Ex-8.7 Quadratic Equations (Part - 1), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

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

Solution:

Let the two consecutive two natural numbers be (x) and ( x + 1)  respectively.

Given,

That the sum of their squares is 85.

Then, by hypothesis, we get,

= x²  + (x + 1)²  = 85

= x² + x² + 2x + 1 = 85

= 2x² + 2x + 1 - 85 = 0

= 2x² + 2x + - 84 = 0

= 2(x² + x + - 42) = 0

Now applying factorization method, we get,

= x² + 7x - 6x - 42 = 0

= x(x + 7) - 6(x + 7) = 0

= (x - 6)(x + 7) = 0

Either,

x - 6 = 0  therefore , x = 6

x + 7 = 0 therefore x = - 7

Hence the consecutive numbers whose sum of squares is 85 are 6 and - 7 respectively.

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

Solution:

Let the two parts be (x) and (29 - x) respectively.

According to the question, the sum of the two parts is 425.

Then by hypothesis,

= x²  + (29 - x)²  = 425

= x² + x² + 841 + - 58x = 425

= 2x² - 58x + 841 - 425 = 0

= 2x² - 58x + 416 = 0

= x² - 29x + 208 = 0

Now, applying the factorization method

= x²  - 13x - 16x + 208 = 0

= x(x - 13) - 16(x - 13) = 0

= (x - 13)(x - 16) = 0

Either x - 13 = 0 therefore x = 13

Or, x - 16 = 0 therefore x = 16

The two parts whose sum of the squares is 425 are 13 and 16 respectively.

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:

Given,

The sum of the sides of the squares are = x cm and (x + 4) cm respectively.

The sum of the areas = 656 cm²

We know that,

Area of the square = side * side

Area of the square = x(x + 4) cm²

Given that the sum of the areas is 656 cm²

Hence by hypothesis,

= x(x + 4) + x(x + 4) = 656

= 2x(x + 4) = 656

= x²  + 4x = 328

Now by applying factorization method,

= x²  + 20x - 16x - 328 = 0

= x(x + 20) - 16(x + 20) = 0

= (x + 20)(x16) = 0

Either x + 20 = 0 therefore x = - 20

Or, x - 16 = 0 therefore x = 16

No negative value is considered as the value of the side of the square can never be negative.

Therefore, the side of the square is 16.

Therefore, x + 4 = 16 + 4 = 20 cm

Hence, the side of the square is 20cm.

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

Solution:

Given the sum of two numbers is 48.

Let the two numbers be x and 48 - x also the sum of their product is 432.

According to the question

= x(48 - x) = 432

= 48x - x² = 432

= x² - 48x + 432 = 0

= x² - 36x - 12x + 432 = 0

= x(x - 36) - 12(x - 36) = 0

= (x - 36)(x - 12) = 0

Either x - 36 = 0 therefore x = 36

Or, x - 12 = 0 therefore x = 12

The two numbers are 12 and 36 respectively.

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

Solution:

Let the integer be x

Given that if an integer is added to its square , the sum is 90

= x + x²  = 90

= x²  + x - 90 = 0

= x²  + 10x - 9x - 90 = 0

= x(x + 10) - 9(x + 10) = 0

= (x + 10)(x - 9) = 0

Either x + 10 = 0

Therefore x = - 10

Or, x - 9 = 0

Therefore x = 9

The values of the integer are 9 and - 10 respectively.

Question 6: Find the whole numbers which when decreased by 20 is equal to69 times the reciprocal of the numbers.

Solution:

Let the whole number be x cm

As it is decreased by 20 = (x - 20) =  69/x

x - 20 = 69/x

x(x - 20) = 69

x²  - 20x - 69 = 0

Now by applying factorization method ,

x² - 23x + 3x - 69 = 0

x(x - 23) + 3(x - 23) = 0

(x - 23)(x + 3) = 0

Either, x = 23

Or, x = - 3

As the whole numbers are always positive x = - 3 is not considered.

The whole number is 23.

Question 7: Find the consecutive natural numbers whose product is 20

Solution:

Let the two consecutive natural number be x and x + 1 respectively.

Given that the product of natural numbers is 20

= x(x + 1) = 20

= x² + x - 20 = 0

= x² + 5x - 4x - 20 = 0

= x(x + 5) - 4(x + 5) = 0

= (x + 5)(x - 4) = 0

Either x + 5 = 0

Therefore x = - 5

Considering the positive value of x.

Or, x - 4 = 0

Therefore x = 4

The two consecutive natural numbers are 4 and 5 respectively.

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

Solution:

Let the consecutive odd positive integer are 2x - 1 and 2x + 1 respectively.

Given, that the sum of the squares is 394.

According to the question,

(2x - 1)² + (2x + 1)²  = 394

4x²  + 1 - 4x + 4x² + 1 + 4x = 394

Now cancelling out the equal and opposite terms  ,

8x² + 2 = 394

8x²  = 392

X²  = 49

X = 7 and - 7

Since the value of the edge of the square cannot be negative so considering only the positive value.

That is 7

Now, 2x - 1 = 14 - 1 = 13

2x + 1 = 14 + 1 = 15

The consecutive odd positive numbers are 13 and 15 respectively.

