Question 1075689
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The solution by "josgarithmetic" is &nbsp;<U>TOTALLY</U>&nbsp; and &nbsp;<U>ABSOLUTELY</U>&nbsp; &nbsp;W&nbsp;R&nbsp;O&nbsp;N&nbsp;G&nbsp;.


Below find the correct solution.



<pre>
Let "a" be one solution of the given equation  {{{2x^2 - kx+ k +2}}} = {{{0}}}.

Then the other solution is {{{(3/2)*a}}}.


If so, then the given polynomial {{{2x^2 - kx+ k +2}}} admits factorization

{{{2x^2 - kx+ k +2}}} = {{{2*(x-a)*(x-(3/2)a)}}} = (x-a)*(2x-3a) = {{{2x^2 - 5ax + 3a^2}}}.


It means that  -k = -5a, 3a^2 = k+2,   or  (substituting)

{{{5a + 2}}} = {{{3a^2}}},

{{{3a^2 - 5a +2}}} = 0  --->  {{{a[1,2]}}} = {{{(5 +- sqrt(25 - 4*3*2))/(2*3)}}} = {{{(5 +- 1)/6}}}.


1.  {{{a[1]}}} = {{{5+1)/6}}} = 1  ---->  k = 5a = 5;


2.  {{{a[2]}}} = {{{(5-1)/6}}} = {{{4/6}}} = {{{2/3}}}.  ---->  k = 5a = {{{10/3}}}.


<U>Answer</U>.  There are TWO solutions for k:  1) k = 5,   and   2)  k = {{{10/3}}}.
</pre>

Solved.



God save you of using the method that "josgarithmetic" uses.