Solving Problems (Based on Quadratic Equations) — Exercise 6(A)

The sum of two natural numbers is 5 and the sum of their reciprocals is $\dfrac{5}{6}$, the numbers are :

1. 2 and 5

2. 4 and 2

3. 2 and 3

4. 3 and 4

Answer

Let two natural numbers be x and y.

Given,

Sum = 5

⇒ x + y = 5

⇒ x = 5 - y ........(1)

Sum of reciprocals = $\dfrac{5}{6}$

⇒ $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{5}{6}$ ...........(2)

Substituting value of x from equation (1) in (2) :

$\Rightarrow \dfrac{1}{5 - y} + \dfrac{1}{y} = \dfrac{5}{6} \\[1em] \Rightarrow \dfrac{y + 5 - y}{y(5 - y)} = \dfrac{5}{6} \\[1em] \Rightarrow \dfrac{5}{5y - y^2} = \dfrac{5}{6} \\[1em] \Rightarrow 5y - y^2 = \dfrac{6 \times 5}{5} \\[1em] \Rightarrow 5y - y^2 = 6 \\[1em] \Rightarrow y^2 - 5y + 6 = 0 \\[1em] \Rightarrow y^2 - 2y - 3y + 6 = 0 \\[1em] \Rightarrow y(y - 2) - 3(y - 2) = 0 \\[1em] \Rightarrow (y - 3)(y - 2) = 0 \\[1em] \Rightarrow y - 3 = 0 \text{ or } y - 2 = 0 \\[1em] \Rightarrow y = 3 \text{ or } y = 2.$

If y = 3,

x = 5 - y = 5 - 3 = 2.

If y = 2,

x = 5 - y = 5 - 2 = 3.

∴ Numbers are 2 and 3.

Hence, Option 3 is the correct option.

The product of two consecutive even whole numbers is 24, the numbers are :

1. -8 and -3

2. 3 and 8

3. -4 and -6

4. 4 and 6

Answer

Let two consecutive even whole numbers be x and x + 2.

Given,

Product of two consecutive even whole numbers = 24.

∴ x(x + 2) = 24

⇒ x² + 2x = 24

⇒ x² + 2x - 24 = 0

⇒ x² + 6x - 4x - 24 = 0

⇒ x(x + 6) - 4(x + 6) = 0

⇒ (x - 4)(x + 6) = 0

⇒ x - 4 = 0 or x + 6 = 0

⇒ x = 4 or x = -6.

Since, x is a whole number so x cannot be equal to -6.

x = 4 and x + 2 = 6.

Hence, Option 4 is the correct option.

The sum of the squares of two consecutive integers is 41. The integers are :

1. 4 and -5 or -4 and 5

2. 4 and 5 or -4 and -5

3. 3 and 4 or -4 and -3

4. 6 and 3 or -6 and -3

Answer

Let two consecutive integers be x and x + 1.

Given,

Sum of squares of two consecutive integers = 41.

⇒ x² + (x + 1)² = 41

⇒ x² + x² + 1 + 2x = 41

⇒ 2x² + 2x = 41 - 1

⇒ 2x² + 2x = 40

⇒ 2(x² + x) = 40

⇒ x² + x = $\dfrac{40}{2}$

⇒ x² + x = 20

⇒ x² + x - 20 = 0

⇒ x² + 5x - 4x - 20 = 0

⇒ x(x + 5) - 4(x + 5) = 0

⇒ (x - 4)(x + 5) = 0

⇒ x - 4 = 0 or x + 5 = 0

⇒ x = 4 or x = -5

If x = 4,

⇒ x + 1 = 4 + 1 = 5.

If x = -5,

⇒ x + 1 = -5 + 1 = -4.

Numbers = 4 and 5 or -4 and -5.

Hence, Option 2 is the correct option.

