Sure, let's solve the given equation [tex]\( f = \frac{L}{4 d^2} \)[/tex] for [tex]\( d \)[/tex] when [tex]\( f = 2 \)[/tex] and [tex]\( L = 288 \)[/tex].
1. Substitute the given values into the equation:
[tex]\[
f = 2 \quad \text{and} \quad L = 288
\][/tex]
So, the equation becomes:
[tex]\[
2 = \frac{288}{4 d^2}
\][/tex]
2. Multiply both sides of the equation by [tex]\( 4 d^2 \)[/tex] to get rid of the fraction:
[tex]\[
2 \cdot 4 d^2 = 288
\][/tex]
Simplifying this, we get:
[tex]\[
8 d^2 = 288
\][/tex]
3. Divide both sides by 8 to isolate [tex]\( d^2 \)[/tex]:
[tex]\[
d^2 = \frac{288}{8}
\][/tex]
Simplifying the right-hand side, we get:
[tex]\[
d^2 = 36
\][/tex]
4. Take the square root of both sides to solve for [tex]\( d \)[/tex]:
[tex]\[
d = \sqrt{36}
\][/tex]
Remember that the square root of a number has both a positive and a negative solution. Therefore:
[tex]\[
d = \pm 6
\][/tex]
5. Final values:
- The positive value of [tex]\( d \)[/tex] is [tex]\( 6 \)[/tex].
- The negative value of [tex]\( d \)[/tex] is [tex]\( -6 \)[/tex].
So, the possible values of [tex]\( d \)[/tex] are [tex]\( 6 \)[/tex] and [tex]\( -6 \)[/tex].
We can also confirm that [tex]\( d^2 \)[/tex] equals [tex]\( 36 \)[/tex], leading to [tex]\( d = 6 \)[/tex] or [tex]\( d = -6 \)[/tex]. Thus, the results are [tex]\( d_1 = 6 \)[/tex], [tex]\( d_2 = -6 \)[/tex], and [tex]\( d^2 = 36 \)[/tex].
