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

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

a. [tex]x = 1[/tex] or [tex]x = 3[/tex]

b. [tex]x = 0[/tex] or [tex]x = -3[/tex]

c. [tex]x = 0[/tex] or [tex]x = 6[/tex]

d. [tex]x = 1[/tex] or [tex]x = -1[/tex]

Please select the best answer from the choices provided:
A
B
C
D


Sagot :

To solve the quadratic equation \(x(x + 3) = 0\) numerically, we'll explore possible values for \(x\) that satisfy the equation \(x^2 + 3x = 0\).

Let's start by understanding the fundamental principle we'll use, called the zero product property. This property states that if the product of two numbers is zero, then at least one of the numbers must be zero. So, for our equation \(x(x + 3) = 0\):

1. Either \(x = 0\)
2. Or \(x + 3 = 0 \implies x = -3\)

These are our two possible solutions. Let's validate them by substituting back into the original equation to make sure they work:

1. For \(x = 0\):
[tex]\[ 0(0 + 3) = 0 \quad \text{which simplifies to} \quad 0 = 0 \][/tex]
This is true, so \(x = 0\) is a valid solution.

2. For \(x = -3\):
[tex]\[ (-3)(-3 + 3) = (-3)(0) = 0 \quad \text{which simplifies to} \quad 0 = 0 \][/tex]
This is also true, so \(x = -3\) is a valid solution.

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

Given the options provided:

a. \(x = 1\) or \(x = 3\)
b. \(x = 0\) or \(x = 6\)
c. \(x = 0\) or \(x = -3\)
d. \(x = 1\) or \(x = -1\)

The correct choice is:

c. [tex]\(x = 0\)[/tex] or [tex]\(x = -3\)[/tex]