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

[tex]\[ x(x+3)=0 \][/tex]

a. \(x=0\) or \(x=-3\)

b. \(x=0\) or \(x=-5\)

c. \(x=1\) or \(x=-1\)

d. \(x=-1\) or \(x=3\)

Please select the best answer from the choices provided:

A. a

B. b

C. c

D. d

Sagot :

GinnyAnswer
Certainly! To solve the quadratic equation \( x(x + 3) = 0 \) numerically, we will find the values of \( x \) that satisfy this equation.

1. The given quadratic equation is \( x(x + 3) = 0 \).

2. To find the solutions to this equation, we set each factor equal to zero.

- For the first factor \( x = 0 \):
[tex]\[ x = 0 \][/tex]

- For the second factor \( x + 3 = 0 \):
[tex]\[ x + 3 = 0 \implies x = -3 \][/tex]

3. Therefore, the solutions to the quadratic equation \( x(x + 3) = 0 \) are \( x = 0 \) and \( x = -3 \).

4. Now we need to select the best answer from the given choices:

a. \( x = -1 \) or \( x = 3 \)
c. \( x = 0 \) or \( x = -3 \)
b. \( x = 0 \) or \( x = -5 \)
d. \( x = 1 \) or \( x = -1 \)

The correct answer is [tex]\(\boxed{c}\)[/tex] [tex]\( x = 0 \)[/tex] or [tex]\( x = -3 \)[/tex].
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