Question

a=√3+√2/√3-√2 , b= √3-√2/√3+√2 then find a^2+b^2​

Answer 1

Answer:

(i) 3√5 (ii) 32√5 (iii) 1√12 (iv) √2√5 (v) √3 +1√2 (vi) √2+√53 (vii) 3√2√5 ... √2=1.414, √3=1.732, √5=2.236 and √10=3.162. (i) 2√ 3 (ii) 3√10 ... (iii) 3+√23-√2=a+b√2

Answer 2

Answer:

If a= (√3+√2) / (√3-√2) and b= (√3-√2) / (√3+√2), then how do I find a^2+b^2?

This problem can be solved in 2 ways.

Now

We know that

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)+(25–20√6+24)]/25–24

= (49+20√6+49–20√6)/ 1

= 98/1

= 98

Hence, a²+b²=98