if a =2+√3+√5,b=3+√3-√5.find(a-2)² + (b-3)²
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Hey There !!
Given that, a = 2+√3+√5 and b = 3+√3–√5
(a–2)² + (b–3)²
= [ (2+√3+√5) – 2 ] + [ (3+√3–√5) – 3 ]
= (√3+√5)² + (√3–√5)²
= ( 3 + 5 + 2√3√5 ) + ( 3 + 5 – 2√3√5 )
= ( 8 + 2√3√5 ) + ( 8 – 2√3√5 )
= 8 + 2√3√5 + 8 – 2√3√5
= 8 + 8
= 16
Given that, a = 2+√3+√5 and b = 3+√3–√5
(a–2)² + (b–3)²
= [ (2+√3+√5) – 2 ] + [ (3+√3–√5) – 3 ]
= (√3+√5)² + (√3–√5)²
= ( 3 + 5 + 2√3√5 ) + ( 3 + 5 – 2√3√5 )
= ( 8 + 2√3√5 ) + ( 8 – 2√3√5 )
= 8 + 2√3√5 + 8 – 2√3√5
= 8 + 8
= 16
kbps:
but where do 2 and 3 go
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