第 10 类 RD Sharma 解决方案 - 第 8 章二次方程 - 练习 8.7 |设置 2
问题 11:一个数与其平方的和是 63/4,求出这些数。
解决方案:
Let the number is = x
so it’s square is = x2
Now according to condition-
⇒(number)+(number)2=63/4
⇒ x+x2 = 63/4
⇒ x2+x-63/4=0
Multiplying by 4-
⇒ 4x2+4x-63=0
⇒ 4x2 +(18-14)x – 63 =0 [because 63*4=252
so 18*14=252 & 18-14=4]
⇒ 4x2+18x-14x-63=0
⇒2x(2x+9)-7(2x+9)=0
⇒(2x+9)(2x-7)=0
either 2x+9=0 or 2x-7=0
x=-9/2 or x=7/2
So number is -9/2 or 7/2.
问题 12:有三个连续的整数,第一个的平方乘以其他两个的乘积得到 154。整数是多少?
解决方案:
Let the first number is = x
so second number is = x+1
and third number is = x+2
now according to the given condition-
⇒ (first number)2 + (second number)*(third number)=154
⇒ x2 + (x+1)(x+2)=154
⇒ x2 + x2 + 3x + 2 = 154
⇒ 2x2+3x-152=0
⇒ 2x2 +(19-16)x-152=0
⇒ 2x2 +19x-16x-152=0
⇒ x(2x+19)-8(2x+19)=0
⇒ (2x+19)(x-8)=0
Either 2x+19=0 or x-8=0
x=-19/2 or x=8
but in question it is said number should be integer
so discard x=-19/2
when x=8
First number is =8
Second number is =9
And Third number is =10
问题 13:5 的两个连续整数倍的乘积是 300。确定乘数。
解决方案:
Let the first integral multiple of 5 is =5x
So next is = 5x+5
Now according to the condition-
⇒ (first integral multiple of 5)*(next integral multiple of 5)=300
⇒ 5x(5x+5)=300
Dividing by 5-
⇒ x(5x+5)=60
again dividing by 5-
⇒ x(x+1)=12
⇒ x2 + x -12=0
⇒ x2+(4-3)x-12=0
⇒ x2+4x-3x-12=0
⇒ x(x+4)-3(x+4)=0
⇒ (x+4)(x-3)=0
Either x+4=0 or x-3=0
x=-4 or x=3
when x=-4
first integral multiple of 5 is = 5*-4 = -20
and next integral multiple of 5 is = 5*-4+5= -15
when x=3
first integral multiple of 5 is = 5*3 = 15
and next integral multiple of 5 is = 5*3+5 = 20
问题 14:两个数的平方和为 233,其中一个数比另一个数的两倍小 3。找到数字。
解决方案:
Let first number is= x
So second number = 2*(first number)-3
= 2x-3
Now according to given condition-
⇒(first number)2+(second number)2 = 233
⇒x2+(2x-3)2=233
⇒x2+4x2-2*2x*3+9=233
⇒5x2-12x+9-233=0
⇒5x2-12x-224=0
⇒5x2-(40-28)x-224=0
⇒5x2-40x+28x-224=0
⇒5x(x-8)+28(x-8)=0
⇒(x-8)(5x+28)=0
Either x-8=0 or 5x+28=0
x=8 or x=-28/5
but when x=-28/3, it doesn’t satisfy given condition.
