Question 1071488
{{{system(y=2x+k,y=(x-1)^2+3)}}} --> {{{system(k=y-2x,(x-1)^2+3)=2x+k)}}} --> {{{system(k=y-2x,x^2-2x+1+3=2x+k)}}} --> {{{system(k=y-2x,x^2-2x+4-2x-k=0)}}} --> {{{system(k=y-2x,x^2-4x+4-k=0)}}}
If we chose a real {{{k}}} so that {{{x^2-4x+4-k=0}}} has 2 real solutions for {{{x}}} ,
we will have the corresponding real solutions for {{{y=2x+k}}} .
For a quadratic equation {{{ax^2+bx+c=0}}} to have 2 real solutions
(which could both be the same)
it must be {{{b^2-4ac>=0}}} .
For the two real solutions to be different we need {{{b^2-4ac>0}}} .
So, in the case of {{{x^2-4x+4-k=0}}} ,
where {{{a=1}}}, {{{b=-4}}} and {{{c=4-k}}} ,
to have 2 real solutions (which could both be the same) we must have
{{{(-4)^2-4*1*(4-k)>=0}}} --> {{{16-4(4-k)>=0}}} --> {{{16-16+4k>=0}}} --> {{{4k>=0}}} --> {{{highlight(k>=0)}}} .
As an interval it would be {{{"[ 0 ,"}}}{{{infinity}}}{{{")"}}} .
 
If we wanted to have two different real solutions,
we would need to have {{{highlight(k>0)}}} ,
and in interval notation that would be {{{"( 0 ,"}}}{{{infinity}}}{{{")"}}} .
 
IF YOU WERE STUDYING CALCULUS,
the answer would be obvious, if somewhat hard to explain the reasoning.
You would know that the slope of the tangent to {{{y=(x-1)^2+3)}}}
is {{{dy/dx=2(x-1)}}} ,
and since the slope of {{{y=2x+k}}} is {{{2}}} ,
the slope of both functions would be the same when
{{{2(x-1)=2}}} <--> {{{x-1=1}}} <--> {{{x=2}}} .
That point could be the point of tangency if
both functions have the same {{{y}}} value for {{{x=2}}} .
That would happen for a real {{{k}}} such that at {{{x=2}}} <--> {{{x-1=1}}}
{{{(x-1)^2+3=2x+k}}} , meaning that
{{{1^2+3=2*2+k}}}
{{{4=4+k}}} <---> {{{k=0}}} .
That point of tangency would be (2,4), with {{{y=2*2+0=4}}} .
For {{{x=2}}} <--> {{{x-1=1}}} ,
the function {{{y=(x-1)^2+3)}}} has {{{y(2)=4}}} at {{{x=2}}} .
For {{{k=0}}} both functions would have the {{{y(2)=4}}} at {{{x=2}}} {{{graph(300,300,-1,4,-2,8,-3,-4,(x-1)^2+3,2x)}}}
If {{{green(k)>0}}} function {{{y=2x+green(k)}}} will have {{{y(2)=2*2+green(k)=4+green(k)>4}}} , and will be passing through a point inside the parabola, crossing it twice:
If {{{red(k)<0}}} function {{{y=2x+red(k)}}} will have {{{y(2)=2*2+red(k)=4+red(k)<4}}} , and will be below the point (2,4) where the slope of {{{y=(x-1)^2+3)}}} is {{{2}}} ,
and since the slope of {{{y=(x-1)^2+3)}}} increase with increasing x,
the function {{{y=2x+red(k)}}} , with its constant {{{slope=2}}} will never be able to catch up.
{{{graph(300,300,-1,4,-2,8,2x-1.2 ,2x+0.8,(x-1)^2+3,2x)}}}