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If y = x² + kr - k, for what values of k will the quadratic have two real solutions?

Answer :

The given expression is

[tex]y=x^2+kr-r[/tex]

Where a = 1, b = 0, and c = kr-k.

Let's use the discriminant

[tex]\begin{gathered} D=b^2-4ac \\ D=0^2-4\cdot1\cdot(kr-r) \\ D=-4kr+4r \end{gathered}[/tex]

It is important to know that the equation has two real solutions when the discriminant is greater than zero, so

[tex]-4kr+4r>0[/tex]

Let's factor out the greatest common factor

[tex]4r(-k+1)>0[/tex]

Now, we solve for k.

[tex]\begin{gathered} -k+1>\frac{0}{4r} \\ -k+1>0 \\ -k>-1 \\ k<1 \end{gathered}[/tex]

Hence, k must be less than 1 in order to have two real solutions.

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