Let the width of the rectangle be $w$ meters. According to the problem, the length $l$ is 5 meters less than twice the width. So, we can express the length as:

$l = 2w - 5$

The area $A$ of a rectangle is given by the formula:

$A = l \times w$

Substituting the given area and the expression for $l$:

$273 = (2w - 5) \times w$

Now, expand and simplify the equation:

$273 = 2w^2 - 5w$

Rearrange it into a standard quadratic form:

$2w^2 - 5w - 273 = 0$

We'll solve this quadratic equation using the quadratic formula:

$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $2w^2 - 5w - 273 = 0$, the coefficients are $a = 2$, $b = -5$, and $c = -273$. Substituting these into the quadratic formula:

$w = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-273)}}{2(2)}$ $w = \frac{5 \pm \sqrt{25 + 2184}}{4}$ $w = \frac{5 \pm \sqrt{2209}}{4}$ $w = \frac{5 \pm 47}{4}$

Now, calculate the two possible solutions:

$w = \frac{5 + 47}{4} = \frac{52}{4} = 13$ $w = \frac{5 - 47}{4} = \frac{-42}{4} = -10.5$

Since the width can't be negative, the width is $w = 13$ meters.

Now, substitute $w = 13$ back into the expression for the length:

$l = 2(13) - 5 = 26 - 5 = 21 \text{ meters}$

Thus, the dimensions of the rectangle are:

• Width: 13 meters
• Length: 21 meters

Would you like further details or have any questions?

Related Questions:

1. How is the quadratic formula derived?
2. What if the width were given, how would we find the length and area?
3. Can the quadratic equation have multiple physical interpretations?
4. What other methods could solve this quadratic equation?
5. How can we use factoring instead of the quadratic formula here?

Tip:

Always check if a quadratic can be factored easily before applying the quadratic formula for faster solutions.