Question

Find all values of c such that

Answer 1

No real solutions.

Explanation:

Step 1: Cross-multiply.

c/c−5=4/c−4

c*(c−4)=4*(c−5)

c^2−4c=4c−20

Step 2: Subtract 4c-20 from both sides.

c^2−4c−(4c−20)=4c−20−(4c−20)

c^2−8c+20=0

For this equation: a=1, b=-8, c=20

1c^2+−8c+20=0

Step 3: Use quadratic formula with a=1, b=-8, c=20.

c= −b±√b^2−4ac/2a

c= −(−8)±√(−8)2−4(1)(20)/2(1)

c= 8±√−16/2

Answer 2

Explanation:

c/(c - 5) = 4/(c - 4)

By Cross-multiplying,

We have c(c - 4) = 4(c - 5).

=> c² - 4c = 4c - 20

=> c² - 8c + 20 = 0

Since the discriminant is negative,

there are no real solutions for c.

However, there exist complex solutions for c.

Using the Quadratic Formula,

c = [8 ± √(-16)]/2

=> c = 4 ± √(-4)

=> c = 4 ± 4i or c = 4(1 ± i).