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Solve the quadratic equation numerically (using tables of [tex]$x[tex]$[/tex]- and [tex]$[/tex]y$[/tex]-values).

[tex]x^2 + 2x + 1 = 0[/tex]

A. [tex]x = -1[/tex]
B. [tex]x = 1[/tex] or [tex]x = -3[/tex]
C. [tex]x = 3[/tex]
D. [tex]x = 2[/tex] or [tex]x = -1[/tex]

Please select the best answer from the choices provided.
A.
B.
C.
D.

Sagot :

GinnyAnsweridn
To solve the quadratic equation \( x^2 + 2x + 1 = 0 \) numerically, we need to find the values of \( x \) for which the equation holds true. Let's follow the steps to solve this problem:

1. Identify the quadratic equation:
[tex]\[ x^2 + 2x + 1 = 0 \][/tex]

2. Rewrite the equation and factorize it (if possible):
Notice that the quadratic expression can be factored:
[tex]\[ x^2 + 2x + 1 = (x + 1)^2 \][/tex]

3. Set the factored form equal to zero:
[tex]\[ (x + 1)^2 = 0 \][/tex]

4. Solve for \( x \) by taking the square root of both sides:
[tex]\[ x + 1 = 0 \][/tex]

5. Isolate \( x \):
[tex]\[ x = -1 \][/tex]

So, the solution to the quadratic equation is \( x = -1 \).

Given the solution, let's check which of the given options includes \( x = -1 \):

- Option A: \( x = -1 \) (correct)
- Option B: \( x = 1 \) or \( x = -3 \) (incorrect)
- Option C: \( x = 3 \) (incorrect)
- Option D: \( x = 2 \) or \( x = -1 \) (incorrect)

The correct answer that matches the solution \( x = -1 \) is:

A
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