The sum of a number and its reciprocal is 5.2. The number is

1. 5 or $\dfrac{1}{5}$

2. 2 or $\dfrac{1}{2}$

3. 4 or $\dfrac{1}{4}$

4. -2 or $-\dfrac{1}{2}$

Answer

Let number be x.

Given,

Sum of a number and its reciprocal is 5.2

$\Rightarrow x + \dfrac{1}{x} = 5.2 \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{52}{10} \\[1em] \Rightarrow 10(x^2 + 1) = 52x \\[1em] \Rightarrow 10x^2 + 10 = 52x \\[1em] \Rightarrow 10x^2 - 52x + 10 = 0 \\[1em] \Rightarrow 10x^2 - 50x - 2x + 10 = 0 \\[1em] \Rightarrow 10x(x - 5) - 2(x - 5) = 0 \\[1em] \Rightarrow (10x - 2)(x - 5) = 0 \\[1em] \Rightarrow 10x - 2 = 0 \text{ or } x - 5 = 0 \\[1em] \Rightarrow 10x = 2 \text{ or } x = 5 \\[1em] \Rightarrow x = \dfrac{2}{10} \text{ or } x = 5 \\[1em] \Rightarrow x = \dfrac{1}{5} \text{ or } x = 5.$

Hence, Option 1 is the correct option.

Two integers differ by 2 and sum of their squares is 52. The integers are :

1. 4 and 6

2. 4 or 6

3. -4 or 6

4. -4 and -6 or 6 and 4

Answer

Let two integers be x and x - 2.

Give,

Sum of squares = 52

⇒ x² + (x - 2)² = 52

⇒ x² + x² + 4 - 4x = 52

⇒ 2x² - 4x + 4 - 52 = 0

⇒ 2x² - 4x - 48 = 0

⇒ 2(x² - 2x - 24) = 0

⇒ x² - 2x - 24 = 0

⇒ x² - 6x + 4x - 24 = 0

⇒ x(x - 6) + 4(x - 6) = 0

⇒ (x + 4)(x - 6) = 0

⇒ x + 4 = 0 or x - 6 = 0

⇒ x = -4 or x = 6.

If x = -4,

x - 2 = -4 - 2 = -6.

If x = 6,

x - 2 = 6 - 2 = 4.

Numbers = -4 and -6 or 6 and 4.

Hence, Option 4 is the correct option.

Divide 15 into two parts such that sum of reciprocals is $\dfrac{3}{10}.$

Answer

Let numbers be x and (15 - x).

According to the question,

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{15 - x} = \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{15 - x + x}{x(15 - x)} = \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{15}{x(15 - x)} = \dfrac{3}{10} \\[1em] \Rightarrow 150 = 3x(15 - x) \\[1em] \Rightarrow 150 = 45x - 3x^2 \\[1em] \Rightarrow 3x^2 - 45x + 150 = 0 \\[1em] \Rightarrow 3(x^2 - 15x + 50) = 0 \\[1em] \Rightarrow x^2 - 10x - 5x + 50 = 0 \\[1em] \Rightarrow x(x - 10) - 5(x - 10) = 0 \\[1em] \Rightarrow (x - 5)(x - 10) = 0 \\[1em] \Rightarrow x - 5 = 0 \text{ or } x - 10 = 0 \\[1em] \Rightarrow x = 5 \text{ or } x = 10.$

Hence, numbers = 5, 10.

The sum of the squares of two positive integers is 208. If the square of the larger number is 18 times the smaller number, find the numbers.

Answer

Let larger number be x and smaller number be y,

According to first part,

x² + y² = 208 ........(i)

x² = 18y ........(ii)

Substituting value of x² from (ii) in (i) we get,

⇒ 18y + y² = 208

⇒ y² + 18y - 208 = 0

⇒ y² + 26y - 8y - 208 = 0

⇒ y(y + 26) - 8(y + 26) = 0

⇒ (y - 8)(y + 26) = 0

⇒ y - 8 = 0 or y + 26 = 0

⇒ y = 8 or y = -26.

