 × x ×$\frac{1}{4}$+ ($\frac{1}{4}$)² = 2 + ($\frac{1}{4}$)²

(x +$\frac{1}{4}$)2 =$\frac{33}{16}$(Using identity: a² + 2ab + b² = (a+b)²)

x +$\frac{1}{4}$= ±$√(\frac{33}{16})$

x =$- \frac{1}{4} ± \frac{√33}{4}$

x =$\frac{-1± √33}{4}$

Hence, x =$\frac{-1+ √33}{4}$or x =$\frac{-1- √33}{4}$

(iii)4x ² +4√3x+ 3 = 0

解决方案：

4x² + 4√3x = -3

Dividing both sides of the equation by 4, we get

x² + √3x = –$\frac{3}{4}$

Now on adding ($\frac{√3}{2}$)² to both sides of the equation, we get,

(x)² + 2 × x ×$\frac{√3}{2}$+ ($\frac{√3}{2}$)² = –$\frac{3}{4}$+ ($\frac{√3}{2}$)²

(x +$\frac{√3}{2}$)² = 0 (Using identity: a² + 2ab + b² = (a+b)²)

Hence, x = –$\frac{√3}{2}$or x = –$\frac{√3}{2}$

(iv)2x ² + x + 4 = 0

解决方案：

2x² + x = -4

Dividing both sides of the equation by 2, we get

x² +$\frac{x}{2}$= -2

Now on adding ($\frac{1}{4}$)² to both sides of the equation, we get,

(x)² + 2 × x ×$\frac{1}{4}$+ ($\frac{1}{4}$)² = – 2 + ($\frac{1}{4}$)²

(x +$\frac{1}{4}$)² =$\frac{-31}{16}$(Using identity: a² + 2ab + b² = (a+b)²)

As we know, the square of numbers cannot be negative.

Hence, there is no real root for the given equation, 2x² + x + 4 = 0.

问题2。通过应用二次方程式，找到上述Q.1中给出的二次方程式的根。

(i)2x ² – 7x + 3 = 0

解决方案：

On comparing the equation with ax² + bx + c = 0, we get,

a = 2, b = -7 and c = 3

Then roots of the quadratic equation =$\frac{-b± √(b^2-4ac)}{2a}$

x =$\frac{-(-7)± √((-7)^2-4(2)(3))}{2(2)}$

x =$\frac{7± √(49-24)}{4}$

x =$\frac{7± √25}{4}$

x =$\frac{7± 5}{4}$

x =$\frac{7+5}{4}$or x =$\frac{7-5}{4}$

x =$\frac{12}{4}$or$\frac{2}{4}$

x = 3 or$\frac{1}{2}$

(ii)2x ² + x – 4 = 0

解决方案：

On comparing the equation with ax² + bx + c = 0, we get,

a = 2, b = 1 and c = -4

Then roots of the quadratic equation =$\frac{-b± √(b^2-4ac)}{2a}$

x =$\frac{-1± √(1^2-4(2)(-4))}{2(2)}$

x =$\frac{-1± √(1+32)}{4}$

x =$\frac{-1± √33}{4}$

x =$\frac{-1+√33}{4}$or x =$\frac{1-√33}{4}$

(iii)4x ² +4√3x+ 3 = 0

解决方案：

On comparing the equation with ax² + bx + c = 0, we get,

a = 4, b = 4√3 and c = 3

Then roots of the quadratic equation =$\frac{-b± √(b^2-4ac)}{2a}$

x =$\frac{-(4√3)± √((4√3)^2-4(4)(3))}{2(4)}$

x =$\frac{-(4√3)± √(48-48)}{8}$

x =$\frac{-4√3}{8}$

x =$\frac{-√3}{2}$or x =$\frac{-√3}{2}$

(iv)2x ² + x + 4 = 0

解决方案：

On comparing the equation with ax² + bx + c = 0, we get,

a = 2, b = 1 and c = 4

Then roots of the quadratic equation =$\frac{-b± √(b^2-4ac)}{2a}$

x =$\frac{-1± √(1^2-4(2)(4))}{2(2)}$

x =$\frac{-1± √(1-32)}{4}$

x =$\frac{-1± √(-31)}{4}$

As we know, the square of a number can never be negative.

Hence, there is no real solution for the given equation.

