Question

if a^2+b^2=41 and ab=4, find the value of a+b

Answer 1

Hey friend ☺ here is the answer __________________________

(a + b)² = a² + b² + 2ab
Now
a² + b² = 41 and ab = 4
(a + b)² = 41 + 2×4 = 41 + 8 = 49

(a + b)² = 49
a + b = √49
a + b = 7

Hope this helps you..

Answer 2

Answer :-

a + b = 7

Solution :-

We know that

(a + b)² = a² + b² + 2ab

Here ab = 4

a² + b² = 41

By substituting the values

⇒ (a + b)² = (41) + 2(4)

⇒ (a + b)² = 41+ 8

⇒ (a + b)² = 49

⇒ a + b = √49

⇒ a + b = 7

Verification :-

(a + b)²= a² + b² + 2ab

⇒ (7)² = (41) + 2(4)

⇒ 49 = 41 - 8

⇒ 49 = 49

Therefore a + b = 7

Identity used :-

• (a + b)² = a² + b² + 2ab

Extra Information :-

Some Important Identities :-

• (x + y)² = x² + y² + 2xy

• (x - y)² = x² + y² - 2xy

• (x + y)(x - y) = x² - y²

• (x + a)(x + b) = x² + (a + b)x + ab

• (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2xz