Question 119097
{{{3kx^2+2(x-1)+k=0}}}


Step 1:  Put the equation into standard form, i.e. {{{ax^2+bx+c=0}}}


{{{3kx^2+2x-2+k=0}}} or more neatly:  {{{3kx^2+2x+k-2=0}}}


Step 2:  Apply the quadratic formula with {{{a=3k}}}, {{{b=2}}}, and {{{c=k-2}}}


{{{x = (-2 +- sqrt( 4-4*(3k)*(k-2) ))/(2*3k) }}} 


Step 3: Simplify


{{{x = (-2 +- sqrt( 4-12k^2+24k) )/(6k) }}}


Step 4: We are looking for k such that {{{x[1]x[2]=1}}}, so:


{{{((-2 + sqrt( 4-12k^2+24k) )/(6k))((-2 - sqrt( 4-12k^2+24k) )/(6k)) =1}}}


Step 5:  This looks like a hideous mess to multiply, but remember that {{{(a+b)(a-b)=a^2-b^2}}}, so:


{{{(4-(4-12k^2+24k))/36k^2=1}}}


Step 6: Simplify and solve


{{{(4-(4-12k^2+24k))/36k^2=1}}}
{{{(12k^2-24k)/36k^2=1}}}
{{{12k^2-24k=36k^2}}}
{{{-24k^2-24k=0}}}
{{{k^2+k=0}}}
{{{k(k+1)=0}}}


So {{{k = 0}}} or {{{k=-1}}}


Step 7:  {{{k=0}}} can be excluded because that would make the lead coefficient on the original quadradic go to zero.  Therefore, {{{k=-1}}}


Step 8:  Check the answer.  Using {{{k=-1}}}, the original quadratic becomes


{{{-3x^2+2x-3=0}}}


{{{x = (-2 +- sqrt( 4-4(-3)(-3)))/(-6)=(-2+-sqrt(-32))/-6=(-2+-sqrt(2)4i)/6 =(-1+-sqrt(2)2i)/3}}}


The product of these two roots should be 1:  Does {{{((-1+sqrt(2)2i)/3)((-1-sqrt(2)2i)/3)=1}}}?


Again, we can simplify something that looks rather messy with the difference of two squares factorization, {{{(a+b)(a-b)=a^2-b^2}}}, 


{{{((-1+sqrt(2)2i)/3)((-1-sqrt(2)2i)/3)=(1-(2)(4)(-1))/9=9/9=1}}}:  Answer checks.