问题3.找到以下方程式的根：

(i)x – $\frac{1}{x}$ = 3，x≠0

解决方案：

After rearranging, we get

x² – 3x -1 = 0

On comparing the equation with ax² + bx + c = 0, we get,

a = 1, b = -3 and c = -1

Then roots of the quadratic equation =$\frac{-b± √(b^2-4ac)}{2a}$

x =$\frac{-(-3)± √((-3)^2-4(1)(-1))}{2(1)}$

x =$\frac{3± √(9+4)}{2}$

x =$\frac{3±√(13)}{2}$

x =$\frac{3+√(13)}{2}$or x =$\frac{3-√(13)}{2}$

(ii) $\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}$ ，x≠-4,7

解决方案：

$\frac{(x-7-x-4)}{(x+4)(x-7)} = \frac{11}{30}$

$\frac{(-11)}{(x+4)(x-7)} = \frac{11}{30}$

After rearranging,

(x+4)(x-7) = -30

x² – 3x – 28 = 30

x² – 3x + 2 = 0

On comparing the equation with ax² + bx + c = 0, we get,

a = 1, b = -3 and c = 2

Then roots of the quadratic equation =$\frac{-b± √(b^2-4ac)}{2a}$

x =$\frac{-(-3)± √((-3)^2-4(1)(2))}{2(1)}$

x =$\frac{3± √(9-8)}{2}$

x =$\frac{3±√1}{2}$

x =$\frac{3±1}{2}$

x =$\frac{4}{2}$or x =$\frac{2}{2}$

x = 2 or x = 1

问题4。3年前和5年后的雷曼年龄的倒数之和(以年为单位)是$\frac{1}{3}$ 。找到他现在的年龄。

解决方案：

Let’s take,

Present age of Rehman is x years.

Three years ago, Rehman’s age was (x – 3) years.

Five years after, his age will be (x + 5) years.

According to the given condition,

$\frac{1}{x-3} + \frac{1}{x-5} = \frac{1}{3}$

$\frac{(x+5+x-3)}{(x-3)(x+5)} = \frac{1}{3}$

$\frac{(2x+2)}{(x-3)(x+5)} = \frac{1}{3}$

3(2x + 2) = (x-3)(x+5)

6x + 6 = x² + 2x – 15

x² – 4x – 21 = 0

x² – 7x + 3x – 21 = 0 (by factorizing)

x(x – 7) + 3(x – 7) = 0

(x – 7)(x + 3) = 0

x = 7, -3

As, age cannot be negative.

Therefore, Rehman’s present age is 7 years.

问题5：在课堂测试中，Shefali在数学和英语中的分数总和为30。如果她在数学中获得2分，而在英语中少得到3分，则它们的分数乘积应为210。两个科目。

解决方案：

Let’s take,

The marks of Shefali in Maths be x.

Then, the marks in English will be 30 – x.

According to the given condition,

(x + 2)(30 – x – 3) = 210

(x + 2)(27 – x) = 210

-x² + 25x + 54 = 210

Multiply the equation by (-1),

x² – 25x + 156 = 0

x² – 12x – 13x + 156 = 0

x(x – 12) -13(x – 12) = 0

(x – 12)(x – 13) = 0

x = 12, 13

Hence, if the marks in Maths are 12, then marks in English will be 30 – 12 = 18

and, the marks in Maths are 13, then marks in English will be 30 – 13 = 17.

问题6.矩形场的对角线比短边大60米。如果较长的一面比较短的一面多30米，请找到视野的两边。

解决方案：

Let’s take,

Breadth = x

Length = x+30

Diagonal = x+60

Diagonal = √(Length² + Breadth²)

According to the given condition,

√((x+30)² + (x)²) = x+60

Squaring both sides,

x² + (x + 30)² = (x + 60)²

x² + x² + 900 + 60x = x² + 3600 + 120x

x² – 60x – 2700 = 0

x² – 90x + 30x – 2700 = 0

x(x – 90) + 30(x -90) = 0

(x – 90)(x + 30) = 0

x = 90, -30

As, side of the field cannot be negative.

Hence, the length of the shorter side will be 90 m.

and, the length of the larger side will be (90 + 30) m = 120 m.

