if x=√2+1 and y=√2-1 find the value of x²+y²+xy
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Hi there!
Given :-
♦ x = √2 + 1
♦ y = √2 - 1
ATQ,
→ x² = (√2 + 1)²
= 2 + 1 + 2(√2)
= 3 + 2√2. ----(i)
→ y² = (√2 - 1)²
= 2 + 1 - 2(√2)
= 3 - 2√2. ----(ii)
→ xy = (√2 + 1) × (√2 - 1)
= (√2)² - (1)²
= 2 - 1
= 1. ----(iii)
Adding eqn. (i), (ii) n' (iii) :-
x² + y² + xy = (3 + 2√2) + (3 - 2√2) + 1
= 3 + 2√2 + 3 - 2√2 + 1
= 3 + 3 + 1
= 7
Hence, The required answer is :-
x² + y² + xy = 7
Hope it helps! :)
Given :-
♦ x = √2 + 1
♦ y = √2 - 1
ATQ,
→ x² = (√2 + 1)²
= 2 + 1 + 2(√2)
= 3 + 2√2. ----(i)
→ y² = (√2 - 1)²
= 2 + 1 - 2(√2)
= 3 - 2√2. ----(ii)
→ xy = (√2 + 1) × (√2 - 1)
= (√2)² - (1)²
= 2 - 1
= 1. ----(iii)
Adding eqn. (i), (ii) n' (iii) :-
x² + y² + xy = (3 + 2√2) + (3 - 2√2) + 1
= 3 + 2√2 + 3 - 2√2 + 1
= 3 + 3 + 1
= 7
Hence, The required answer is :-
x² + y² + xy = 7
Hope it helps! :)
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