Question

6/3√2-2√3=3√2-a√3 find value a​

Answer 1

Required Answer:-

Given:

$\dag\ \boxed{\tt \dfrac{6}{3\sqrt{2}-2\sqrt{3}}=3\sqrt{2}-a\sqrt{3}}$

To Find:

• The value of a.

Solution:

Given that,

$\tt\implies \dfrac{6}{3\sqrt{2}-2\sqrt{3}}=3\sqrt{2}-a\sqrt{3}$

Multiplying both numerator and denominator by (3√2 + 2√3), we get,

$\tt\implies \dfrac{6 \times (3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}+2\sqrt{3})}=3\sqrt{2}-a\sqrt{3}$

Simplifying the result,

$\tt\implies \dfrac{6 \times (3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2})^{2}-(2\sqrt{3})^{2}}=3\sqrt{2}-a\sqrt{3}$

$\tt\implies \dfrac{6 \times (3\sqrt{2}+2\sqrt{3})}{18-12}=3\sqrt{2}-a\sqrt{3}$

$\tt\implies \dfrac{6 \times (3\sqrt{2}+2\sqrt{3})}{6}=3\sqrt{2}-a\sqrt{3}$

$\tt\implies(3\sqrt{2}+2\sqrt{3})=3\sqrt{2}-a\sqrt{3}$

$\tt\implies 2\sqrt{3}=-a\sqrt{3}$

Dividing both sides by √3, we get,

$\tt\implies 2=-a$

$\tt\implies a=-2$

★ Hence, the value of a is -2.

Answer:

• The value of a is -2.

•••♪

Answer 2

Required Answer :

• The value of a is -2.

Concept :

Here, we'll use the method of rationalisation to find the value of a. After rationalising, we'll simplify the resulting number to obtain the appropriate value of a.

Step by step explanation :

Given :

$\frac{6}{3 \sqrt{2 } - 2 \sqrt{3} } = 3 \sqrt{2 } - a \sqrt{3}$

To find :

• Value of a.

Explanation :

We have given,

$\frac{6}{3 \sqrt{2 } - 2 \sqrt{3} } = 3 \sqrt{2 } - a \sqrt{3}$

Now, rationalising the denominator by multiplying both denominator and numerator by (3√2 + 2√3),

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

$→$ $\frac{6 \times (3 \sqrt{2} + 2 \sqrt{3}) }{18 - 12} = 3 \sqrt{2} - a \sqrt{3}$

$→$ $\frac{6 \times (3 \sqrt{2} + 2 \sqrt{3}) }{6} = 3 \sqrt{2} - a \sqrt{3}$

• Cancelling 6 in the LHS.

$→$ $(3 \sqrt{2} + 2 \sqrt{3} ) = 3 \sqrt{2} - a \sqrt{3}$

• Cancelling $3 \sqrt{2}$ on both the sides.

$→$ $2 \sqrt{3} = -a \sqrt{3}$

• Dividing both sides by √3 for getting the value of a.

$→$ $2 = -a$

∴ $a = - 2$ ✔️