(1) \( \sqrt{2 x^{2}-x-1}+1=x \) (2) \( \sqrt{x+2}-x+4=0 \) (3) \( 6=\sqrt{x+12}-x \) (4) \( x+\sqrt{-4 x-3}=0 \)

Solve the equation \( \sqrt{2x^{2}-x-1}+1=x \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sqrt{2x^{2}-x-1}+1=x\) - step1: Find the domain: \(\sqrt{2x^{2}-x-1}+1=x ,x \in \left(-\infty,-\frac{1}{2}\right]\cup \left[1,+\infty\right)\) - step2: Move the constant to the right-hand side: \(\sqrt{2x^{2}-x-1}=x-1\) - step3: Evaluate: \(\sqrt{2x^{2}-x-1}=x-1,x-1\geq 0\) - step4: Evaluate: \(\sqrt{2x^{2}-x-1}=x-1,x\geq 1\) - step5: Solve the equation: \(\begin{align}&x=1\\&x=-2\end{align},x\geq 1\) - step6: Find the intersection: \(x=1\) - step7: Check if the solution is in the defined range: \(x=1,x \in \left(-\infty,-\frac{1}{2}\right]\cup \left[1,+\infty\right)\) - step8: Find the intersection: \(x=1\) - step9: Check the solution: \(x=1\) Solve the equation \( \sqrt{x+2}-x+4=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sqrt{x+2}-x+4=0\) - step1: Find the domain: \(\sqrt{x+2}-x+4=0,x\geq -2\) - step2: Move the expression to the right-hand side: \(\sqrt{x+2}=x-4\) - step3: Evaluate: \(\sqrt{x+2}=x-4,x-4\geq 0\) - step4: Evaluate: \(\sqrt{x+2}=x-4,x\geq 4\) - step5: Solve the equation: \(\begin{align}&x=7\\&x=2\end{align},x\geq 4\) - step6: Find the intersection: \(x=7\) - step7: Check if the solution is in the defined range: \(x=7,x\geq -2\) - step8: Find the intersection: \(x=7\) - step9: Check the solution: \(x=7\) Solve the equation \( 6=\\sqrt{x+12}-x \). Solve the equation by following steps: - step0: Solve for \(x\): \(6=\sqrt{x+12}-x\) - step1: Find the domain: \(6=\sqrt{x+12}-x,x\geq -12\) - step2: Swap the sides: \(\sqrt{x+12}-x=6\) - step3: Move the variable to the right-hand side: \(\sqrt{x+12}=6+x\) - step4: Evaluate: \(\sqrt{x+12}=6+x,6+x\geq 0\) - step5: Evaluate: \(\sqrt{x+12}=6+x,x\geq -6\) - step6: Solve the equation: \(\begin{align}&x=-3\\&x=-8\end{align},x\geq -6\) - step7: Find the intersection: \(x=-3\) - step8: Check if the solution is in the defined range: \(x=-3,x\geq -12\) - step9: Find the intersection: \(x=-3\) - step10: Check the solution: \(x=-3\) Solve the equation \( x+\\sqrt{-4x-3}=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(x+\sqrt{-4x-3}=0\) - step1: Find the domain: \(x+\sqrt{-4x-3}=0,x\leq -\frac{3}{4}\) - step2: Move the variable to the right-hand side: \(\sqrt{-4x-3}=-x\) - step3: Evaluate: \(\sqrt{-4x-3}=-x,-x\geq 0\) - step4: Evaluate: \(\sqrt{-4x-3}=-x,x\leq 0\) - step5: Solve the equation: \(\begin{align}&x=-1\\&x=-3\end{align},x\leq 0\) - step6: Find the intersection: \(\begin{align}&x=-1\\&x=-3\end{align}\) - step7: Check if the solution is in the defined range: \(\begin{align}&x=-1\\&x=-3\end{align},x\leq -\frac{3}{4}\) - step8: Find the intersection: \(\begin{align}&x=-1\\&x=-3\end{align}\) - step9: Check the solution: \(\begin{align}&x=-1\\&x=-3\end{align}\) - step10: Rewrite: \(x_{1}=-3,x_{2}=-1\) The solutions to the given equations are: 1. \( x = 1 \) 2. \( x = 7 \) 3. \( x = -3 \) 4. \( x = -3 \) or \( x = -1 \)