rectangle has length \( 2 x \mathrm{~cm} \), perimeter 18 cm and area \( 10 \mathrm{~cm}^{2} \). i) Show that \( 2 x^{2}-9 x+5=0 \) \( x \)

To find \( x \) based on the given conditions, let's first express the width \( w \) in terms of \( x \). The perimeter \( P \) of a rectangle is given by the formula \( P = 2(l + w) \). Substituting the given length, we have: \[ 18 = 2(2x + w) \] Dividing both sides by 2, we get: \[ 9 = 2x + w \quad \Rightarrow \quad w = 9 - 2x \] Next, the area \( A \) of the rectangle is given by \( A = l \cdot w \). Using the length and the expression for the width, we have: \[ 10 = (2x)(9 - 2x) \] Expanding the equation: \[ 10 = 18x - 4x^2 \] Rearranging terms results in: \[ 4x^2 - 18x + 10 = 0 \] Dividing the entire equation by 2 yields: \[ 2x^2 - 9x + 5 = 0 \] Thus, we have shown that \( 2 x^{2}-9 x+5=0 \). Now, we can solve for \( x \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting \( a = 2 \), \( b = -9 \), and \( c = 5 \): \[ x = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 2 \cdot 5}}{2 \cdot 2} = \frac{9 \pm \sqrt{81 - 40}}{4} = \frac{9 \pm \sqrt{41}}{4} \] Therefore, the values for \( x \) can be further evaluated for numerical solutions!