Question

2. Solve the problems given in Example 1.
3. Find two numbers whose sum is 27 and product is 182.
4. Find two consecutive positive integers, sum of whose squares is 365.​

Answer 1

S O L U T I O N :

$\underline{\bf{Given\::}}$

3. Two numbers whose sum is 27 & product is 182 .

$\underline{\bf{Explanation\::}}$

Let the two numbers be x & y respectively.

A/q

→ x + y = 27

→ x = 27 - y................(1)

Now,

→ xy = 182

→ (27 - y)y = 182  [from(1)]

→ 27y - y² = 182

→ y² - 27y + 182 = 0

→ y² - 14y - 13y + 182 = 0

→ y(y - 14) -13(y - 14) = 0

→ (y - 14) (y - 13) = 0

→ y - 14 = 0  Or  y - 13 = 0

→ y = 14  Or  y = 13

∴ Putting the value of both y in equation (1),we get;

→ x = 27 - 14   Or   x = 27 - 13

→ x = 13  Or  x = 14

Thus,

The two numbers will be 13 & 14 .

S O L U T I O N 4:

Let the two consecutive positive numbers be x & x + 1;

A/q

⇒ (x)² + (x+1)² = 365

⇒ x² + (x)² + (1)² + 2 × x × 1 = 365  [∴using (a+b)² ]

⇒ x² + x² + 1 + 2x = 365

⇒ 2x² + 1 + 2x = 365

⇒ 2x² + 2x = 365 - 1

⇒ 2x² + 2x = 364

⇒ 2x² + 2x - 364 = 0

⇒ 2(x² + x - 182) = 0

⇒ x² + x - 182 = 0/2

⇒ x² + x - 182 = 0

⇒ x² + 14x - 13x - 182 = 0

⇒ x(x + 14) -13(x + 14) = 0

⇒ (x + 14) (x - 13) = 0

⇒ x + 14 = 0  Or  x - 13 = 0

⇒ x = -14  Or  x = 13

As we know that negative value isn't acceptable .

So, x = 13

Thus,

• First number = x = 13
• Second number = (x+1) = (13 + 1) = 14