Question

find the value of b root2+root3/3root2-2root3=2-b root6

Answer 1

Step-by-step explanation:

Given:-

(√2+√3)/(3√2-2√3) = a-b√6

To find:-

Rationalised the denominator and find out the value of a and b.

Solution:-

Given that

(√2+√3)/(3√2-2√3) = a-b√6

Denominator = 3√2-2√3

We know that

Rationalising factor of a√b-a√b = a√b+a√b

Rationalising factor of 3√2-2√3 = 3√2+2√3

On Rationalising the denominator then

=> [(√2+√3)/(3√2-2√3)]×[(3√+2√3)/(3√2+2√3)]

=> [(√2+√3)(3√2+2√3)]/[(3√2-2√3)(3√2+2√3)]

Now, multiplying numerator left side to right side we get,

=> [3(√2×2)+2(√3×2)+3(√2×3)+2(√3×3)]/[(3√2-2√3)(3√2+2√3)]

Now, applying algebraic identity in denominator

We know that

(a-b)(a+b)=a^2-b^2

Where a = 3√2 and b=2√3

=> (3×2+2√6+3√6+2×3)/[(3√2)^2-(2√3)^2]

=> (12+5√6)/(18-12)

=> 12+5√6/6

=> 2+⅚√6

∴ a-b√6 = 2+⅚√6

On, comparing with R.H.S

a = 2,

b = -⅚√6 = -⅚

Answer:-

Hence, the value of a = 2 and b = -5/6.

Used formulae:-

• Rationalising factor of a√b-a√b = a√b+a√b

• (a-b)(a+b)=a^2-b^2

:)

Answer 2

The correct answer is  $\frac{-5}{6}$.

Given: The equation = $\frac{\sqrt{2}+\sqrt{3} }{3\sqrt{2}-2\sqrt{3} } =2-b\sqrt{6}$.

To Find: Value of b.

Solution:

$\frac{\sqrt{2}+\sqrt{3} }{3\sqrt{2}-2\sqrt{3} } =2-b\sqrt{6}$

Rationalize the term.

$\frac{\sqrt{2}+\sqrt{3} }{3\sqrt{2} -2\sqrt{3} } *\frac{3\sqrt{2} +2\sqrt{3} }{3\sqrt{2} +2\sqrt{3} }$

= $\frac{6+2\sqrt{6}+3\sqrt{6}+6 }{(3\sqrt{2})^{2}- (2\sqrt{3} )^{2} }$

= $\frac{12+5\sqrt{6} }{18- 12 }$

= $\frac{12+5\sqrt{6} }{6 }$

= $2- (\frac{-5}{6} )\sqrt{6}$

Compare with RHS.

b = $\frac{-5}{6}$

Hence, the value of b is  $\frac{-5}{6}$.

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