Math, asked by sunny9482, 8 months ago

√3+√2/√3-√2=a+b√c find a and b​

Answers

Answered by ssd435156
0

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.

Answered by jainakash445
0

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

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