√3+√2/√3-√2=a+b√c find a and b
Answers
Answer:
3+2√6+2
Step-by-step explanation:
√3+√2.√3+√2/√3-√2.√3+√2
=√3(√3+√2)+√2(√3+√2)/√3(√3+√2)-√2(√3+√2)
=3+√6+√6+2/3-2
=3+√6+√6+2/1
= 3+2√6+2
= a=3 and b=2√6
Hope this answer is helpful for you.
1st :-
a=3–√+2–√3–√−2–√
=(3–√+2–√)(3–√+2–√)(3–√−2–√)(3–√+2–√)
=(3–√+2–√)23−2
=((3–√)2+(2–√)2+2(3–√)(2–√)
=3+2+26–√
=5+26–√
b=3–√−2–√3–√+2–√
=(3–√−2–√)(3–√−2–√)(3–√+2–√)(3–√−2–√)
=(3–√−2–√)23−2
=((3–√)2+(2–√)2+−2(3–√)(2–√)
=3+2−26–√
=5−26–√
Now
a2+b2
We know that
(x+y)2=x2+y2+2xy
⟹x2+y2=(x+y)2−2xy
If we take x=a and y=b then
a2+b2
=(a+b)2−2ab
=(5+26–√+5−26–√)2−2((5+26–√)(5−26–√))
=(10)2−2(25−24)
Here we use identity (a+b)(a−b)=a2−b2
=100−2
=98
2nd :-
a=3–√+2–√3–√−2–√
b=3–√−2–√3–√+2–√
Now
ab=3–√+2–√3–√−2–√×3–√−2–√3–√+2–√
=1
a+b=3–√+2–√3–√−2–√+3–√−2–√3–√+2–√
=(3–√+2–√)2+(3–√−2–√)2(3–√+2–√)(3–√−2–√)
=5+26–√+5−26–√3−2
=10
Now
(x+y)2=x2+y2+2xy
⟹x2+y2=(x+y)2−2xy
Replacing x=a,y=b
a2+b2=(a+b)2−2ab
=(10)2−2×1
Replacing (x+y)=10 and xy=1
=100−2
=98
Therefore a2+b2=98
It totally depends on you which method to use but I recommend the second one .
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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)+(2
Rationalize RHS of both a and b
a=(3√+2√)(3√+2√)(3√+2√)(3√−2√)
a=(3–√+2–√)2
a=5+26–√
Similary
b=5−26–√
a2+b2=(a+b)2−2ab
=(5+26–√+5−26–√)2−2(5+2–√)(5−26–√)
=100−2(25−24)
=98