2.1 Solve for \( x \) \( 2.1 .1 \quad x(2 x-5)=0 \) 2.1.2 \( 3 x^{3}-2 x-8=0 \) \( 2.1 .3 \quad 5^{2 n-1}-1=0 \) 2,1.4 \( \quad x=y+x y \) \( 2.1 .5 \quad \frac{8 x^{3}-1}{2 x-1}=1 \) 2.2 The following inequality is given; \( -11<-2 x+1<-9 \); where \( x \in R \). 2.2.1 Solve for \( x \). 2.2.2 Hence, and without the use of a calculator, show that \( x=\sqrt{29} \) would satisfy the above inequality.

To solve \( 2.1.1 \), we set each factor of the equation \( x(2x - 5) = 0 \) to zero: \( x = 0 \) or \( 2x - 5 = 0 \) (which gives us \( x = 2.5 \)). So the solutions are \( x = 0 \) and \( x = 2.5 \). For \( 2.1.2 \), we can use techniques like synthetic division or the Rational Root Theorem to test possible rational roots. After checking values, you might find that \( x = 2 \) is a root, leading to the factorization and solving the quadratic that follows. Now onto \( 2.2 \), start by breaking the compound inequality into two parts: \( -11 < -2x + 1 \) and \( -2x + 1 < -9 \). Solving these separately, you would arrive at \( -5 < x < -4 \). To check if \( x = \sqrt{29} \) falls within this range, note that \( \sqrt{29} \approx 5.385\), which doesn’t satisfy the inequality. It's essential to ensure you plug \( x = \sqrt{29} \) back into the original inequalities to confirm that it holds for those bounds.