To solve the problem, we need to find the number [tex]\( x \)[/tex] that satisfies the equation where the square of [tex]\( x \)[/tex] is 16 less than eight times [tex]\( x \)[/tex].
The equation can be set up as follows:
[tex]\[ x^2 = 8x - 16 \][/tex]
First, we bring all the terms to one side of the equation to set it to 0:
[tex]\[ x^2 - 8x + 16 = 0 \][/tex]
This is a quadratic equation in the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex]. Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -8 \)[/tex], and [tex]\( c = 16 \)[/tex].
To solve this quadratic equation, we can factor it. We look for two numbers that multiply to [tex]\( 16 \)[/tex] (the constant term) and add up to [tex]\( -8 \)[/tex] (the coefficient of the linear term):
Those numbers are [tex]\( -4 \)[/tex] and [tex]\( -4 \)[/tex]. Thus, the equation can be factored as:
[tex]\[ (x - 4)(x - 4) = 0 \][/tex]
or
[tex]\[ (x - 4)^2 = 0 \][/tex]
Setting each factor to 0 gives:
[tex]\[ x - 4 = 0 \][/tex]
Solving for [tex]\( x \)[/tex], we find:
[tex]\[ x = 4 \][/tex]
Therefore, the number is [tex]\( 4 \)[/tex].
So, the correct answer is:
[tex]\[ \boxed{4} \][/tex]
