IDNLearn.com makes it easy to find the right answers to your questions. Ask anything and receive well-informed answers from our community of experienced professionals.
Sagot :
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]
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For clear and precise answers, choose IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.