Sehyr solved the equation w2−7=11 . She seems to have made an error in her work because the solution she found does not work. Observe Sehyr’s work to find her error and correct it to find the solution to the equation. Sehyr’s work: w2−7=11 w2−7+7=11+7 Using the Addition Property of Equality w22=182 Using the Division Property of Equality w=9 (1 point) The solution to the equation is

Answer 1

Let's go through Sehyr's work step by step to identify where she made the error and find the correct solution to the equation \( w^2 - 7 = 11 \).

1. Original equation: \[ w^2 - 7 = 11 \]

2. Adding 7 to both sides: Sehyr correctly applies the Addition Property of Equality: \[ w^2 - 7 + 7 = 11 + 7 \] This simplifies to: \[ w^2 = 18 \]

3. Error in division: After getting \( w^2 = 18 \), Sehyr incorrectly wrote: \[ w^2 = 182 \] This is not correct; she should have written \( w^2 = 18 \).

4. Taking the square root: To solve for \( w \), we take the square root of both sides: \[ w = \sqrt{18} \quad \text{or} \quad w = -\sqrt{18} \]

5. Simplifying \(\sqrt{18}\): \[ \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2} \]

Thus, the solutions to the equation are: \[ w = 3\sqrt{2} \quad \text{or} \quad w = -3\sqrt{2} \]

Therefore, the correct solution to the equation \( w^2 - 7 = 11 \) is: \[ w = \pm 3\sqrt{2} \]

Answer 2

what is the answer?

Answer 3

The correct solutions to the equation \( w^2 - 7 = 11 \) are:

\[ w = 3\sqrt{2} \quad \text{and} \quad w = -3\sqrt{2} \]

So, the answer can be summarized as \( w = \pm 3\sqrt{2} \).

Answer 4

what is sqrt

Answer 5

The square root (often denoted as \(\sqrt{}\)) of a number \(x\) is a value that, when multiplied by itself, gives the original number \(x\).

For example:

• \(\sqrt{9} = 3\) because \(3 \times 3 = 9\).
• \(\sqrt{16} = 4\) because \(4 \times 4 = 16\).
• \(\sqrt{25} = 5\) because \(5 \times 5 = 25\).

If the number is not a perfect square, the square root is an irrational number. For example:

• \(\sqrt{2} \approx 1.414\)
• \(\sqrt{3} \approx 1.732\)

In the case of the solutions \( w = 3\sqrt{2} \) and \( w = -3\sqrt{2} \):

• The expression \( 3\sqrt{2} \) means \( 3 \) multiplied by the square root of \( 2 \).

The square root can also have two values: a positive and a negative. Therefore, for \( w^2 = 18 \), the solutions are \( w = \sqrt{18} \) and \( w = -\sqrt{18} \). This is why we express \( w \) as \( \pm 3\sqrt{2} \).