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Solve the quadratic equation.

[tex]\[
(x+4)(x+1)=0
\][/tex]

A. [tex]\(x=4\)[/tex] or [tex]\(x=1\)[/tex]

B. [tex]\(x=-4\)[/tex] or [tex]\(x=1\)[/tex]

C. [tex]\(x=-4\)[/tex] or [tex]\(x=-1\)[/tex]

D. [tex]\(x=4\)[/tex] or [tex]\(x=-1\)[/tex]

Sagot :

GinnyAnswer
To solve the quadratic equation [tex]\((x+4)(x+1)=0\)[/tex], we need to find the values of [tex]\(x\)[/tex] that make this equation true.

Start by setting each factor equal to zero:

1. [tex]\(x + 4 = 0\)[/tex]
2. [tex]\(x + 1 = 0\)[/tex]

Solve each equation for [tex]\(x\)[/tex]:

1. [tex]\(x + 4 = 0\)[/tex]
[tex]\[ x = -4 \][/tex]

2. [tex]\(x + 1 = 0\)[/tex]
[tex]\[ x = -1 \][/tex]

Thus, the solutions to the equation [tex]\((x+4)(x+1)=0\)[/tex] are [tex]\(x = -4\)[/tex] and [tex]\(x = -1\)[/tex].

The correct answer is:
[tex]\[ x = -4 \text{ or } x = -1 \][/tex]
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