<PRE><B>Question 93556</B>
&lt;pre&gt;&lt;font size = 5 color =&quot;indigo&quot;&gt;&lt;b&gt;
Could you help me with this?

{{{8sqrt(x-1)+1}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{x}}}

Thank-you

{{{8sqrt(x-1)+1}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{x}}}

Add -1 to both sides to get the radical 
term alone on the left:

  {{{8sqrt(x-1)}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{x-1}}}

Square both sides of the equation:

 {{{ ( 8sqrt(x-1) )^2}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{ (x-1)^2 }}}

{{{8^2(sqrt(x-1))^2}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{(x-1)(x-1)}}}

{{{64(x-1)}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{x^2-x-x+1}}}

{{{64x-64}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{x^2-2x+1}}}

Switch left and right sides for convenience:

{{{x^2-2x+1}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{64x-64}}}

Get 0 on the right 

{{{x^2-66x+65}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{0}}}

Factor left side by thinking of two whole numbers
which have product 65 and sum 66, whch are 65 and 1:

{{{(x-65)(x-1)}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; 0

Set each factor &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; 0

{{{x-65=0}}} gives {{{x=65}}}

{{{x-1=0}}} gives {{{x=1}}}

However we must check these answers because radical
equations often have extraneous answers, which are
not solutions.  

Checking {{{x=65}}}

{{{8sqrt(x-1)+1}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{x}}}

{{{8sqrt(65-1)+1}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{65}}}

{{{8sqrt(64)+1}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{65}}}

{{{8(8)+1}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{65}}}

{{{64+1}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{65}}}

{{{65}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{65}}}

So we know {{{x=65}}} is a valid solution.

Checking {{{x=1}}}

{{{8sqrt(x-1)+1}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{x}}}

{{{8sqrt(1-1)+1}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{1}}}

{{{8sqrt(0)+1}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{1}}}

{{{8(0)+1}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{1}}}

{{{0+1}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{1}}}

{{{1}}} &lt;/font&gt;&lt;/b&gt;=&lt;font size = 5 color = &quot;indigo&quot;&gt;&lt;b&gt; {{{1}}}

So as it turns out, both {{{x=1}}} and {{{x=65}}}
are valid solutions.

But be aware that this is not always the case.
Sometimes when you get two answers, one of them
will be a solution and the other will not check,
and thus must be discarded; sometimes neither one 
is a solution and the equation has no solution.
Just be sure to ALWAYS CHECK a RADICAL EQUATION.

Edwin&lt;/pre&gt;


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