Question 170220
<font size = 7 color = "red"><b>Stanbon's solution is 
incorrect because he did
not check for extraneous 
solutions:</font></b>
<pre><font size = 4 color = "indigo"><b>
{{{sqrt(9x+67)=x+5}}} (solve)

Square both sides:

{{{(sqrt(9x+67))^2=(x+5)^2}}}

{{{9x+67=(x+5)(x+5)}}}
{{{9x+67=x^2+10x+25}}}
Get 0 on the left:
{{{0=x^2+x-42}}}
Switch sides:
{{{x^2+x-42=0}}}
Factor
{{{(x-6)(x+7)=0}}}
Use the zero-factor principle:

{{{matrix(2,5,    
         x-6=0,  "",  "",  "", x+7=0,
           x=6,  "",  "",  "",  x=-7)}}} 

Stanbon quit here, but you MUST check
radical equations for extraneous solutions:

Checking {{{x=6}}}

{{{sqrt(9x+67)=x+5}}}
{{{sqrt(9(6)+67)=(6)+5}}}
{{{sqrt(54+67)=6+5}}}
{{{sqrt(121)=11}}}
{{{11=11}}}

That checks.

Checking {{{x=-7}}}

{{{sqrt(9x+67)=x+5}}}
{{{sqrt(9(-7)+67)=(-7)+5}}}
{{{sqrt(-63+67)=-7+5}}}
{{{sqrt(4)=-2}}}
{{{2=-2}}}

This does not check so we
must discard {{{x=-7}}} as
an extraneous solution. The
only solution is {{{x=6}}}.

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</pre></b></font>
Rewrite with a rational exponent {{{root(6,19)}}}.
<pre><font size = 4 color = "indigo"><b>
Use the rule {{{root(N,X^M)=X^(M/N)}}}

{{{root(6,19)=6^(1/19)}}}.

</pre></b></font>
Convert to decimal notation {{{matrix(1,3,8.87,"×",10^7)}}}. I had {{{8870000000}}}. (I know this is incorrect)
<pre><font size = 4 color = "indigo"><b>
Yes, you don't put on 7 zeros. You put on just enough
zeros so that you can move the decimal 7 places right.
You are to move the decimal of {{{8.87}}} 7 places right.
There are only 2 decimal places showing so you will
need to put on 5 zeros so there will be 7 places. So you
consides {{{8.87}}} as {{{8.8700000}}} and then when you
move the decimal 7 places to the right, it will go at the
end and be invisible:

So {{{matrix(1,3,8.87,"×",10^7)=matrix(1,3,8.8700000,"×",10^7)=88700000}}} 

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Use rational exponents to write xy small 5*y small 1|6 (1&6 are small numbers)*2 small 1|3 (1&3 are small numbers)

I think you meant this:

{{{xy^5*y^(1/6)*2^(1/3)}}}

Write the {{{1/3}}} as {{{2/6}}}

{{{xy^5*y^(1/6)*2^(2/6)}}}

Write {{{y^(1/6)}}} as {{{root(6,y)}}}
Write {{{2^(2/6)}}} as {{{root(6,2^2)}}}

{{{xy^5*root(6,y)*root(6,2^2)}}}
{{{xy^5*root(6,y)*root(6,4)}}}

Then multiply under the radicals:

{{{xy^5*root(6,4y)}}}

Edwin</pre>