1. **State the problem:** We are given the 5th term in the expansion of $ (1 + kx)^8 $ as $ 11720x^4 $. We need to find the value of $ k $. 2. **Recall the binomial expansion formula:** The general term (r+1) in the expansion of $ (a + b)^n $ is given by: $$ T_{r+1} = \binom{n}{r} a^{n-r} b^r $$ 3. **Apply the formula to our problem:** Here, $ a = 1 $, $ b = kx $, and $ n = 8 $. The 5th term corresponds to $ r = 4 $ (since term number = r+1). $$ T_5 = \binom{8}{4} (1)^{8-4} (kx)^4 = \binom{8}{4} k^4 x^4 $$ 4. **Calculate the binomial coefficient:** $$ \binom{8}{4} = \frac{8!}{4!4!} = \frac{40320}{24 \times 24} = 70 $$ 5. **Write the term explicitly:** $$ T_5 = 70 k^4 x^4 $$ 6. **Set the term equal to the given value:** $$ 70 k^4 x^4 = 11720 x^4 $$ 7. **Cancel $ x^4 $ from both sides:** $$ 70 k^4 = 11720 $$ 8. **Divide both sides by 70:** $$ \cancel{70} k^4 = \frac{11720}{\cancel{70}} $$ $$ k^4 = 167.4285714 $$ 9. **Find the fourth root of both sides:** $$ k = \sqrt[4]{167.4285714} $$ 10. **Calculate the value:** $$ k \approx 3.55 $$ **Final answer:** $$ \boxed{k \approx 3.55} $$