Question

Rationalize the denominator of $\frac{2}{3\sqrt{5} + 2\sqrt{11}}$ and write your answer in the form $\displaystyle \frac{A\sqrt{B} + C\sqrt{D}}{E}$, where $B < D$, the fraction is in lowest terms and all radicals are in simplest radical form. What is $A+B+C+D+E$? PLEASE HELP ME I WILL DO ANYTHING

Answer 1

Answer:

19

Step-by-step explanation:

Given the surdic expression $\frac{2}{3\sqrt{5} + 2\sqrt{11}}$, to rationalize the expression, we will have to multiply the numerator and denominator of the expression by the conjugate of the denominator as shown;

$= \frac{2}{3\sqrt{5} + 2\sqrt{11}} * \frac{3\sqrt{5} - 2\sqrt{11}}{3\sqrt{5} - 2\sqrt{11}}\\\\= \frac{2(3\sqrt{5} - 2\sqrt{11})}{(3\sqrt{5} + 2\sqrt{11})(3\sqrt{5} - 2\sqrt{11})}\\\\= \frac{6\sqrt{5} - 4\sqrt{11} }{9\sqrt{25}+6\sqrt{55}- 6\sqrt{55}-4\sqrt{121} } \\\\= \frac{6\sqrt{5} - 4\sqrt{11} }{9(5)-4(11) }\\\\= \frac{6\sqrt{5} - 4\sqrt{11} }{45-44 }\\\\= \frac{6\sqrt{5} - 4\sqrt{11} }{1}$

Comparing the result $\frac{6\sqrt{5} - 4\sqrt{11} }{1}$ with the expression $\frac{A\sqrt{B} + C\sqrt{D}}{E}$, it can be seen that A = 6, B = 5, C = -4, D = 11 and E = 1

A+B+C+D+E = 6+5+(-4)+11+1

A+B+C+D+E = 11-4+12

A+B+C+D+E = 19

Hence the value of A+B+C+D+E is 19