if a^2+b^2=41 and ab =4, find the value of a-b
Answers
Answer :-
a - b = √33
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 = 33
⇒ a - b = √33
Verification :-
(a - b)²= a² + b² - 2ab
⇒ (√33)² = (41) - 2(4)
⇒ 33 = 41 - 8
⇒ 33 = 33
Therefore a - b = √33
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
Answer:
Step-by-step explanation:
SOLUTION:
(a-b)^2 = a^2 + b^2 + 2ab
REMEMBER:- a^2 - b^2 = 41, ab = 4
(a - b)^2 = 41 - 2 * 4
(a - b)^2 = 41 - 8 = 33
a-b = under root 33
a-b = approximately 6.
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