If a= √3-√2/√3+√2 and b= √3+√2/√3-√2. Find the value od a²+b²-5ab?
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a = (√3-√2)/(√3+√2)
= (√3-√2)/(√3+√2)× (√3-√2)/(√3-√2) [ rationalizing it]
=(√3 -√2)²/1 = (√3 -√2)² = 5- 2√6
b= (√3+√2)/(√3-√2)×(√3+√2)/(√3+√2) [ rationalizing it]
= (√3+√2)²/1 =(√3+√2)² = 5+2√6
now, a²+b² - 5ab
= (5-2√6)² + (5+2√6)² - 5(5-2√6)(5+2√6)
=25 +24 -20√6 +25 +24 + 20√6 - 5(25-24)
=98 -5 = 93
= (√3-√2)/(√3+√2)× (√3-√2)/(√3-√2) [ rationalizing it]
=(√3 -√2)²/1 = (√3 -√2)² = 5- 2√6
b= (√3+√2)/(√3-√2)×(√3+√2)/(√3+√2) [ rationalizing it]
= (√3+√2)²/1 =(√3+√2)² = 5+2√6
now, a²+b² - 5ab
= (5-2√6)² + (5+2√6)² - 5(5-2√6)(5+2√6)
=25 +24 -20√6 +25 +24 + 20√6 - 5(25-24)
=98 -5 = 93
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