Respuesta :
We will have the following:
[tex]\begin{gathered} x^2-6x+7=k(2x-3)\Rightarrow x^2-6x+7=2kx-3k \\ \\ \Rightarrow x^2-6x-2kx+7+3k=0 \end{gathered}[/tex]Now, we want to operate like terms and take to "solve for x" using the quadratic expression, that is:
[tex]\Rightarrow x^2+(-6-2k)x+(7+3k)=0\Rightarrow x=\frac{-(-6-2k)\pm\sqrt{(-6-2k)^2-4(1)(7+3k)}}{2(1)}[/tex]No, in order to determine the possible real values for "k" we have to analyze the value under the root, so:
[tex]\begin{gathered} \sqrt{(-6-2k)^2-4(1)(7+3k)}\Rightarrow(-6-2k)^2-4(1)(7+3k)\ge0 \\ \\ \Rightarrow36+24k+4k^2-28-12k\ge0\Rightarrow8+12k+4k^2\ge0 \end{gathered}[/tex]Now, we will have to use the quadratic expression again to determine the values for "k" that make the inequality true, that is:
[tex]\begin{gathered} k=\frac{-(12)\pm\sqrt{(12)^2-4(4)(8)}}{2(4)} \\ \\ \Rightarrow k\leq-2 \\ \\ and \\ \\ \Rightarrow k\ge-1 \end{gathered}[/tex]So, the values for which the inequality and thus the expression under the root are true are then given by:
[tex]k\leq-2\wedge x\ge-1[/tex]In other notation:
[tex](-\infty,2]\cup[-1,\infty)[/tex]