If p=√3-√2/√3+√2 q=√3+√2/√3-√2 then find value …
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Posted by Abhishek Gupta Rahul Gupta 6 years, 9 months ago
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Gaurav Seth 6 years, 9 months ago
P = √3-√2 ÷ √3+√2
→ √3-√2/√3+√2 × √3-√2/√3-√2
→ (√3-√2)(√3-√2)/(√3+√2)(√3-√2)
→ [√3²+√2²-2(√3)(√2)]/(√3²-√2²)
→ (3+2-2√6)/(3-2)
→ (5-2√6)
q = √3+√2÷√3-√2
→ √3+√2/√3-√2×√3+√2/√3+√2
→ (√3+√2)(√3+√2)/(√3-√2)(√3+√2)
→ [√3²+√2²+2(√3)(√2)]/(√3²-√2²)
→ (3+2+2√6)/(3-2)
→ 5+2√6
⇨p²+q²
⇨ (5-2√6)²+(5+2√6)²
⇨[5²+(2√6)²-2(5)(2√6)] + [5²+(2√6)²+2(5)(2√6)]
⇨(25+4(6)-20√6)+(25+4(6)+20√6)
⇨25+24-20√6+25+24+20√6
⇨49+49
⇨p²+q² = 98
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