To solve the quadratic equation [tex]\(-3 = x^2 + 4x + 1\)[/tex], we first need to rewrite it in standard quadratic form:
[tex]\[ x^2 + 4x + 1 + 3 = 0 \][/tex]
Simplifying the constants on the left side gives us:
[tex]\[ x^2 + 4x + 4 = 0 \][/tex]
In the standard quadratic form [tex]\( ax^2 + bx + c = 0 \)[/tex], the coefficients are:
[tex]\( a = 1 \)[/tex],
[tex]\( b = 4 \)[/tex],
[tex]\( c = 4 \)[/tex].
We calculate the discriminant using the formula:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = 4^2 - 4 \cdot 1 \cdot 4 = 16 - 16 = 0 \][/tex]
The discriminant [tex]\(\Delta\)[/tex] is 0.
Based on the value of the discriminant, we determine the number of real number solutions for the quadratic equation:
- If the discriminant is greater than 0, there are 2 distinct real solutions.
- If the discriminant is 0, there is exactly 1 real solution.
- If the discriminant is less than 0, there are no real solutions.
Since the discriminant is 0, the quadratic equation has exactly 1 real solution.
So, the discriminant is 0, and there is 1 real number solution for the equation [tex]\(-3 = x^2 + 4x + 1\)[/tex].
