Question

Find the value of a
in the following
equation.
. ​

Answer 1

ANSWER

Given:

$\,\,\,\,\, \bullet \dfrac{6}{3\sqrt{2}-2\sqrt{3}} = 3\sqrt{2} - a\sqrt{3}$

To Find:

• Value of a

Solution:

$:\longrightarrow \dfrac{6}{3\sqrt{2}-2\sqrt{3}} = 3\sqrt{2} - a\sqrt{3} \\\\\text{Rationalising the denominator,} \\\\:\implies \dfrac{6}{3\sqrt{2}-2\sqrt{3}} * \dfrac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} = 3\sqrt{2} - a\sqrt{3} \\\\:\implies \dfrac{6*(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2})^2-(2\sqrt{3})^2} = 3\sqrt{2} - a\sqrt{3} \\\\:\implies \dfrac{6*(3\sqrt{2}+2\sqrt{3})}{18-12} = 3\sqrt{2} - a\sqrt{3} \\\\:\implies \dfrac{6*(3\sqrt{2}+2\sqrt{3})}{6} = 3\sqrt{2} - a\sqrt{3} \\\\:\implies 3\sqrt{2}+2\sqrt{3} = 3\sqrt{2} - a\sqrt{3} \\\\:\implies 2\sqrt{3} = - a\sqrt{3} \\\\:\implies 2 = -a \\\\\bf{:\implies a = -2}$

Formula Used:

• (a+b)(a-b) = a² - b²

Learn More:

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identities}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\bf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\bf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\bf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) - B^{3}\\\\8)\bf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\9)\bf\: A^{3} - B^{3} = (A-B)(A^{2} + AB + B^{2})\\\\ \end{minipage}}$

Answer 2

don't know !!!!!!!!!!!!!!!!!!!!!!!!!???!!!!