第 10 类 RD Sharma 解决方案 - 第 8 章二次方程 - 练习 8.7 |设置 1
问题 1:找出两个平方和为 85 的连续数。
解决方案:
Let first number is = x
⇒ Second number = (x+1)
Now according to given condition—
⇒ Sum of squares of the numbers = 85
⇒ x2 + (x+1)2 = 85
⇒ x2 + x2 + 2x + 1 = 85 [ because (a+b)2 = a2 + 2ab + b2]
⇒ 2x2 + 2x + 1 – 85 = 0
⇒ 2x2 + 2x – 84 = 0
⇒ x2 + x – 42 = 0 [ dividing by 2 both sides]
now for factorization, convert coefficient of x in difference form of two numbers such that product of those numbers
be 42-
⇒ 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
Now when x = -7
⇒ First number = x = -7 and Second number = x+1 = -7+1
= -6
So numbers are -7, -6.
Now when x = 6
⇒ First number = x = 6 and second number = x+1 = 7
So numbers are 6, 7.
问题 2:将 29 分成两部分,使得两部分的平方和为 425。
解决方案:
Let first part is = x
so second part will be = (29 – x)
Now coming to the condition-
⇒ x2 + (29-x)2 = 425
⇒ x2 + 292 – 2*29*x + x2 = 425 [because (a+b)2 = a2 + 2ab + b2]
⇒ 2x2 + 841 – 58x = 425 [because 292=841]
⇒ 2x2 -58x + 841-425 = 0
⇒ 2x2 – 58x + 416 = 0
⇒ x2 – 29x + 208 = 0
by factorization method—
⇒ x2 – (16+13)x + 208 = 0
⇒ x2 -16x – 13x + 208 = 0
⇒ x(x-16) – 13(x-16) = 0
⇒ (x-16)(x-13) = 0
Either x-16 = 0 or x-13 = 0
x = 16 or x = 13
when first part = 16 then second part = 29 – x
= 29-16
= 13
and when first part = 13 then second part = 29-13
= 16
So parts will be 13, 16.
问题 3:两个正方形的边分别为 x cm 和 (x + 4) cm。它们的面积之和为656 cm 2 。找到正方形的边。
解决方案:
It is given that-
the side of first square = x cm
and that of second is = (x+4) cm
And we know that area of a square = (side)2
so area of first square = x2
and area of second square = (x+4)2
Now according to the given condition—
⇒ (Area of first square) + (Area of second square) = 656
⇒ x2 + (x+4)2 = 656
⇒ x2 + x2 + 2*x*4 + 42 = 656 [because (a+b)2 = a2 + 2*a*b + b2]
⇒ 2x2 + 8x + 16 – 656 = 0
⇒ 2x2 + 8x – 640 = 0
⇒ x2 + 4x – 320 = 0 [dividing by 2 both sides]
By factorization method—
⇒ x2 +(20-16)x – 320 = 0
⇒ x2 + 20x -16x – 320 = 0
⇒ x(x+20) – 16(x+20) = 0
⇒ (x+20)(x-16) = 0
Either x+20=0 or x-16 = 0
x = -20 or x=16
but x = -20 is invalid because length can never be negative,
So on discarding x=-20 and taking x=16 —
side of first square is = x = 16 cm
and the side of second square is = x+4
= 20 cm
问题 4:两个数之和是 48,它们的乘积是 432。找出这些数。
解决方案:
Let the first number = x
So second number = (48 – x) [because sum of numbers is 48]
Now it is also given-
Product of number is = 432
⇒ x*(48-x) = 432
⇒ 48x -x2 = 432
⇒ x2 – 48x + 432 = 0
By factorization method–
⇒ x2 – (36+12)x + 432 = 0
⇒ x2 – 36x – 12x + 432 = 0
⇒ x(x-36) – 12(x-36) = 0
⇒ (x-36)(x-12) = 0
Either x-36 = 0 or x-12 = 0
x = 36 or x = 12
When x=36 then –
First number = x = 36
and second number = 48-x = 12
And when x=12 then –
First number = x = 12
and second number = 48-x = 36
⇒ Means One number is 12 and another is 36.
问题 5:如果将一个整数与其平方相加,则和为 90。借助二次方程求整数。
解决方案:
Let the number is = x
So it’s square is = x2
Now according to the given condition-
Number + Square of number = 90
⇒ x + x2 = 90
⇒ x2 + x – 90 = 0
By factorization method-
⇒ x2 +(10-9)x – 90 = 0
⇒ x2 + 10x – 9x – 90 = 0
⇒ x(x+10) – 9(x+10) = 0
⇒ (x+10)(x-9) = 0
⇒ Either x+10=0 or x-9 = 0
x = -10 or x = 9
Now on taking any value of x satisfies given condition so –
Required Integer can be -10 or 9.
