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a rectangle has a length that is 5 inches greater than its width, and i…

Question

a rectangle has a length that is 5 inches greater than its width, and its area is 104 square inches. the equation ((x + 5)x = 104) represents the situation, where (x) represents the width of the rectangle.
((x + 5)x = 104)
(x^2 + 5x - 104 = 0)
determine the solutions of the equation. what solution makes sense for the situation?
(x = square)
what are the dimensions of the rectangle?
width = square inches
length = square inches

Explanation:

Step1: Identify quadratic coefficients

For $x^2 + 5x - 104 = 0$, we have $a=1$, $b=5$, $c=-104$.

Step2: Apply quadratic formula

Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Substitute values:

$$ x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-104)}}{2(1)} $$

Step3: Calculate discriminant

Compute $\sqrt{25 + 416} = \sqrt{441} = 21$

Step4: Find both solutions

First solution: $x = \frac{-5 + 21}{2} = \frac{16}{2} = 8$
Second solution: $x = \frac{-5 - 21}{2} = \frac{-26}{2} = -13$

Step5: Select valid solution

Width cannot be negative, so $x=8$.

Step6: Calculate rectangle length

Length = $x + 5 = 8 + 5 = 13$

Answer:

$x = 8$ or $x = -13$; the valid solution for the situation is $x=8$
width = 8 inches
length = 13 inches