Since, numbers are positive integers,

∴ y ≠ -26.

Substituting value of y = 8 in (ii),

⇒ x² = 18(8) = 144

⇒ x = $\sqrt{144} = \pm 12$.

Since, numbers are positive integers,

∴ x ≠ -12.

Hence, numbers are 12 and 8.

Find two consecutive positive odd numbers, the sum of whose squares is 74.

Answer

Let two consecutive positive odd numbers be x and (x + 2).

⇒ (x)² + (x + 2)² = 74

⇒ x² + x² + 4 + 4x = 74

⇒ 2x² + 4x + 4 - 74 = 0

⇒ 2x² + 4x - 70 = 0

⇒ 2(x² + 2x - 35) = 0

⇒ x² + 2x - 35 = 0

⇒ x² + 7x - 5x - 35 = 0

⇒ x(x + 7) - 5(x + 7) = 0

⇒ (x + 7)(x - 5) = 0

⇒ (x + 7) = 0 or (x - 5) = 0

⇒ x = -7 or x = 5.

Since, numbers are positive and even,

∴ x ≠ -7.

∴ x = 5 and (x + 2) = 7.

Hence, numbers are 5, 7.

Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.

Answer

Let two parts be x and (20 - x).

According to question,

3x² - (20 - x) = 10

3x² + x - 20 = 10

3x² + x - 30 = 0

3x² + 10x - 9x - 30 = 0

x(3x + 10) - 3(3x + 10) = 0

(x - 3)(3x + 10) = 0

x - 3 = 0 or (3x + 10) = 0

x = 3 or x = $-\dfrac{10}{3}$.

20 - x = 20 - 3 = 17.

Hence, numbers are 3 and 17.

Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60.

Assume the middle number to be x and form a quadratic equation satisfying the above statement. Hence; find the three numbers.

Answer

Let three consecutive numbers be (x - 1), x and (x + 1).

According to question,

⇒ x² - [(x + 1)² - (x - 1)²] = 60

⇒ x² - [x² + 1 + 2x - (x² + 1 - 2x)] = 60

⇒ x² - [x² + 1 + 2x - x² - 1 + 2x] = 60

⇒ x² - 4x = 60

⇒ x² - 4x - 60 = 0

⇒ x² - 10x + 6x - 60 = 0

⇒ x(x - 10) + 6(x + 10) = 0

⇒ (x + 6)(x - 10) = 0

⇒ x + 6 = 0 or x - 10 = 0

⇒ x = -6 or x = 10.

Since, numbers are natural numbers,

∴ x = 10.

∴ x - 1 = 9 and x + 1 = 11.

Hence, the numbers are 9, 10, 11 and quadratic equation = x² - 4x - 60 = 0.

Out of three consecutive positive integers, the middle number is p. If three times the square of the largest is greater than the sum of the squares of the other two numbers by 67; calculate the value of p.

Answer

Let the numbers be (p - 1), p and (p + 1).

According to question,

⇒ 3(p + 1)² = (p - 1)² + p² + 67

⇒ 3(p² + 1 + 2p) = (p² + 1 - 2p) + p² + 67

⇒ 3p² + 3 + 6p = 2p² - 2p + 68

⇒ 3p² - 2p² + 6p + 2p + 3 - 68 = 0

⇒ p² + 8p - 65 = 0

⇒ p² + 13p - 5p - 65 = 0

⇒ p(p + 13) - 5(p + 13) = 0

⇒ (p - 5)(p + 13) = 0

⇒ p - 5 = 0 or p + 13 = 0

⇒ p = 5 or p = -13

Since, numbers are natural numbers,

∴ p ≠ -13.

∴ p - 1 = 4 and p + 1 = 6.

Hence, p = 5.

A positive number is divided into two parts such that the sum of the squares of the two parts is 20. The square of the larger part is 8 times the smaller part. Taking x as the smaller part of the two parts, find the number.