问题 6:求减 20 后等于该数倒数的 69 倍的整数。
解决方案:
Let the whole number is=x
So reciprocal of the number is = 1/x
now according to condition-
⇒ Number-20=69*(Reciprocal of the number)
⇒ x-20=69*(1/x);
⇒ x-(69/x)-20=0
by taking LCM-
⇒ (x2-69-20x)/x = 0
but denominator can’t be equal to 0 so-
⇒ (x2-20x-69)=0
⇒ x2 – (23-3)x -69=0
⇒ x2 – 23x + 3x -69=0
⇒ x(x-23)+3(x-23)=0
⇒(x-23)(x+3)=0
Either x+3=0 or x-23=0
x=-3 or x=23
but x=-3 is not a whole number
so taking x=23, it is a whole number
⇒ So our answer is x=23
问题 7:找到两个连续的自然数,其乘积为 20。
解决方案:
Let the first number is = x
So next number is = x+1
Now according to condition-
⇒(first number)*(second consecutive number) = 20
⇒ x(x+1)=20
⇒ x2 + x = 20
⇒ x2 + x – 20=0
⇒ x2 +(5-4)x – 20=0
⇒ x2 + 5x – 4x – 20=0
⇒ x(x+5)-4(x+5)=0
⇒(x+5)(x-4)=0
⇒Either x+5=0 or x-4=0
x=-5 or x=4
But x=-5 is not a natural number,
So taking x=4,
First number=4
and Second number=x+1
Second number=5
问题8:两个连续奇数正整数的平方和为394。求它们。
解决方案:
Let the first positive odd number is = x
so Second positive odd number is = x+2
Now according to the condition-
⇒(First number)2+(Second number)2 = 394
⇒ x2 + (x+2)2 = 394
⇒ x2 + x2 + 2*x*2 + 4 = 394 [because (a+b)2 = a2 + 2*a*b + b2]
⇒ 2x2 + 4x + 4 = 394
⇒ 2x2 + 4x – 390 = 0
Dividing by 2 –
⇒ x2 + 2x – 195 = 0
⇒ x2 + (15-13)x – 195 = 0
⇒ x2 + 15x – 13x -195 = 0
⇒ x(x+15) – 13(x+15) = 0
⇒ (x+15)(x-13) = 0
Either x+15=0 or x-13=0
x=-15 or x=13
but x=-15 is not a positive odd number
So on taking x=13
First positive odd number=13
and Second number = x+2
Second number = 15
问题 9:两个数之和是 8,它们的倒数之和也是 8 的 15 倍。求这些数。
解决方案:
Let first number is = x
and sum of numbers is given which is = 8
So second number is = (8-x)
Now,
Reciprocal of first number is = 1/x
And reciprocal of second number is = 1/(8-x)
Given condition is-
⇒15[(1/x)+(1/(8-x))]=8
⇒15[((8-x)+x)/(x(8-x))]=8 [by taking LCM]
⇒15*[8-x+x]=8[x(8-x)]
⇒15*8 = 8[8x-x2]
Dividing by 8–
⇒ 15 = 8x-x2
⇒x2-8x+15=0
⇒x2-(5+3)x+15=0
⇒x2-5x-3x+15=0
⇒x(x-5)-3(x-5)=0
⇒(x-5)(x-3)=0
Either x-5=0 or x-3=0
x=5 or x=3
So numbers are 3,5.
问题 10:一个数与其正平方根之和为 6/25。找到号码。
解决方案:
Let the number is = x
So it’s square root is = √x
Now according to condition-
⇒x+√x=6/25
⇒x-6/25=-√x
Now squaring both sides-
⇒(x-6/25)2=(-√x)2
⇒x2 – 2*x*(6/25) + (6/25)2 = x [because (a+b)2 = a2 + 2*a*b + b2]
⇒x2-(12/25)x+(36/625) = x [because (6/25)2=36/625]
⇒ x2 – (12/25)x -x +(36/625) = 0
⇒x2 -[(12/25)+1]x + (36/625) = 0
⇒ x2 -(37/25)x + (36/625) = 0
now for making factor-
⇒ x2 -[(36/25)+(1/25)]x + (36/625) = 0
⇒x2 – (36/25)x – (1/25)x + (36/625) = 0
⇒x[x-(36/25)] – (1/25)[x-(36/25)]=0
⇒[x-(1/25)][x-(36/25)]=0
Either x-(1/25) = 0 or x-(36/25)=0
x=(1/25) or x=(36/25)
but when x=(36/25)
then (36/25)+√(36/25)=(36/25)+(6/5)
= (36+30)/25
= 66/25
So when x=36/25 , it doesn’t fulfill given condition.
So required number = 1/25.