Question 156386
<pre>
show that the equations {{{y=kx-1}}} and {{{y=x^2+(k-1)x+k^2}}}
have no real common solution for all values of k.
<pre><font size = 4 color = "indigo"><b>
{{{system(y=kx-1, y=x^2+(k-1)x+k^2)}}}

Set the right side of the first equation equal to
the right side of the second equation, since both
equal to y:

{{{kx-1=x^2+(k-1)x+k^2}}}

Distribute to remove the parentheses:

{{{kx-1=x^2+kx-x+k^2}}}

Get 0 on the left side by adding {{{-kx+1}}}
to both sides:

{{{kx-1-kx+1=x^2+kx-x+k^2-kx+1}}}

{{{0=x^2-x+k^2+1}}}

Swap sides:

{{{x^2-x+k^2+1=0}}}

Group the last two terms on the left in 
parentheses:

We need to find the DISCRIMINANT

{{{DISCRIMINANT_OF_ax^2+bx+c = b^2-4ac}}}
{{{x^2-x+(k^2+1)=0}}}

is the same as

{{{1x^2-1x+(k^2+1)=0}}}

So {{{a=1}}}, {{{b=-1}}}, {{{c=k^2+1}}}

{{{DISCRIMINANT=b^2-4ac}}}
{{{DISCRIMINANT=(-1)^2-4(1)(k^2+1)}}}
{{{DISCRIMINANT=1-4(k^2+1)}}}
{{{DISCRIMINANT=1-4k^2-4}}}
{{{DISCRIMINANT=-3-4k^2}}}

There are no real solutions when the
{{{DISCRIMINANT<0}}}

and {{{-3-4k^2}}} is ALWAYS negative,
since {{{k^2}}} is never negative.

So there can be no real common solutions 
for any real value of {{{k}}}.

Edwin</pre></font></b>