Question 943236

{{{y=kx^2-5x+2}}} that will result in the intersection of the line {{{y=-3x+4 }}}with the quadratic at

remember that if
{{{b^2 - 4ac > 0}}} you will have two solutions
{{{b^2 - 4ac = 0}}} you will have 1 solution (2 equal ones)
{{{b^2 - 4ac < 0}}} , there is no solution
so apply those conditions for each of your questions 

{{{y=kx^2-5x+2  }}}

{{{y=-3x+4}}}
-------------

{{{kx^2-5x+2 =-3x+4}}}

{{{kx^2-5x+3x+2 -4=0}}}

{{{kx^2-2x -2=0}}} => {{{a=k}}}, {{{b=-2}}},{{{ c=-2}}}

{{{b^2 - 4ac > 0}}} you will have two solutions

{{{(-2)^2 - 4k(-2)> 0 }}}

{{{4+ 8k> 0 }}}

 {{{8k> -4 }}}

{{{ k> -4/8 }}}

{{{k> -1/2}}}

{{{b^2 - 4ac = 0}}} you will have 1 solution (2 equal ones)


{{{(-2)^2 - 4k(-2)=0 }}}

{{{4+ 8k= 0 }}}

 {{{8k= -4 }}}

 {{{k=-4/8 }}}

{{{k=-1/2}}}

{{{b^2 - 4ac < 0}}} , there is no solution

{{{(-2)^2 - 4k(-2)<0 }}}

{{{4+ 8k<0 }}}

 {{{8k< -4}}} 

 {{{k<-4/8 }}}

{{{k<-1/2}}}


{{{y=kx^2-5x+2}}}    for {{{k > -1/2}}}, 2 solutions ; if {{{k=1}}} and {{{1 > -1/2}}}, we have

{{{y=x^2-5x+2}}}

then
{{{y=x^2-5x+2}}}

{{{y=-3x+4}}}
----------------
{{{x^2-5x+2=-3x+4}}}

{{{x^2-5x+3x+2-4=0}}}

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

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

{{{x = (-(-2) +- sqrt((-2)^2-4*1*(-2) ))/(2*1) }}}

{{{x = (2 +- sqrt(4+8 ))/2 }}}

{{{x = (2 +- sqrt(12 ))/2 }}}

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

{{{x = 1+sqrt(3)}}}

{{{x = 1-sqrt(3)}}}

if {{{k=-1/2}}}, we will have

{{{y=x^2-5x+2}}
{{{y=-3x+4}}}
----------------
{{{(-1/2)x^2-5x+2=-3x+4}}}

{{{-x^2-10x+4=-6x+8}}}

{{{0=x^2+10x-6x+8-4}}}

{{{x^2+4x+4=0}}}

{{{x^2+4x+2^2=0}}}

{{{(x+2)^2=0}}}

{{{(x+2)=0}}} if {{{x=-2}}} => one double solution



if {{{k=-1}}}, we will have

{{{y=x^2-5x+2}}
{{{y=-3x+4}}}
----------------

{{{-1x^2-5x+2=-3x+4}}}

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

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

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

{{{x = (-(2) +- sqrt((2)^2-4*1*(2) ))/(2*1) }}}

{{{x = (-2 +- sqrt(4-8 ))/2 }}}

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

{{{x = (-2 +- 2i)/2 }}}

solutions:

{{{x = -1 + 2i }}}

{{{x = -1 - 2i }}}