Question

3√8 - 5√6 by 2√8 + 3 √6 rationalize​

Answer 1

Step-by-step explanation:

Here's your answer

Hope it helps

Answer 2

Given:

The expression -

$\bf \to \dfrac{3\sqrt{8}-5\sqrt{6}}{2\sqrt{8}+3\sqrt{6}}$

What To Find:

We have to -

• Rationalise the expression.

Solution:

$\rm \to \dfrac{3\sqrt{8}-5\sqrt{6}}{2\sqrt{8}+3\sqrt{6}}$

Here the rationalising factor of the denominator is,

$\rm \to 2\sqrt{8} - 3\sqrt{6}$

Multiply it with the expression,

$\rm \to \dfrac{3\sqrt{8}-5\sqrt{6}}{2\sqrt{8}+3\sqrt{6}} \times \dfrac{2\sqrt{8}-3\sqrt{6}}{2\sqrt{8}-3\sqrt{6}}$

Take them as common,

$\rm \to \dfrac{(3\sqrt{8}-5\sqrt{6}) \times (2\sqrt{8}-3\sqrt{6})}{(2\sqrt{8}+3\sqrt{6}) \times (2\sqrt{8}-3\sqrt{6})}$

Solve the numerator,

$\rm \to \dfrac{3\sqrt{8}(2\sqrt{8}-3\sqrt{6}) - 5\sqrt{6}(2\sqrt{8}-3\sqrt{6})}{(2\sqrt{8}+3\sqrt{6}) \times (2\sqrt{8}-3\sqrt{6})}$

Solve the first brackets,

$\rm \to \dfrac{48 - 9\sqrt{48} - 5\sqrt{6}(2\sqrt{8}-3\sqrt{6})}{(2\sqrt{8}+3\sqrt{6}) \times (2\sqrt{8}-3\sqrt{6})}$

Solve the second brackets,

$\rm \to \dfrac{48 - 9\sqrt{48} - 10\sqrt{48}+90}{(2\sqrt{8}+3\sqrt{6}) \times (2\sqrt{8}-3\sqrt{6})}$

Rearrange the terms,

$\rm \to \dfrac{48+90 - 9\sqrt{48} - 10\sqrt{48}}{(2\sqrt{8}+3\sqrt{6}) \times (2\sqrt{8}-3\sqrt{6})}$

Solve the terms,

$\rm \to \dfrac{138 - 19\sqrt{48}}{(2\sqrt{8}+3\sqrt{6}) \times (2\sqrt{8}-3\sqrt{6})}$

Solve the denominator by using (a - b) (a + b) = a² - b²,

$\rm \to \dfrac{138 - 19\sqrt{48}}{(2\sqrt{8})^2-(3\sqrt{6})^2}$

Find the squares,

$\rm \to \dfrac{138 - 19\sqrt{48}}{32-54}$

Subtract the terms,

$\rm \to \dfrac{138 - 19\sqrt{48}}{-22}$

Can be written as,

$\rm \to -\dfrac{138 - 19\sqrt{48}}{22}$

Final Answer:

∴ Thus, the answer after rationalising the expression is -

• $\bf -\dfrac{138 - 19\sqrt{48}}{22}$