QUESTION IMAGE
Question
solve each polynomial equation by completing the square and match it to the solution set.
| equation | solution set |
|---|---|
| 2. $n^2 - 18n = 40$ | b. $\\{-1, 3\\}$ |
| c. $\\{-2, 20\\}$ | |
| d. $\\{-20, 2\\}$ |
- choose
- choose
options for choosing: c, a, d, b
Step1: Solve \( n^2 - 2n - 3 = 0 \) by completing the square
First, move the constant term to the right: \( n^2 - 2n = 3 \)
Then, take half of the coefficient of \( n \) (which is -2), square it: \( (\frac{-2}{2})^2 = 1 \)
Add this to both sides: \( n^2 - 2n + 1 = 3 + 1 \)
This factors to \( (n - 1)^2 = 4 \)
Take square roots: \( n - 1 = \pm 2 \)
So \( n = 1 + 2 = 3 \) or \( n = 1 - 2 = -1 \). Solution set: \(\{-1, 3\}\) (matches option B)
Step2: Solve \( n^2 - 18n = 40 \) by completing the square
Move the constant term (40 is already on the right). Take half of -18: \( \frac{-18}{2} = -9 \), square it: \( (-9)^2 = 81 \)
Add to both sides: \( n^2 - 18n + 81 = 40 + 81 \)
Factor: \( (n - 9)^2 = 121 \)
Take square roots: \( n - 9 = \pm 11 \)
So \( n = 9 + 11 = 20 \) or \( n = 9 - 11 = -2 \). Solution set: \(\{-2, 20\}\) (matches option C)
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- B. \(\{-1, 3\}\)
- C. \(\{-2, 20\}\)