Question 177767
<FONT FACE="Times New Roman" SIZE="+2">


A.  *[tex \Large (3 - sqrt{x+2})^2 = 4]


<FONT COLOR=#FF000> Step 1: <FONT COLOR=#00000>Expand the binomial on the left:


*[tex \Large \text{        } \math 9 - 6 sqrt{x + 2} + (x + 2) = 4]


<FONT COLOR=#FF000>Step 2:<FONT COLOR=#00000> A simple substitution will allow us to deal with this ugliness in a fairly straightforward manner.


Let *[tex \Large t = sqrt{x + 2}], then:


*[tex \Large \text{        } \math 9 - 6t + t^2 = 4]


<FONT COLOR=#FF000>Step 3:<FONT COLOR=#00000> Put the quadratic into standard form:


*[tex \Large \text{        } \math t^2 - 6t + 5 = 0]


Giving us a factorable quadratic in <i>t</i>, so:


<FONT COLOR=#FF000>Step 4:<FONT COLOR=#00000> Factor the quadratic:


*[tex \Large \text{        } \math (t - 1)(t - 5) = 0]


Therefore *[tex \Large \text{    } \math  t = 1 \text{     }] or *[tex \Large \text{    } \math t = 5 \text{     }]


But we aren't done yet.


<FONT COLOR=#FF000>Step 5:<FONT COLOR=#00000> Since *[tex \Large t = sqrt{x + 2}],


*[tex \Large \text{        } \math sqrt{x + 2} = 1]*[tex \Large \text{   or        } \math sqrt{x + 2} = 5]


*[tex \Large \text{        } \math x + 2 = 1]*[tex \Large \text{   or        } \math x + 2 = 25]


*[tex \Large \text{        } \math x = -1]*[tex \Large \text{   or        } \math x = 23]


But we are <i>still</i> not done.


<FONT COLOR=#FF000>Step 6:<FONT COLOR=#00000>  Check the potential roots to ensure that we have not introduced extraneous roots in the process of solving the problem:


If *[tex \Large x = -1] then,
*[tex \Large \text{        } \math (3 - sqrt{-1+2})^2 = 4]


*[tex \Large \text{        } \math (3 - 1)^2 = 4]


*[tex \Large \text{        } \math (2)^2 = 4]


<FONT COLOR=#00C00>True statement, potential solution *[tex \Large x = -1] confirmed.<FONT COLOR=#00000>


If *[tex \Large x = 23] then,
*[tex \Large \text{        } \math (3 - sqrt{23+2})^2 = 4]


*[tex \Large \text{        } \math (3 - sqrt{25})^2 = 4]


*[tex \Large \text{        } \math (3 - 5)^2 = 4]


*[tex \Large \text{        } \math (-2)^2 = 4]


<FONT COLOR=#00C00>True statement, potential solution *[tex \Large x = 23] confirmed.<FONT COLOR=#00000>


<i>Now</i> we are done.


=======================================================================

B.  *[tex \Large 3 + 5 sqrt{1 + 2x^2} = 13]


<FONT COLOR=#FF000>Step 1:<FONT COLOR=#00000> Add -3 to both sides and then multiply by *[tex \large \frac {1}{5}] to obtain:


*[tex \Large \text{        } \math sqrt{1 + 2x^2} = 2]


<FONT COLOR=#FF000>Step 2:<FONT COLOR=#00000> Square both sides:


*[tex \Large \text{        } \math 1 + 2x^2 = 4]


<FONT COLOR=#FF000>Step 3:<FONT COLOR=#00000> Add -1 to both sides and then multiply by *[tex \large \frac {1}{2}] to obtain:


*[tex \Large \text{        } \math x^2 = \frac {3}{2}]


<FONT COLOR=#FF000>Step 4:<FONT COLOR=#00000> Take the square root of both sides, considering both the positive and negative roots, and then rationalize the denominators:


*[tex \Large \text{        } \math x = \pm sqrt{\frac {3}{2}}]


*[tex \Large \text{        } \math x = \pm \frac {sqrt{6}}{2}]


<FONT COLOR=#FF000>Step 5:<FONT COLOR=#00000> Check the potential solutions.


Since *[tex \Large x^2 = \frac {3}{2}] regardless of whether *[tex \Large x =  \frac {sqrt{6}}{2}] or *[tex \Large x =  -\frac {sqrt{6}}{2}], if one solution is good, then they both are:



*[tex \Large \text{        } \math 3 + 5 sqrt{1 + 2 (\frac {3}{2})} = 13]


*[tex \Large \text{        } \math 3 + 5 sqrt{1 + 3} = 13]


*[tex \Large \text{        } \math 3 + 10 = 13]


<FONT COLOR=#00C00>True statement, both potential solutions *[tex \Large x = \pm \frac{sqrt{6}}{2}] confirmed.<FONT COLOR=#00000>
</FONT>