问题7.两个数字的平方差是180。较小数字的平方是较大数字的8倍。找到两个数字。

解决方案：

Let’s take,

The larger number = x

and, smaller number = y

According to the given condition,

x² – y² = 180 and y² = 8x (It means x has to be positive, because it is obtained by squaring a number)

x² – 8x = 180

x² – 8x – 180 = 0

x² – 18x + 10x – 180 = 0

x(x – 18) +10(x – 18) = 0

(x – 18)(x + 10) = 0

x = 18, -10

As x cannot be negative,

Hence, the larger number will be 18.

x = 18

So, As y² = 8x

= 8 × 18

= 144

y = ±√144 = ±12

So, Smaller number = ±12

Hence, the numbers are 18 and 12 or 18 and -12.

问题8.火车以匀速行驶360公里。如果速度提高了5 km / h，则相同的旅程将少花1个小时。找到火车的速度。

解决方案：

Let’s take

The speed of the train = x km/hr.

As, Speed =$\frac{Distance}{Time}$

Time taken to cover 360 km =$\frac{360}{x}$hr.

As per the question given,

(x + 5)($\frac{360}{x}$– 1) = 360

(x + 5)($\frac{360 - x}{x}$) = 360

(x + 5)(360 – x) = 360x

x² + 5x -1800 = 0

x² + 45x – 40x + 1800 = 0

x(x + 45) -40(x + 45) = 0

(x + 45)(x – 40) = 0

x = 40, -45

As we know, the value of speed cannot be negative.

Hence, the speed of train is 40 km/h.

问题9.两个水龙头可以一起装满9个水箱$\frac{3}{8}$小时。较大直径的水龙头要比较小的水龙头少花10个小时，以便分别填充水箱。找出每个水龙头可以分别填充水箱的时间。

解决方案：

Let’s take

The time taken by the smaller pipe to fill the tank = x hr.

Time taken by the larger pipe = (x – 10) hr

Part of tank filled by smaller pipe in 1 hour =$\frac{1}{x}$

Part of tank filled by larger pipe in 1 hour =$\frac{1}{x-10}$

According to the given condition,

9$\frac{3}{8}$hrs taken to fill with both the pipe.

So,

$\frac{1}{x} + \frac{1}{x-10} = \frac{8}{75}$

$\frac{x-10+x}{x(x-10)} = \frac{8}{75}$

$\frac{2x-10}{x(x-10)} = \frac{8}{75}$

75(2x – 10) = 8x² – 80x

150x – 750 = 8x² – 80x

8x² – 230x +750 = 0

8x² – 200x – 30x + 750 = 0

8x(x – 25) -30(x – 25) = 0

(x – 25)(8x -30) = 0

x = 25,$\frac{30}{8}$

Time taken by the smaller pipe cannot be$\frac{30}{8}$hours, as the time taken by the larger pipe will become negative.

Hence, time taken by the smaller pipe = 25hours

and, by the larger pipe =15 hours

问题10：在迈索尔和班加罗尔之间行驶132公里(不考虑中间站的停留时间)，快车比旅客列车少花1个小时。如果快车的平均速度比旅客列车的平均速度高11 km / h，请找到两列火车的平均速度。

解决方案：

Let’s take

The average speed of passenger train = x km/h.

Average speed of express train = (x + 11) km/h

According to the given condition,

$\frac{132}{x} + \frac{132}{x+11}$= 1

$\frac{132(x+11-x)}{x(x+11)}$= 1

$\frac{132 × 11}{x(x+11)}$= 1

132 × 11 = x(x + 11)

x² + 11x – 1452 = 0

x² + 44x -33x -1452 = 0

x(x + 44) -33(x + 44) = 0

(x + 44)(x – 33) = 0

x = – 44, 33

As we know, Speed cannot be negative.

Hence, the speed of the passenger train will be 33 km/h

and, the speed of the express train will be 33 + 11 = 44 km/h.

问题11：两个正方形的面积之和为468 m ² 。如果它们的周长差为24 m，请找到两个正方形的边。

解决方案：

Let the sides of the two squares be x and y meter.

Perimeter = 4x and 4y respectively

Area = x² and y² respectively.

According to the given condition,

4x – 4y = 24

x – y = 6

x = y + 6 ………………….(I)

and,

x² + y² = 468

(6 + y)² + y² = 468 (From (I))

36 + y² + 12y + y² = 468

2y² + 12y + 432 = 0

y² + 6y – 216 = 0

y² + 18y – 12y – 216 = 0

y(y +18) -12(y + 18) = 0

(y + 18)(y – 12) = 0

y = -18, 12

As we know, the side of a square cannot be negative.

Hence, the sides of the squares are 12 m

and, (12 + 6) m = 18 m.