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

Solution:

Let the numbers be x and 8 - x respectively.

Given that the sum of the numbers is 8 and 15 times the sum of their reciprocals.

According to the question,

= 120 = 8(8x - x²)

= 120 = 64x - 8x²

= 8x² - 64x + 120 = 0

= 8(x² - 8x + 15) = 0

= x² - 8x + 15 = 0

= x² - 5x - 3x + 15 = 0

= x(x - 5) - 3(x - 5) = 0

= (x - 5)(x - 3) = 0

Either x - 5 = 0 therefore x = 5

Or, x - 3 = 0 therefore x = 3

The two numbers are 5 and 3 respectively.

Question 10: The sum of a number and its positive square root is 625. Find the numbers.

Solution:

Let the number be x

By the hypothesis, we have

x + √ x = 6/25

Let us assume that x = y² , we get

y + y² = 6/25

= 25y² + 25y - 6 = 0

The value of y can be determined by:

Where a = 25, b = 25 , c = - 6

The number x is 1/25

Question 11: There are three consecutive integers such that the square of the first increased by the product of the other two integers gives 154.  What are the integers?

Solution:

Let the three consecutive numbers be x, x + 1, x + 2 respectively.

X² + (x + 1)(x + 2) = 154

= x² + x² + 3x + 2 = 154

= 3x² + 3x - 152 = 0

The value of x can be obtained by the formula

Here a = 3 , b = 3 , c = 152

x = 8andx = −19/2

Considering the value of x

If x = 8

x + 1 = 9

x + 2 = 10

The three consecutive numbers are 8 , 9 , 10 respectively.

Question 12: The product of two successive integral multiples of 5 is 300. Determine the multiples.

Solution:

Given that the product of two successive integral multiples of 5 is 300

Let the integers be 5x and 5(x + 1)

According to the question,

5x[5(x + 1)] = 300

= 25x(x + 1) = 300

= x² + x = 12

= x² + x - 12 = 0

= x² + 4x - 3x - 12 = 0

= x(x + 4) - 3(x + 4) = 0

= (x + 4)(x - 3) = 0

Either x + 4 = 0

Therefore x = - 4

Or, x - 3 = 0

Therefore x = 3

x = - 4

5x = - 20

5(x + 1) = - 15

x = 3

5x = 15

5(x + 1) = 20

The two successive integral multiples are 15,20 and - 15 and - 20 respectively.

Question 13: The sum of the squares of two numbers is 233 and one of the numbers is 3 less than the other number. Find the numbers.

Solution:

Let the number is x

Then the other number is 2x - 3

According to the question:

x² + (2x - 3)²  = 233

= x² + 4x² + 9 - 12x = 233

= 5x² - 12x - 224 = 0

The value of x can be obtained by x =

Here a = 5 , b = - 12 , c = - 224

x =  x =

x = 8 and x = −28/5

Considering the value of x = 8

2x - 3 = 15

The two numbers are 8 and 15 respectively.

Question 14: The difference of two number is 4 . If the difference of the reciprocal is 4/21 . find the numbers.

Solution:

Lethe two numbers be x and x - 4 respectively.

Given, that the difference of two numbers is 4 .

By the given hypothesis we have,

= 84 = 4x(x - 4)

= x² - 4x - 21 = 0

Applying factorization theorem,

= x²  - 7x + 3x - 21 = 0

= (x - 7)(x + 3) = 0

Either x - 7 = 0 therefore x = 7

Or, x + 3 = 0 therefore x = - 3

Hence the required numbers are - 3 and 7 respectively.

Question 15: Let us find two natural numbers which differ by 3 and whose squares have the sum 117.

Solution:

Let the numbers be x and x - 3

According to the question

x² + (x - 3)² = 117

= x² + x² + 9 - 6x - 117 = 0

= 2x² - 6x - 108 = 0

= x² - 3x - 54 = 0

= x² - 9x + 6x - 54 = 0

= x(x - 9) + 6(x - 9) = 0

= (x - 9)(x + 6) = 0

Either x - 9 = 0 therefore x = 9

Or ,x + 6 = 0 therefore x = - 6

Considering the positive value of x that is 9

x = 9

x - 3 = 6

The two numbers are 6 and 9 respectively.

Question 16: The sum of the squares of these consecutive natural numbers is 149. Find the numbers.

Solution:

Let the numbers be x , x + 1, and x + 2 respectively.

According to given hypothesis

X² + (x + 1)² + (x + 2)²  = 149

X² + X²  + X²  + 1 + 2x + 4 + 4x = 149

3x²  + 6x - 144 = 0

X² + 2x - 48 = 0

Now applying factorization method,

X²  + 8x - 6x - 48 = 0

X(x + 8) - 6(x + 8) = 0

(x + 8)(x - 6) = 0

Either x + 8 = 0 therefore x = - 8

Or, x - 6 = 0 therefore x = 6

Considering only the positive value of x that is 6 and discarding the negative value.

x = 6

x + 1 = 7

x + 2 = 8The three consecutive numbers are 6 , 7 , and 8 respectively.