Answer :
To solve the quadratic equation [tex]\(x^2 + 2x + 1 = 0\)[/tex], we can use the quadratic formula which states that for any quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex], the solutions for [tex]\(x\)[/tex] are given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Step-by-step, we apply this formula to our specific equation [tex]\(x^2 + 2x + 1 = 0\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = 1\)[/tex].
1. Calculate the discriminant:
The discriminant ([tex]\(\Delta\)[/tex]) is given by [tex]\(b^2 - 4ac\)[/tex]. Substituting the values, we get:
[tex]\[ \Delta = 2^2 - 4(1)(1) \][/tex]
[tex]\[ \Delta = 4 - 4 \][/tex]
[tex]\[ \Delta = 0 \][/tex]
The discriminant is 0, which indicates that the quadratic equation has one real repeated root.
2. Calculate the solutions:
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{\Delta}}{2a} \)[/tex], we substitute the values [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(\Delta = 0\)[/tex]:
[tex]\[ x = \frac{-2 \pm \sqrt{0}}{2 \times 1} \][/tex]
This simplifies to:
[tex]\[ x = \frac{-2 \pm 0}{2} \][/tex]
[tex]\[ x = \frac{-2}{2} \][/tex]
[tex]\[ x = -1 \][/tex]
Since the discriminant is 0, the two solutions are not distinct and both are [tex]\(x = -1\)[/tex].
Therefore, the quadratic equation [tex]\(x^2 + 2x + 1 = 0\)[/tex] has a discriminant of 0, and the single (repeated) solution is:
[tex]\[ x = -1 \][/tex]
Thus, the roots of the equation are:
[tex]\[ x_1 = -1 \quad \text{and} \quad x_2 = -1 \][/tex]
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Step-by-step, we apply this formula to our specific equation [tex]\(x^2 + 2x + 1 = 0\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = 1\)[/tex].
1. Calculate the discriminant:
The discriminant ([tex]\(\Delta\)[/tex]) is given by [tex]\(b^2 - 4ac\)[/tex]. Substituting the values, we get:
[tex]\[ \Delta = 2^2 - 4(1)(1) \][/tex]
[tex]\[ \Delta = 4 - 4 \][/tex]
[tex]\[ \Delta = 0 \][/tex]
The discriminant is 0, which indicates that the quadratic equation has one real repeated root.
2. Calculate the solutions:
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{\Delta}}{2a} \)[/tex], we substitute the values [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(\Delta = 0\)[/tex]:
[tex]\[ x = \frac{-2 \pm \sqrt{0}}{2 \times 1} \][/tex]
This simplifies to:
[tex]\[ x = \frac{-2 \pm 0}{2} \][/tex]
[tex]\[ x = \frac{-2}{2} \][/tex]
[tex]\[ x = -1 \][/tex]
Since the discriminant is 0, the two solutions are not distinct and both are [tex]\(x = -1\)[/tex].
Therefore, the quadratic equation [tex]\(x^2 + 2x + 1 = 0\)[/tex] has a discriminant of 0, and the single (repeated) solution is:
[tex]\[ x = -1 \][/tex]
Thus, the roots of the equation are:
[tex]\[ x_1 = -1 \quad \text{and} \quad x_2 = -1 \][/tex]
