if a=√3-√2/√3+√2 and b=√3+√2/√3-√2 , find the value of a²+b²-5ab
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Answer:
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Answer:
a=√3-√2/√3+√2 & b=√3+√2/√3-√2.
a²+b²-5ab=?.
a=√3-√2/√3+√2.
Rationalising denominator of √3+√2 is √3-√2.
Multiply numerator & denominator by √3-√2.
a=√3-√2/√3+√2×√3-√2/√3-√2.
=(√3-√2)²/(√3)²-(√2)². (Because a+b×a-b=a²-b² where a=√3 & b=√2).
=(√3)²+(√2)²-2(√3)(√2)/3-2. (Because (a-b)²=a²+b²-2ab where a=√3 & b=√2).
=3+2-2(√6)/1. (Because √3×√2=√6).
=5-2√6.
b=√3+√2/√3-√2.
Rationalising denominator of √3-√2 is √3+√2.
Multiply numerator & denominator by √3+√2.
b=√3+√2/√3-√2×√3+√2/√3+√2.
=(√3+√2)²/(√3)²-(√2)². (Because a-b×a+b=a²-b² where a=√3 & b=√2).
=(√3)²+(√2)²+2(√3)(√2)/3-2. (Because (a+b)²=a²+b²+2ab where a=√3 & b=√2).
=3+2+2(√6)/1. (Because √3×√2=√6).
=5+2√6.
a²+b²-5ab=(5-2√6)²+(5+2√6)²-5(5-2√6)(5+2√6). (Because a=5-2√6 & b=5+2√6).
=((5)²+(2√6)²-2(5)(√6))+((5)²+(2√6)²+2(5)(2√6))-5((5)²-(2√6)²). (Because (a-b)²=a²+b²-2ab, (a+b)²=a²+b²+2ab where a=5 & b=√6 & (a-b)(a+b)=a²-b² where a=5 & b=√6).
=(25+24-20√6)+(25+24+20√6)-5(25-24).
=49-20√6+49+20√6-5(1)
=49+49-5.
=98-5.
=93.
a²+b²-5ab=93.
I think this is your answer.