Question

question 11 noah’s teacher gave him incomplete information about a square’s area. noah says there are two squares that would match this criteria. what are the dimensions and area of the two possible squares? (chart with 4r² and 49 and some? marks)

Explanation:

Step1: Analyze the given terms

We have two terms: \(4x^{2}\) and \(49\). Notice that \(4x^{2}=(2x)^{2}\) and \(49 = 7^{2}\). This seems to be related to the area of a square, and also, if we consider the product of two binomials (maybe a square of a sum or difference), we can use the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\) or \((a - b)^{2}=a^{2}-2ab + b^{2}\).

Step2: First possible square (sum)

Let's assume the square is \((2x + 7)^{2}\). Expanding this using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), where \(a = 2x\) and \(b=7\):
\[

$$\begin{align*} (2x + 7)^{2}&=(2x)^{2}+2\times(2x)\times7+7^{2}\\ &=4x^{2}+28x + 49 \end{align*}$$

\]
So the area of the square is \(4x^{2}+28x + 49\), and the side length is \(2x + 7\).

Step3: Second possible square (difference)

Let's assume the square is \((2x-7)^{2}\). Expanding this using the formula \((a - b)^{2}=a^{2}-2ab + b^{2}\), where \(a = 2x\) and \(b = 7\):
\[

$$\begin{align*} (2x-7)^{2}&=(2x)^{2}-2\times(2x)\times7+7^{2}\\ &=4x^{2}-28x + 49 \end{align*}$$

\]
So the area of the square is \(4x^{2}-28x + 49\), and the side length is \(2x - 7\).

Answer:

The two possible squares have side lengths \(2x + 7\) (area \(4x^{2}+28x + 49\)) and \(2x - 7\) (area \(4x^{2}-28x + 49\)).