Question

The difference between 2 positive no.s is 5 and the sum of their squares is 71. Find the product of these no.s​

Answer 1

Hey
Try using the algebraic identity
(a-b)^2 = a^2 + b^2 - 2ab
Replace a with x and b with y as I have done
I hope this helps
(:

Answer 2

Given :-

• Difference between 2 numbers is 5
• The sum of their squares is 71

Aim :-

• To find the product of these numbers.

Answer :-

• let the numbers be x and y respectively.

Method 1 :-

By using the identity (a-b)² = a²+ b² - 2ab

• a = x
• b = y

Substituting,

⇒ (x-y)² = (x)² + (y)² - 2(x)(y)

Here, it's given, the difference is 5 and the sum of their squares is 71.

• x - y = 5
• x² + y² = 71

Substituting these values,

⇒ (5)² = 71 - 2xy

⇒ 25 = 71 - 2xy

Transposing 71,

⇒ 25 - 71 = -2xy

⇒ -46 = -2xy

Transposing (-2),

$\implies \sf \dfrac{-46}{-2} = xy$

$\implies \sf \dfrac{\not- 46}{\not- 2} = xy$

Cancelling the negative sign as (-) ÷ (-) = (+)

$\implies \sf \dfrac{46}{2} = xy$

Reducing to the lowest terms,

⇒ 23 = xy

Hence the product of the two numbers is 23.

Identities :-

• (a+b)² = a² + 2ab + b²
• (a-b)² = a² - 2ab + b²
• (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (x+a)(x+b) = x² + x(a+b) + ab
• a²-b² = (a+b)(a-b)
• (a+b)³ = a³ + 3a²b + 3ab² + b³
• (a-b)³ = a³ - 3a²b + 3ab² - b³
• a³+b³ = (a+b)(a² - ab + b²)
• a³-b³ = (a-b)(a² + ab + b²)
• a³+b³+c³ - 3abc = (a+b+c)(a² + b² + c² - ab - bc - ca)