Question 132645
Their slopes will be perpendicular when their product is -1. So we have the equation


{{{(2k-4)*(k+6)=-1}}}



{{{2k^2+8k-24=-1}}} Foil



{{{2k^2+8k-23=0}}} Add 1 to both sides


Let's use the quadratic formula to solve for k:



Starting with the general quadratic


{{{ak^2+bk+c=0}}}


the general solution using the quadratic equation is:


{{{k = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{2*k^2+8*k-23=0}}} ( notice {{{a=2}}}, {{{b=8}}}, and {{{c=-23}}})





{{{k = (-8 +- sqrt( (8)^2-4*2*-23 ))/(2*2)}}} Plug in a=2, b=8, and c=-23




{{{k = (-8 +- sqrt( 64-4*2*-23 ))/(2*2)}}} Square 8 to get 64  




{{{k = (-8 +- sqrt( 64+184 ))/(2*2)}}} Multiply {{{-4*-23*2}}} to get {{{184}}}




{{{k = (-8 +- sqrt( 248 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{k = (-8 +- 2*sqrt(62))/(2*2)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{k = (-8 +- 2*sqrt(62))/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{k = (-8 + 2*sqrt(62))/4}}} or {{{k = (-8 - 2*sqrt(62))/4}}}



Now break up the fraction



{{{k=-8/4+2*sqrt(62)/4}}} or {{{k=-8/4-2*sqrt(62)/4}}}



Simplify



{{{k=-2+sqrt(62)/2}}} or {{{k=-2-sqrt(62)/2}}}



So these expressions approximate to


{{{k=1.93700393700591}}} or {{{k=-5.93700393700591}}}



So our solutions are:

{{{k=1.93700393700591}}} or {{{k=-5.93700393700591}}}