Question 896267
<pre>
{{{sqrt(2z+1)+sqrt(3z+4)}}}{{{""=""}}}{{{7}}}

Isolate one of the radical terms

{{{sqrt(3z+4)}}}{{{""=""}}}{{{7}}}{{{""-""}}}{{{sqrt(2z+1)}}}

We square both sides:

{{{(sqrt(3z+4))^2}}}{{{""=""}}}{{{(7-sqrt(2z+1))^2}}}

Squaring the square root on the left takes away the radical:
Write the right side twice multiplied by itself, so we can use FOIL:

{{{3z+4}}}{{{""=""}}}{{{(7-sqrt(2z+1))(7-sqrt(2z+1))}}}

{{{3z+4}}}{{{""=""}}}{{{49-7sqrt(2z+1)-7sqrt(2z+1)+2z+1}}}

Combine the radical terms on the right

{{{3z+4}}}{{{""=""}}}{{{50-14sqrt(2z+1)+2z}}}

Isolate the radical term on the left

{{{14sqrt(2z+1)}}}{{{""=""}}}{{{46-z}}}

Square both sides:

{{{196(2x+1)}}}{{{""=""}}}{{{(46-z)^2}}}

{{{392z+196}}}{{{""=""}}}{{{2116-92z+z^2}}}

{{{"0"}}}{{{""=""}}}{{{z^2-484z+1920}}}

Factor the right side:

{{{"0"}}}{{{""=""}}}{{{(z-4)(z-480)}}}

Use the zero factor property:

        z - 4 = 0;   z-480 = 0
            z = 4;       z = 480

We must always check radical equations with square
or even roots for extraneous solutions:

Checking z=4
{{{sqrt(2z+1)+sqrt(3z+4)}}}{{{""=""}}}{{{7}}}
{{{sqrt(2(4)+1)+sqrt(3(4)+4)}}}{{{""=""}}}{{{7}}}
{{{sqrt(8+1)+sqrt(12+4)}}}{{{""=""}}}{{{7}}}
{{{sqrt(9)+sqrt(16)}}}{{{""=""}}}{{{7}}}
{{{3+4}}}{{{""=""}}}{{{7}}}
{{{7}}}{{{""=""}}}{{{7}}}



So 4 is a solution.

Checking x=480
{{{sqrt(2z+1)+sqrt(3z+4)}}}{{{""=""}}}{{{7}}}
{{{sqrt(2(480)+1)+sqrt(3(480)+4)}}}{{{""=""}}}{{{7}}}
{{{sqrt(960+1)+sqrt(1440+4)}}}{{{""=""}}}{{{7}}}
{{{sqrt(961)+sqrt(1441)}}}{{{""=""}}}{{{7}}}
{{{31+38}}}{{{""=""}}}{{{7}}}
{{{69}}}{{{""=""}}}{{{7}}}

So 480 is an extraneous answer, not a solution.

The only solution is z = 4

Edwin</pre>