Question 1188994
<pre><font color="red"><b>
Ikleyn has made an error.  I will reproduce her solution here, and point
out her error in red.</font></pre></b>
Question 1188994
.
The negative reciprocal of the sum of all values of x that satisfy the
equation, (x-sqrt(x+1))/(x+sqrt(x+1)) = 11/5 is...
A) 7 and 1/9 B) -(9/64) C) 1 and 1/8 D) -(1/8) E) 8/9
~~~~~~~~~~~~~~~~~


<pre>
I will simply solve the given equation 

    {{{(x-sqrt(x+1))/(x+sqrt(x+1))}}} = {{{11/5}}}


and will find its roots. Then I will calculate the negative reciprocal of the sum of the roots.


To solve equation, I will introduce new variable  u = {{{sqrt(x+1)}}},  so  {{{u^2}}} = x+1,  x = {{{u^2 - 1}}}.<pre><font color="red"><b>
Right here, she should have observed that since u = {{{sqrt(x+1)}}},
that u must be non-negative, since a square root radical [or 
ANY EVEN-ROOT radical], by convention, is never taken to be 
negative, so below when she gets a negative value for u, she 
should have taken it to be extraneous.</font></b>

With new variable, equation (1)  takes the form

   {{{(u^2 - 1 - u)/(u^2 -1 + u)}}} = {{{11/5}}},

    5*(u^2 - 1 - u) = 11*(u^2 - 1 + u)

    5u^2 - 5 - 5u = 11u^2 - 11 + 11u

    6u^2 + 16u - 6 = 0

    3u^2 + 8u - 3 = 0

<pre><font color="red"><b>
This is correct, but when she gets here, she uses the quadratic 
formula to solve an easily factorable quadratic.  It is perfectly 
correct to do so, but normally we should try to factor first, 
especially when the coefficients are small (in absolute value) as 
in this case.  Factoring gives: 

(x + 3)(3x - 1) = 0
x + 3 = 0;  3x - 1 = 0
    x = -3;  3x = 1
              x = 1/3</font></b>
    {{{u[1,2]}}} = {{{(-8 +- sqrt((-8)^2 - 4*3*(-3)))/(2*5)}}} = {{{(-8 +- sqrt(64 + 36))/6}}} = {{{(-8 +- 10)/6}}}.


So,  {{{u[1]}}} = -3;  {{{u[2]}}} = {{{1/3}}}.

<pre><font color="red"><b>
As you see here, she got the same values using the quadratic formula 
as was obtained above by factoring.  Again, this was not a mistake, 
but just an opportunity to point out that we should always try to 
factor a quadratic before using the quadratic formula.</font></b>

Thus, there are two solutions for x, and they are  {{{x[1]}}} = {{{u[1]^2-1}}} = {{{(-3)^2-1}}} = 9 - 1 = 8;

<pre><font color="red"><b>
Her error here was that she should have discarded as extraneous 
the false solution u = -3, by observing above that u must not be 
negative.  So x cannot be 8. So there is no need for subscription 
u or x.</font></b>
                                                   {{{x[2]}}} = {{{u[2]^2-1}}} = {{{(1/3)^2-1}}} = {{{1/9 - 1}}} = {{{-8/9}}}.

<pre><font color="red"><b>
That is correct, and is the ONE and ONLY solution for x, x=-8/9.</font></b>

The sum of the roots of the original equation is  {{{x[1]}}} + {{{x[2]}}} = 8 - {{{8/9}}} = 7 {{{1/9}}} = {{{64/9}}},

    and the negative reciprocal of it is  {{{-9/64}}}.    <U>ANSWER</U>
</pre>
<pre><font color="red"><b>This is incorrect.  Her error above caused her to select the wrong 
answer. Since -8/9 is the ONLY solution, the sum of all values of 
x that satisfy the equation is -8/9 itself.  To get the negative 
reciprocal of -8/9, we invert it and change the sign.  So the 
correct solution is +9/8 and if we change that to a mixed number 
we get {{{1&1/8}}}.  So the correct choice is C).</font></b>

Solved.

<pre><font color="red"><b>Corrected.

No doubt, the author of this problem listed -(9/64) as an incorrect 
choice to catch students who would make the very error that Ikleyn 
made -- the error of failing to look for extraneous answers when 
dealing with square root radicals, or any EVEN-ROOT radicals.  
EVEN-ROOT radicals are NEVER negative.  [But be careful, for this 
is NOT SO with ODD-ROOT radicals, such as CUBE ROOTS!]

Edwin</font></b></pre>