Question 988996
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(1) In general, to find the square root of an expression of the form {{{sqrt(a+b*sqrt(c))}}}<br>
Consider this:<br>
{{{(sqrt(x)+sqrt(y))^2=x+2sqrt(xy)+y=(x+y)+2sqrt(xy)}}} [1]<br>
Consider the form of the expression on the right above.  This required form has the sum of two integers as the rational part and the product of those two integers as the radicand, with multiplier 2 on the radical.<br>
If we can put our given square root expression in that form, then the equivalent expression is {{{sqrt(x)+sqrt(y)}}}<br>
In this problem, we have the expression<br>
{{{sqrt(7+4sqrt(3))}}}<br>
To put this in the form in [1] above, we take 2 out of the 4 and put it back inside the radical, leaving the required "2" outside the radical:<br>
{{{sqrt(7+4sqrt(3))=sqrt(7+2sqrt(12))}}}<br>
Then we see that, since 4+3=7 and 4*3=12, the expression is equivalent to {{{sqrt(4)+sqrt(3)}}} or {{{2+sqrt(3)}}}<br>
So we have {{{sqrt(7+4sqrt(3))=2+sqrt(3)}}}.<br>
Similarly, we will find {{{sqrt(7-4sqrt(3))=2-sqrt(3)}}}.<br>
And so the given expression is equivalent to {{{(2+sqrt(3))+(2-sqrt(3))=4}}}<br>
(2) And for an expression in the exact form of the given expression, there is a very different way to evaluate the expression.<br>
Square the given expression and watch how it simplifies.<br>
{{{(sqrt(7+4sqrt(3))+sqrt(7-4sqrt(3)))^2}}}<br>
{{{(7+4sqrt(3))+2sqrt((7+4sqrt(3))(7-4sqrt(3)))+(7-4sqrt(3))}}}<br>
{{{14+2sqrt(49-48)=14+2=16}}}<br>
Then, since the square of the given expression is 16, the given expression is equal to 4.<br>