Question

Rationalise
5√3 -4√2÷ 4√3+3√2​

Answer 1

GIVEN :-

$\large \bold{5 \sqrt{3} - 4 \sqrt{2} \div 4 \sqrt{3} + 3 \sqrt{2} }$

$\tiny \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

SOLUTION :-

$\orange\looparrowright\large \bold \orange{5 \sqrt{3} - 1 \sqrt{2} \sqrt{3} + 3 \sqrt{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\orange\looparrowright\large \bold \orange{5 \sqrt{3} - \sqrt{6} + 3 \sqrt{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\red\looparrowright\large \bold \red{10.4534} \huge \red \checkmark$

Answer 2

Given:

The expression -

$\bf \dfrac{5\sqrt{3}-4}{4\sqrt{3}+3\sqrt{2}}$

What To Find:

We have to -

• Rationalise the denominator.

Solution:

Here, the conjugate of the denominator is,

$\sf \implies 4\sqrt{3} - 3\sqrt{2}$

Multiply the conjugate with the expression,

$\sf \implies \dfrac{5\sqrt{3}-4}{4\sqrt{3}+3\sqrt{2}} \times \dfrac{4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}-3\sqrt{2}}$

Take them as common,

$\sf \implies \dfrac{5\sqrt{3}-4 \times 4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}+3\sqrt{2} \times 4\sqrt{3}-3\sqrt{2}}$

Solving the numerator,

$\sf \implies \dfrac{5\sqrt{3}(4\sqrt{3}-3\sqrt{2})-4(4\sqrt{3}-3\sqrt{2})}{4\sqrt{3}+3\sqrt{2} \times 4\sqrt{3}-3\sqrt{2}}$

Solving the numerator further,

$\sf \implies \dfrac{60 - 15\sqrt{6} - 16\sqrt{3}+12\sqrt{2}}{4\sqrt{3}+3\sqrt{2} \times 4\sqrt{3}-3\sqrt{2}}$

Solving the denominator using the identities (a - b) (a + b) = a² - b²,

$\sf \implies \dfrac{60 - 15\sqrt{6} - 16\sqrt{3}+12\sqrt{2}} {(4\sqrt{3})^2-(3\sqrt{2})^2}$

Solve the brackets,

$\sf \implies \dfrac{60 - 15\sqrt{6} - 16\sqrt{3}+12\sqrt{2}} {48 - 18}$

Subtract 18 from 48,

$\sf \implies \dfrac{60 - 15\sqrt{6} - 16\sqrt{3}+12\sqrt{2}} {30}$

Final Answer:

∴ Thus, the answer is $\sf \dfrac{60 - 15\sqrt{6} - 16\sqrt{3}+12\sqrt{2}} {30}$ after rationalising the denominator.