Let the width of the "super size" color print be denoted by $w$, and the length by $l$. According to the problem, the length is 2 inches more than the width, so we can express the length as:

$l = w + 2$

The area of the rectangle is given by:

$\text{Area} = l \times w = 24 \, \text{square inches}$

Substitute $l = w + 2$ into the area equation:

$(w + 2) \times w = 24$

Expanding the equation:

$w^2 + 2w = 24$

Rearranging it to form a quadratic equation:

$w^2 + 2w - 24 = 0$

Now, solve this quadratic equation using the quadratic formula:

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

Where $a = 1$, $b = 2$, and $c = -24$. Substituting these values:

$w = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2(1)}$ $w = \frac{-2 \pm \sqrt{4 + 96}}{2}$ $w = \frac{-2 \pm \sqrt{100}}{2}$ $w = \frac{-2 \pm 10}{2}$

Thus, the two possible values for $w$ are:

$w = \frac{-2 + 10}{2} = \frac{8}{2} = 4$ $w = \frac{-2 - 10}{2} = \frac{-12}{2} = -6$

Since the width cannot be negative, $w = 4$. Therefore, the width is 4 inches.

Now, substitute $w = 4$ into the expression for the length:

$l = w + 2 = 4 + 2 = 6$

Thus, the length is 6 inches.

Final Answer:

• Width: 4 inches
• Length: 6 inches

Would you like further details or explanations?

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Here are five related questions you could explore:

1. What would the dimensions be if the area were 48 square inches with the same condition on length and width?
2. How would the equation change if the length were 3 inches more than the width instead of 2?
3. Can you solve the quadratic equation by factoring instead of using the quadratic formula?
4. What is the perimeter of the "super size" print with the dimensions found?
5. How can you express the area formula for a rectangle in terms of length only?

Tip: When solving quadratic equations, always check if the quadratic can be factored for a simpler solution before using the quadratic formula.