In a transformation, an object of area 64cm² is mapped onto an image whose area is 32cm². Given that the determinant of the matrix of transformation is (x+4 x), Find the value of x.

Uwiano wa eneo la picha kwa eneo la kitu ni sawa na thamani kamili ya kiamua cha matriksi ya mabadiliko.

$$\frac{\text{Eneo la picha}}{\text{Eneo la kitu}} = |\det(M)|$$

Tunapewa:
Eneo la kitu = $64 \text{cm}^2$
Eneo la picha = $32 \text{cm}^2$
Matriksi ya mabadiliko $M = \begin{pmatrix} x+4 & x \\ 2 & 1 \end{pmatrix}$

Kwa matriksi $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, kiamua ni $ad - bc$.
Hapa, $a = x+4$, $b = x$, $c = 2$, $d = 1$.

$$\det(M) = (x+4)(1) - (x)(2)$$ $$\det(M) = x+4 - 2x$$ $$\det(M) = 4 - x$$

$$\frac{32}{64} = |4 - x|$$ $$\frac{1}{2} = |4 - x|$$

Kuna uwezekano mbili kwa $4 - x$:
Kesi 1: $4 - x = \frac{1}{2}$

$$x = 4 - \frac{1}{2}$$ $$x = \frac{8}{2} - \frac{1}{2}$$ $$x = \frac{7}{2}$$

Kesi 2: $4 - x = -\frac{1}{2}$

$$x = 4 + \frac{1}{2}$$ $$x = \frac{8}{2} + \frac{1}{2}$$ $$x = \frac{9}{2}$$

Kwa hivyo, kuna thamani mbili zinazowezekana za $x$.

Thamani za $x$ ni $\boxed{\frac{7}{2} \text{ au } \frac{9}{2}}$.