a=√3-√2/√3+√2 then a=1/a=?
Answers
Answer:
If a= (√3+√2) / (√3-√2) and b= (√3-√2) / (√3+√2), then how do I find a^2+b^2?
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9 Answers
Naihrik Debnath, studied at GOLDEN VALLEY H.S SCHOOL Dharmanagar, North Tripura
Answered Apr 23, 2019
1.1k views
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OTHER ANSWERS
Sudam Katkar, B. Sc, BA, CAIIB PART 1 Chemistry & Mathematics, Shivaji University (1973)
Answered Apr 6, 2019 · Author has 1.1k answers and 774.8k answer views
a=(√3+√2)/(√3-√2)……b=(√3-√2)/(√3+√2)
a^2=(3+2+2√6)/(3+2–2√6)
…=(5+2√6)/(5-2√6)……b^2=(5–2√6)/(5+2√6)
a^2+b^2=(5+2√6)/5–2√6)+(5–2√6)/5+2√6)
…………={(5+2√6)^2+(5–2√6)^2}/5^2–4*6
…………..=(25+4*6+20√6+25+4*6–20√6)/25–24
……”……=49+49=98
2.2k views
Budha Chandra Singha, Quantitative Aptitude Trainer
Answered Aug 26, 2017 · Author has 181 answers and 165.8k answer views
Rationalize RHS of both a and b
[math]a=\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}[/math]
[math]a=(\sqrt{3}+\sqrt{2})^2[/math]
[math]a=5+2\sqrt{6}[/math]
Similary
[math]b=5-2\sqrt{6}[/math]
[math]a^2+b^2=(a+b)^2-2ab[/math]
[math]=(5+2\sqrt{6}+5-2\sqrt{6})^2-2(5+2\sqrt{6})(5-2\sqrt{6})[/math]
[math]=100-2(25-24)[/math]
[math]=98[/math]
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Manan Raheja, Maths has been my grinding stone since forever
Answered Mar 12, 2018 · Author has 630 answers and 816.5k answer views
There are two ways to approach this question.
First is to rationalize both a and b, and then find out the answer by squaring and adding them.
We get a = 5 + 2.sqrt(6), b = 5 - 2.sqrt(6).
So a^2 + b^2 = 25 + 24 + 25 + 24 = 98.
The second approach, is to observe that a = 1/b.
So a^2 + b^2 = a^2 + (1/a)^2 = (a + 1/a)^2 - 2.a.(1/a) = (10)^2 - 2 = 98.
2.4k views
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Dhrubajyoti Bhattacharjee
Answered May 2, 2018 · Author has 463 answers and 386.5k answer views
First of all, we see, a×b = 1.
Also, a+b = 10 [after simplifying].
Now, we have,
a^2 + b^2 = (a+b)^2 - 2ab
= 10^2 - 2×1
= 98.
Hope, you'll understand.
Thank You!
1.7k views · View 3 Upvoters
Deepti Mohanty, BSC 2 nd yr PhD in Mathematics & Mathematics, Indira Gandhi National Open University
Answered Jul 3, 2018
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Revathi, I have experience in teaching to my juniors & answered in Quora
Answered May 19, 2018 · Author has 128 answers and 195.2k answer views
Soln: Here, just we have to substitute the values of 'a' & 'b'. That's it! Come, let us solve.
= a²+b²
= [(√3+√2)/(√3-√2)]² + [(√3-√2)/(√3+√2)]²
= (√3+√2)²/(√3-√2)² + (√3-√2)²/(√3+√2)² [b/z, (a/b)² = a²/b²]
= [(√3)²+2(√3)(√2)+(√2)²]/[(√3)²-2(√3)(√2)+(√2)²] + [(√3)²-2(√3)(√2)+(√2)²]/[(√3)²+2(√3)(√2)+(√2)²] [b/z, (a+b)²=(a²+2ab+b²) & (a-b)²=(a²-2ab+b²)]
= (3+2√6+2)/(3–2√6+2)+(3–2√6+2)/(3+2√6+2)
= (5+2√6)/(5–2√6)+(5–2√6)/(5+2√6)
Now, take LCM,
= [(5+2√6)(5+2√6)+(5–2√6)(5–2√6)]/(5–2√6)(5+2√6)
= [(5+2√6)²+(5–2√6)²]/(5)²-(2√6)²
= {[(5)²+2(5)(2√6)+(2√6)²]+[(5)²-2(5)(2√6)+(2√6)²]}/25–4(√6)(√6)
= [(25+20√6+24)...
Answer:
answer is 9
Step-by-step explanation:
I can't understand