so on taking x=8
first number is=x= 8
and second number is=2x-3=13
问题 15:求两个平方和为 340 的连续偶数。
解决方案:
Let first even integer=2x
so second even integer=2x+2
According to given condition-
⇒(first integer)2+(second integer)2=340
⇒(2x)2+(2x+2)2=340
⇒4x2+4x2+2*2x*2+4=340
⇒8x2+8x-336=0
Dividing by 8-
⇒x2+x-42=0
⇒x2+(7-6)x-42=0
⇒x2+7x-6x-42=0
⇒x(x+7)-6(x+7)=0
⇒(x+7)(x-6)=0
Either x+7=0 or x-6=0
x=-7 or x=6
When x=-7
then first integer=2*x= -14
and second integer=2x+2=-12
when x=6
then first integer=2*x= 12
and second integer=2x+2=14
问题 16:两个数的差是 4,如果它们的倒数差是 4/21,求这两个数。
解决方案:
Let first number is=x
So second number is=x-4
Reciprocal of first number is=1/x
and reciprocal of second number is=1/x-4
According to given condition-
⇒(reciprocal of first number)-(reciprocal of second number)=4/21
⇒(1/(x-4))-(1/x)=4/21
on taking LCM-
⇒(x-x+4)/(x(x-4))=4/21
⇒21(4)=4(x(x-4)
⇒21=(x2-4x)
⇒x2-4x-21=0
⇒x2-(7-3)-21=0
⇒x2-7x+3x-21=0
⇒x(x-7)+3(x-7)=0
⇒(x-7)(x+3)=0
Either x-7=0 or x+3=0
x=7 or x=-3
When x=7
numbers are= 3,7
and when x=-3
numbers are=-7,-3
问题 17:找出两个相差 3 且平方和为 117 的自然数。
解决方案:
Let the first number is=x
So second number is=x-3
Now according to given condition-
⇒(first number)2+(second number)2=117
⇒x2+(x-3)2=117
⇒x2+x2-2*x*3+9=117
⇒2x2-6x+9-117=0
⇒2x2-6x-108=0
Dividing by 2-
⇒x2-3x-54=0
⇒x2-(9-6)x-54=0
⇒x2-9x+6x-54=0
⇒x(x-9)+6(x-9)=0
⇒(x-9)(x+6)=0
Either x-9=0 or x+6=0
x=9 or x=-6
But x=-6 is not a natural number.
So when x=9
⇒first number is=9
⇒second number is=9-3=6
问题 18:三个连续自然数的平方和为 149。求这些数。
解决方案:
Let first number is=x
so second is=x+1
and third is=x+2
Now according to given condition-
⇒(first number)2+(second number)2+(third number)2=149
⇒x2+(x+1)2+(x+2)2=149
⇒x2+x2+2*x*1+1+x2+2*x*2+4=149
⇒3x2+6x+5-149=0
⇒3x2+6x-144=0
Dividing by 3-
⇒x2+2x-48=0
⇒x2+(8-6)x-48=0
⇒x2+8x-6x-48=0
⇒x(x+8)-6(x+8)=0
⇒(x+8)(x-6)=0
Either x+8=0 or x-6=0
x=-8 or x=6
But x=-8 is not a natural number.
so when x=6
⇒first number is=x=6
⇒second number is=x+1=7
⇒third number is=x+2=8
问题19:两个数的和是16,它们的倒数和是1/3。找到数字。
解决方案:
Let first number is=x
So second number is=16-x
Reciprocal of first number is=1/x
and Reciprocal of second number is=1/(16-x)
Now according to given condition-
⇒(Reciprocal of first number)+(Reciprocal of second number)=1/3
⇒(1/x)+(1/(16-x))=1/3
on taking LCM-
⇒(16-x+x)/(x(16-x))=1/3
⇒16*3=x(16-x)
⇒48=16x-x2
⇒x2-16x+48=0
⇒x2-(12+4)x+48=0
⇒x2-12x-4x+48=0
⇒x(x-12)-4(x-12)=0
⇒(x-12)(x-4)=0
Either x-12=0 or x-4=0
x=12 or x=4
so when x=12
numbers are = 12,4
when x=4
numbers are=4,12
means required numbers are=4,12
问题 20:确定乘积为 270 的 3 的两个连续倍数。
解决方案:
Let first multiple of 3 is=3x
so second is=3x+3
Now according to given condition-
⇒ (first multiple of 3)*(second multiple of 3)=270
⇒3x(3x+3)=270
⇒9x2+9x=270
on dividing by 9-
⇒x2+x=30
⇒x2+x-30=0
⇒x2+(6-5)x-30=0
⇒x2+6x-5x-30=0
⇒x(x+6)-5(x+6)=0
⇒(x+6)(x-5)=0
Either x+6=0 or x-5=0
x=-6 or x=5
when x=-6
first multiple is=3x=-18
second multiple is=3x+3=-15
when x=5
first multiple is=3x=15
second multiple is=3x+3=18
问题 21. 一个数和它的倒数之和是 17/4。找到号码。
解决方案:
Let the number be x.
Then from the question, we have
x + 1/x = 17/4
(x2 + 1)/x = 17/4
⇒ 4(x2+1) = 17x
⇒ 4x2 + 4 – 17x = 0
⇒ 4x2 + 4 – 16x – x = 0
⇒ 4x(x – 4) – 1(x – 4) = 0
⇒ (4x – 1)(x – 4) = 0
Now, either x – 4 = 0 ⇒ x = 4
Or, 4x – 1 = 0 ⇒ x = 1/4
Thus, the value of x is 4.