Solve for [tex]$x$[/tex] using the quadratic formula: [tex]$x^2 + 2x + 1 = 0$[/tex]

[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]

A. [tex]$x = 2$[/tex]
B. [tex]$x = 1$[/tex]
C. [tex][tex]$x = 0$[/tex][/tex]
D. [tex]$x = -1$[/tex]



Answer :

To solve the quadratic equation [tex]\( x^2 + 2x + 1 = 0 \)[/tex] using the quadratic formula, let's follow these steps carefully:

1. Identify the coefficients:
The quadratic equation is given in the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex]. By comparing, we can identify:
[tex]\[ a = 1 \quad \text{(coefficient of \(x^2\))} \\ b = 2 \quad \text{(coefficient of \(x\))} \\ c = 1 \quad \text{(constant term)} \][/tex]

2. Write the quadratic formula:
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

3. Calculate the discriminant:
The discriminant is given by [tex]\( b^2 - 4ac \)[/tex]. Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \text{Discriminant} = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 1 = 4 - 4 = 0 \][/tex]

4. Substitute the values into the quadratic formula:
Since the discriminant is 0, we will have one real solution (a repeated root):
[tex]\[ x = \frac{-b \pm \sqrt{0}}{2a} = \frac{-2 \pm 0}{2 \cdot 1} = \frac{-2}{2} = -1 \][/tex]

Thus, the solution to the quadratic equation [tex]\( x^2 + 2x + 1 = 0 \)[/tex] is:
[tex]\[ x = -1 \][/tex]

This is a repeated root, meaning the quadratic equation has a double root at [tex]\( x = -1 \)[/tex].

So the answer is:
[tex]\[ x = -1 \][/tex]