Answer:
c = 5
k = 3
Explanation:
We have the expression:
(3+√c) (2√c - 3) = 1 + k √c
Expanding the LHS using the FOIL method =>
(3 + √c) (2√c -3)
= (3)(2√c) + 3(-3) + (√c)(2√c) -√c(3)
= 6√c -9 + 2(√c)² -3√c
= 3√c + 2c - 9 [(√c)² = c]
Setting this equal to the RHS of 1 + k√c gives us
3√c + 2c - 9 = 1 + k √c
Since c and k are prime numbers, √c is irrational.
Therefore we match the irrational terms to get
3√c = k√c
=> k = 3 [divide both sides by √c]
Substitute k = 3 into 3√c + 2c - 9 = 1 + k √c
=> 3√c + 2c - 9 = 1 + 3√c
Add 9 to both sides
=> 3√c + 2c - 9 + 9 = 1 + 9 + 3√c
=> 3√c + 2c - 0 = 10 + 3√c
Subtract 3√c from both sides to eliminate this term on both sides
=> 3√c - 3√c + 2c - 0 = 10 + 3√c 3√c
=> 0 + 2c = 10 + 0
=> 2c = 10
Dividing by 2 on both sides gives us
c = 5
Answer: c = 5, k = 3