# Finding values of k for different points of intersection

Find the values of K for which $y=kx-3$ does not intersect $y=x^2-2x+1$

To solve it my textbook says to do this.

$x^2-2x+1=kx-3$

$x^2-(2+k)x+4=0$

Then we use the discriminant.

$(2+k)^2-4\times 1\times 4<0$

$k^2+4k-12<0$

$(k+6)(k-2)<0$

So -6 < k < 2

My question is why does this work? What is actually happening for this to work? I've used Desmos and I can't figure out why the -6 < k < 2 part is equal to the resulting equation $k^2+4k-12<0$ between the two points of intersection on the x-axis.

Math Gnome

87

## Answer

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Erdos

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The answer is accepted.

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