Certainly! Let's solve the given quadratic equation step-by-step and find the value of [tex]\( r \)[/tex], the sum of the solutions.
The quadratic equation given is:
[tex]\[ x^2 - 14 = 2 \][/tex]
First, we need to rearrange this equation into standard quadratic form [tex]\( ax^2 + bx + c = 0 \)[/tex].
[tex]\[ x^2 - 14 - 2 = 0 \][/tex]
[tex]\[ x^2 - 16 = 0 \][/tex]
We identify that the equation is now in standard quadratic form, where:
[tex]\[ a = 1, \quad b = 0, \quad c = -16 \][/tex]
Next, we factor the quadratic equation:
[tex]\[ x^2 - 16 = (x - 4)(x + 4) = 0 \][/tex]
Setting each factor equal to zero gives us the solutions:
[tex]\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \][/tex]
[tex]\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \][/tex]
Now we have the solutions [tex]\( x = 4 \)[/tex] and [tex]\( x = -4 \)[/tex].
To find the sum of the solutions, we simply add them together:
[tex]\[ 4 + (-4) = 0 \][/tex]
Therefore, the value of [tex]\( r \)[/tex], the sum of the solutions, is:
[tex]\[ r = 0 \][/tex]
So, the correct answer is:
[tex]\[ \boxed{0} \][/tex]
This confirms that the value of [tex]\( r \)[/tex] is indeed [tex]\( 0 \)[/tex].
