QUESTION 4 4.1 Solve for \( x \) : 4.1.1 \( x(x-1)=0 \) 4.1.2 \( 2 x^{2}-4 x-5=0 \) (correct to TWO decimal places) \( 4.13 \quad 5^{\prime}=\frac{1}{125} \) \( 4.14(x-3)(2-x)>0 \) 4.2 Given \( f(x)=x+1 \) and \( g(x)=\frac{-4}{x-3} \) 4.2.1 For which values of \( x \) will \( g(x) \) be undefined? 4.2.2 Solve for \( x \) if \( f(x)=g(x) \)

Solve the equation \( x(x-1)=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x\left(x-1\right)=0\) - step1: Separate into possible cases: \(\begin{align}&x-1=0\\&x=0\end{align}\) - step2: Solve the equation: \(\begin{align}&x=1\\&x=0\end{align}\) - step3: Rewrite: \(x_{1}=0,x_{2}=1\) Solve the equation \( 2x^{2}-4x-5=0 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(2x^{2}-4x-5=0\) - step1: Solve using the quadratic formula: \(x=\frac{4\pm \sqrt{\left(-4\right)^{2}-4\times 2\left(-5\right)}}{2\times 2}\) - step2: Simplify the expression: \(x=\frac{4\pm \sqrt{\left(-4\right)^{2}-4\times 2\left(-5\right)}}{4}\) - step3: Simplify the expression: \(x=\frac{4\pm \sqrt{56}}{4}\) - step4: Simplify the expression: \(x=\frac{4\pm 2\sqrt{14}}{4}\) - step5: Separate into possible cases: \(\begin{align}&x=\frac{4+2\sqrt{14}}{4}\\&x=\frac{4-2\sqrt{14}}{4}\end{align}\) - step6: Simplify the expression: \(\begin{align}&x=\frac{2+\sqrt{14}}{2}\\&x=\frac{4-2\sqrt{14}}{4}\end{align}\) - step7: Simplify the expression: \(\begin{align}&x=\frac{2+\sqrt{14}}{2}\\&x=\frac{2-\sqrt{14}}{2}\end{align}\) - step8: Rewrite: \(x_{1}=\frac{2-\sqrt{14}}{2},x_{2}=\frac{2+\sqrt{14}}{2}\) Solve the equation \( 5^{x}=\frac{1}{125} \). Solve the equation by following steps: - step0: Solve for \(x\): \(5^{x}=\frac{1}{125}\) - step1: Rewrite in exponential form: \(5^{x}=5^{-3}\) - step2: Set the exponents equal: \(x=-3\) The inequality \( (x-3)(2-x)>0 \) cannot be solved using the provided method. Given \( f(x)=x+1 \) and \( g(x)=\frac{-4}{x-3} \): 1. For which values of \( x \) will \( g(x) \) be undefined? \( g(x) \) will be undefined when the denominator \( x-3 \) is equal to zero. Therefore, \( g(x) \) will be undefined when \( x = 3 \). 2. Solve for \( x \) if \( f(x)=g(x) \): \( f(x) = g(x) \) implies \( x+1 = \frac{-4}{x-3} \). To solve for \( x \), we can cross multiply and solve the resulting equation. Solve the equation by following steps: - step0: Solve for \(x\): \(x+1=\frac{-4}{x-3}\) - step1: Find the domain: \(x+1=\frac{-4}{x-3},x\neq 3\) - step2: Rewrite the fraction: \(x+1=-\frac{4}{x-3}\) - step3: Multiply both sides of the equation by LCD: \(\left(x+1\right)\left(x-3\right)=-\frac{4}{x-3}\times \left(x-3\right)\) - step4: Simplify the equation: \(x^{2}-2x-3=-4\) - step5: Move the expression to the left side: \(x^{2}-2x-3-\left(-4\right)=0\) - step6: Subtract the numbers: \(x^{2}-2x+1=0\) - step7: Factor the expression: \(\left(x-1\right)^{2}=0\) - step8: Simplify the expression: \(x-1=0\) - step9: Move the constant to the right side: \(x=0+1\) - step10: Remove 0: \(x=1\) - step11: Check if the solution is in the defined range: \(x=1,x\neq 3\) - step12: Find the intersection: \(x=1\) Therefore, the solution to the equation \( f(x)=g(x) \) is \( x = 1 \).