To solve the given problem, we need to determine the constant that allows [tex]\( r = 5\sqrt{3} \)[/tex] to be a solution for the equation [tex]\( r^2 + \square = 0 \)[/tex]. Let's work through this step-by-step.
Firstly, if [tex]\( r = 5\sqrt{3} \)[/tex], we need to substitute this value into the equation [tex]\( r^2 + c = 0 \)[/tex] to find the value of [tex]\( c \)[/tex].
1. Compute [tex]\( r^2 \)[/tex]:
[tex]\[
r = 5\sqrt{3}
\][/tex]
[tex]\[
r^2 = (5\sqrt{3})^2
\][/tex]
[tex]\[
r^2 = 25 \times 3
\][/tex]
[tex]\[
r^2 = 75
\][/tex]
2. Now, substitute [tex]\( r^2 = 75 \)[/tex] into the equation [tex]\( r^2 + c = 0 \)[/tex]:
[tex]\[
75 + c = 0
\][/tex]
3. Solve for [tex]\( c \)[/tex]:
[tex]\[
c = -75
\][/tex]
So, the equation [tex]\( r^2 + \square = 0 \)[/tex] becomes:
[tex]\[
r^2 - 75 = 0
\][/tex]
To find the solutions to the equation [tex]\( r^2 - 75 = 0 \)[/tex], we solve for [tex]\( r \)[/tex]:
[tex]\[
r^2 = 75
\][/tex]
[tex]\[
r = \pm\sqrt{75}
\][/tex]
Simplifying the square root:
[tex]\[
r = \pm\sqrt{75} = \pm\sqrt{25 \times 3} = \pm 5\sqrt{3}
\][/tex]
Therefore, the two solutions to the equation are:
[tex]\[
r = 5\sqrt{3} \quad \text{and} \quad r = -5\sqrt{3}
\][/tex]
Summarizing, the completed problem and the solutions are:
The missing number is:
[tex]\[
-75
\][/tex]
and the two solutions to the equation are:
[tex]\[
r = 5\sqrt{3} \quad \text{and} \quad r = -5\sqrt{3}
\][/tex]
