Given \[ x^{2}+(k-1) x=-k+2 \] Show that the roots are real and rational for all

Solve the equation \( x^{2}+(k-1)x=-k+2 \). Solve the equation by following steps: - step0: Solve for \(x\): \(x^{2}+\left(k-1\right)x=-k+2\) - step1: Move the expression to the left side: \(x^{2}+\left(k-1\right)x-\left(-k+2\right)=0\) - step2: Remove the parentheses: \(x^{2}+\left(k-1\right)x+k-2=0\) - step3: Expand the expression: \(x^{2}+kx-x+k-2=0\) - step4: Factor the expression: \(\left(x+k-2\right)\left(x+1\right)=0\) - step5: Separate into possible cases: \(\begin{align}&x+k-2=0\\&x+1=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=-k+2\\&x=-1\end{align}\) The roots of the equation \(x^{2}+(k-1)x=-k+2\) are \(x=-k+2\) and \(x=-1\). To show that the roots are real and rational for all values of \(k\), we need to analyze the nature of the roots. 1. Real roots: The roots are real because they are in the form of \(x=a\) and \(x=b\), where \(a\) and \(b\) are real numbers. 2. Rational roots: The roots are rational because they are in the form of \(x=\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\). In this case, the roots are \(x=-k+2\) and \(x=-1\), which are rational for all values of \(k\). Therefore, the roots of the equation \(x^{2}+(k-1)x=-k+2\) are real and rational for all values of \(k\).