Question

tomorrow is my maths test ....plzzz....answer fast

Answer 1

Answer:

$a+b=\sqrt{22}$

Step-by-step explanation:

Given,

$a=\sqrt{6-\sqrt{11}}\\\;\\b=\sqrt{6+\sqrt{11}}$

$a+b=\sqrt{6-\sqrt{11}}+\sqrt{6+\sqrt{11}}\\\;\\\text{Making square of both sides}\\\;\\(a+b)^2=(\sqrt{6-\sqrt{11}}+\sqrt{6+\sqrt{11}})^2\\\;\\(a+b)^2=(\sqrt{6-\sqrt{11}}\;)^2+(\sqrt{6+\sqrt{11}}\;)^2+2\times(\sqrt{6-\sqrt{11}})\times(\sqrt{6+\sqrt{11}})\\\;\\(a+b)^2=6-\sqrt{11}\;+6+\sqrt{11}+2\times(\sqrt{(6-\sqrt{11})(6+\sqrt{11}})\\\;\\(a+b)^2=6+6+2\times(\sqrt{(6^2-(\sqrt{11})^2)}\\\;\\(a+b)^2=12+2\times(\sqrt{(36-11})}\\\;\\(a+b)^2=12+2\times(\sqrt{25})\\\;\\(a+b)^2=12+2\times5\\\;\\(a+b)^2=12+10$

$(a+b)^2=22\\\;\\a+b=\sqrt{22}$

$Note:\\\;\\1.\;(a+b)^2=a^2+b^2+2ab\\\;\\2.\;(a+b)(a-b)=a^2-b^2\\\;\\3.\;\sqrt{a}\times\sqrt{b}=\sqrt{ab}$