Answer

Let smaller part be x.

According to second part of question,

Larger part = $\sqrt{8x}$

According to first part of question,

x² + $(\sqrt{8x})^2$ = 20

x² + 8x = 20

x² + 8x - 20 = 0

x² + 10x - 2x - 20 = 0

x(x + 10) - 2(x + 10) = 0

(x - 2)(x + 10) = 0

x - 2 = 0 or x + 10 = 0

x = 2 or x = -10.

Since number is positive,

∴ x ≠ -10

Larger part = $\sqrt{8x} = \sqrt{8(2)} = \sqrt{16} = 4.$

Number = Smaller part + Larger part = 2 + 4 = 6.

Hence, number = 6.

The difference between two natural numbers is 5 and the difference of their reciprocals is $\dfrac{1}{10}$. Find the numbers.

Answer

It is given that the difference between two natural numbers = 5.

Let one number be x. So, the other number = x + 5

And, the difference of their reciprocals is $\dfrac{1}{10}$.

$\Rightarrow \dfrac{1}{x} - \dfrac{1}{x + 5} = \dfrac{1}{10}\\[1em] \Rightarrow \dfrac{x + 5}{x(x + 5)} - \dfrac{x}{x(x + 5)} = \dfrac{1}{10}\\[1em] \Rightarrow \dfrac{x + 5 - x}{x(x + 5)} = \dfrac{1}{10}\\[1em] \Rightarrow \dfrac{5}{x(x + 5)} = \dfrac{1}{10}\\[1em] \Rightarrow 5 \times 10 = x(x + 5)\\[1em] \Rightarrow 50 = x^2 + 5x\\[1em] \Rightarrow x^2 + 5x - 50 = 0\\[1em] \Rightarrow x^2 + 10x - 5x - 50 = 0\\[1em] \Rightarrow x(x + 10) - 5(x + 10) = 0\\[1em] \Rightarrow (x + 10)(x - 5) = 0\\[1em] \Rightarrow (x + 10) = 0 \text{ or }(x - 5) = 0\\[1em] \Rightarrow x = -10 \text{ or }x = 5$

It is given that numbers are natural numbers. So, number cannot be -10.

∴ x = 5

Other number = 5 + 5 = 10

Hence, the two numbers are 5 and 10.

The sum of the squares of two consecutive odd natural numbers is 394. Find the numbers.

Answer

Let the two consecutive odd natural numbers be x and x + 2.

It is given that the sum of the squares of the numbers = 394.

⇒ x² + (x + 2)² = 394

⇒ x² + x² + 2² + 4x = 394

⇒ 2x² + 4 + 4x - 394 = 0

⇒ 2x² + 4x - 390 = 0

⇒ x² + 2x - 195 = 0

⇒ x² + 15x - 13x - 195 = 0

⇒ x(x + 15) - 13(x + 15) = 0

⇒ (x + 15)(x - 13) = 0

⇒ (x + 15) = 0 or (x - 13) = 0

⇒ x = -15 or x = 13

As the number is natural number, x cannot be equal to -15. So, x = 13.

Other odd natural number = 13 + 2 = 15

Hence, the natural numbers are 13 and 15.

The sum of the squares of two consecutive multiplies of 7 is 637.

Taking the bigger number x a positive number, find the smaller of these two angles.

Answer

Let the multiplies of 7 be 7x and 7(x + 1).

It is given that the sum of the squares of two consecutive multiplies of 7 is 637.

⇒ (7x)² + [7(x + 1)]² = 637

⇒ 49x² + [7x + 7]² = 637

⇒ 49x² + 49x² + 7² + 98x - 637 = 0

⇒ 98x² + 49 + 98x - 637 = 0

⇒ 98x² + 98x - 588 = 0

⇒ x² + x - 6 = 0

⇒ x² + 3x - 2x - 6 = 0

⇒ x(x + 3) - 2(x + 3) = 0

⇒ (x + 3)(x - 2) = 0

⇒ (x + 3) = 0 or (x - 2) = 0

⇒ x = -3 or x = 2

It is given that x is bigger positive number.

