Class 10 RD Sharma Solutions- Chapter 8 Quadratic Equations – Exercise 8.7 | Set 1

• Last Updated : 03 Jan, 2021

Question 1: Find two consecutive numbers whose squares have the sum 85.

Solution:

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.

Question 2: Divide 29 into two parts so that the sum of the squares of the parts is 425.

Solution:

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.

Question 3: Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm2. Find the sides of the squares.

Solution:

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

Question 4: The sum of two numbers is 48 and their product is 432. Find the numbers.

Solution:

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.

Question 5: If an integer is added to its square, the sum is 90. Find the integer with the help of a quadratic equation.

Solution:

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.

Question 6: Find the whole number which when decreased by 20 is equal to 69 times the reciprocal of the number.

Solution:

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

Question 7: Find two consecutive natural numbers whose product is 20.

Solution:

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

Question 8: The sum of the squares of two consecutive odd positive integers is 394. Find them.

Solution:

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

Question 9:The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.

Solution:

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.

Question 10:The sum of a number and its positive square root is 6/25. Find the number.

Solution:

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.

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