第 10 类 RD Sharma 解——第 8 章二次方程——练习 8.2

问题1：两个连续正整数的乘积是306。如果x表示较小的整数，则形成二次方程以求整数。

解决方案：

Given that the smallest integer of the two consecutive positive integer is denoted by x. Hence, the two integers are x and (x+1). Now its also given that the product of the two consecutive positive integers is 306.

Hence,        x*(x+1)= 306

or               x²+ x=306

or               x²+x-306=0

hence, the required quadratic equation is x²+x-306=0 .

问题 2： John 和 Jivanti 共有 45 个弹珠。他们两人各丢了 5 颗弹珠，他们现在拥有的弹珠数量的乘积是 128。如果约翰有 x 颗弹珠，则建立二次方程以找出他们必须从多少颗弹珠开始。

解决方案：

Given that the John and Jivanti together have 45 marbles and john had x marbles. Now , let us assume that the number of marbles Jivanti had is y. Hence, according to question,      x+y=45

or                                                 y=45-x

So, Jivanti had (45-x) marbles.  Now , it is given in the question that both of them lost 5 marbles. 

Hence, the number of marbles John has after losing 5 marbles is = x-45

and, the number of marbles Jivanti has after losing 5 marbles is = 45-x-5= 40-x

Now, given that the product of the number of marbles they have now is = 128 

Hence, (x-5)*(40-x)=128

or         40x – x²– 200 + 5x – 128 = 0

or         -x² + 45x – 328 = 0

or         x² – 45x + 328 = 0

Therefore, the required quadratic equation is x² – 45x + 328 = 0.

问题 3：家庭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（以卢比计）为 55 减去一天生产的物品数量。在某一天，生产的总成本为卢比。 750. 如果 x 表示当天生产的玩具数量，则形成求 x 的二次方程。

解决方案：

Given that x denotes the number of toys produced that day. 

Hence, according to question the cost of production of each toy is= 55- Number of toys produced in a day = 55-x

now, we can say that the total cost of production on a particular day is  Rs x * (55-x)

Given that the on a particular day, the total cost of production was Rs. 750

hence,   x * (55 – x) = 750

or          x² – 55x + 750 = 0

Therefore, the required quadratic a equation is x² – 55x + 750 = 0 .

问题4：直角三角形的高度比它的底边少7厘米。如果斜边是 13 厘米，请形成二次方程以找到三角形的底。

解决方案：

Let us assume that the base of the right triangle is x. Now, given that the height of the right triangle is 7 cm less than its base.

Hence, the height of the right triangle is = (x-7) cm

Given that the hypotenuse of the right triangle is 13 cm.

Hence, according to Pythagoras Theorem,     

(Hypotenuse)² = (Base)² + (Height)²

or               (13 )² = x² + (x – 7)²

or              169 = x² + x² – 14x + 49

or              2x² – 14x + 49 – 169 = 0

or              2x² – 14x – 120 = 0

or              x²– 7x – 60 = 0

Therefore, the required quadratic equation is x² – 7x – 60 = 0.

问题5：特快列车在迈索尔和班加罗尔之间行驶132公里，比客运列车少1小时。如果特快列车的平均速度比客运列车的平均速度快 11 公里/小时，则由二次方程求快车的平均速度。

解决方案：

Let us assume that the average speed of the express train is x km/hr. Now, given that the average speed of the express train is 11 km/hr more than that of the passenger train. We can say that,

The average speed of the passengers train is = (x-11) Km/hr

Now, according to the question the total distance travelled by each train is 132km.

We know that,

Time taken to travel= Distance travelled/average speed

Time taken by the express train (t1) = distance travelled/average speed of the express train

                                                       = $\frac{132}{x}$  hr

Time taken by the passengers train (t2) = distance travelled/average speed of the passengers train

                                                               = $\frac{132}{x-11}$  hr

Given that the time taken by the express train is 1 h less than the passengers train,

Hence,          t2-t1=1

or                 $\frac{132}{x-11}-\frac{132}{x}=1$

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

or                 132*(x-x+11)=x*(x-11)

or                 132*11=x²-11x

or                 x²-11x=1452  

or                 x²-11x-1452=0

Therefore, the required quadratic equation is x²-11x-1452=0.

问题6：一列火车匀速行驶360公里。如果速度再提高 5 公里/小时，同样的旅程将减少 1 小时。形成二次方程以找到火车的速度。

解决方案：

Let us assume that the speed of the train is x km/hr. Given that the distance covered by the train at a uniform speed is 360 km.

hence, the time travelled by the train is = Distance covered/speed of the train (t1)= $\frac{360}{x}$  hr

now if the speed of the train is 5 km/hr more then the speed of the should be = (x+5) km/hr

hence, the time travelled by the train when it’s speed is 5 km/h more is (t2)= $\frac{360}{x+5}$  hr

Now , given that the train would have taken 1 hour less if it’s speed was 5 km/hr more.

hence,          t1-t2=1

or                 $\frac{360}{x}-\frac{360}{x+5}=1$

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

or                 360*5=x*(x+5)

or                 x²+5x=1800

or                 x²+5x=1800

Therefore, the required quadratic equation is  x²+5x=1800.