So, the multiplies of 7 is 7 x 2 = 14 and 7(2 + 1) = 7 x 3 = 21

Hence, the smaller angle is 14.

Two positive numbers difference by 5. Three times the square of the larger number exceeds twice the square of smaller number by 334, find the larger of these two numbers.

Answer

It is given that difference between two positive numbers = 5.

Let the one number be x and other number be 5 + x.

Three times the square of the larger number exceeds twice the square of smaller number by 334.

⇒ 3 $\times$ (5 + x)² - 2 $\times$ x² = 334

⇒ 3 $\times$ (25 + x² + 10x) - 2 $\times$ x² = 334

⇒ 75 + 3x² + 30x - 2x² - 334 = 0

⇒ x² + 30x - 259 = 0

⇒ x² + 37x - 7x - 259 = 0

⇒ x(x + 37) - 7(x + 37) = 0

⇒ (x + 37)(x - 7) = 0

⇒ (x + 37) = 0 or (x - 7) = 0

⇒ x = -37 or x = 7

Larger number = 5 + x = 5 + 7 = 12

Hence, the larger number = 12.

Find three consecutive positive odd integers, the sum of whose squares is 155.

Answer

Let the three consecutive positive odd integers be : x, x + 2, x + 4

Given,

The sum of squares of three consecutive positive odd integers is 155.

⇒ x² + (x + 2)² + (x + 4)² = 155

⇒ x² + x² + 4x + 4 + x² + 8x + 16 = 155

⇒ 3x² + 12x + 20 = 155

⇒ 3x² + 12x + 20 - 155 = 0

⇒ 3x² + 12x - 135 = 0

⇒ 3(x² + 4x - 45) = 0

⇒ x² + 4x - 45 = 0

⇒ x² + 9x - 5x - 45 = 0

⇒ x(x + 9) - 5(x + 9) = 0

⇒ (x - 5)(x + 9) = 0

⇒ (x - 5) = 0 or (x + 9) = 0

⇒ x = 5 or x = -9.

Since the integers are positive, x = 5

The three consecutive positive odd integers,

x = 5,

x + 2 = 5 + 2 = 7,

x + 4 = 5 + 4 = 9.

Hence, three consecutive positive odd integers are 5, 7, 9.

Find three consecutive positive even integers such that three times the square of least integer plus seven times the square of middle integer minus five time the square of the third integer is 328.

Answer

Let the three consecutive positive even integers be: x, x + 2, x + 4.

Given,

Three times the square of least integer plus seven times the square of middle integer minus five time the square of the third integer is 328

⇒ 3x² + 7(x + 2)² - 5(x + 4)² = 328

⇒ 3x² + 7(x² + 4x + 4) - 5(x² + 8x + 16) = 328

⇒ 3x² + 7x² + 28x + 28 - 5x² - 40x - 80 = 328

⇒ 3x² + 7x² - 5x² + 28x - 40x - 80 + 28 = 328

⇒ 5x² - 12x - 52 = 328

⇒ 5x² - 12x - 52 - 328 = 0

⇒ 5x² - 12x - 380 = 0

⇒ 5x² - 50x + 38x - 380 = 0

⇒ 5x(x - 10) + 38(x - 10) = 0

⇒ (5x + 38)(x - 10) = 0

⇒ (5x + 38) = 0 or (x - 10) = 0

⇒ x = $-\dfrac{38}{5}$ or x = 10

Since integers must be positive even numbers, x = 10.

Three consecutive positive even integers are :

x = 10

x + 2 = 10 + 2 = 12

x + 4 = 10 + 4 = 14

Hence, three consecutive positive even integers are 10, 12, 14.