Let the width of the strips removed be \( x \) cm. When the strips of width \( x \) are cut off one side and one end of the rectangle, the dimensions of the remaining piece of cardboard will be:
- Length: \( 17 - x \) cm (after cutting \( x \) from one side)
- Width: \( 14 - x \) cm (after cutting \( x \) from one end)
The area \( A \) of the remaining piece is given to be \( 108 \) cm². Thus, we can write the equation:
\[ (17 - x)(14 - x) = 108 \]
Expanding the left-hand side:
\[ 17 \cdot 14 - 17x - 14x + x^2 = 108 \]
Calculating \( 17 \cdot 14 \):
\[ 238 - 31x + x^2 = 108 \]
Now, rearranging the equation gives us:
\[ x^2 - 31x + 238 - 108 = 0 \]
This simplifies to:
\[ x^2 - 31x + 130 = 0 \]
Next, we apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -31 \), and \( c = 130 \):
Calculating the discriminant:
\[ b^2 - 4ac = (-31)^2 - 4 \cdot 1 \cdot 130 = 961 - 520 = 441 \]
Now, we find the square root of the discriminant:
\[ \sqrt{441} = 21 \]
Applying the quadratic formula:
\[ x = \frac{31 \pm 21}{2} \]
Calculating the two possible values for \( x \):
- \( x = \frac{31 + 21}{2} = \frac{52}{2} = 26 \)
- \( x = \frac{31 - 21}{2} = \frac{10}{2} = 5 \)
Since \( x \) represents the width of the strips removed, it must be less than both dimensions of the cardboard (17 cm and 14 cm). Thus, we reject \( x = 26 \) cm as it is greater than both dimensions.
Hence, the width of the strips removed is:
\[ \boxed{5 \text{ cm}} \]
To verify, if \( x = 5 \), the dimensions of the remaining piece of cardboard will be:
- Length: \( 17 - 5 = 12 \) cm
- Width: \( 14 - 5 = 9 \) cm
Calculating the area:
\[ 12 \times 9 = 108 \text{ cm}^2 \]
Thus, the solution is confirmed that the width of the strips removed is indeed \( \boxed{5} \text{ cm} \).