How many solutions are there to this equation?

Question

  2       +     4      = 1

x+1          x^2-1
 

A) 0

B) 1

C) 2

D) infinite number

Answer 1

I found the same two answers in the same ways, but when I put the answers into the original equation to check, the x=-1 answer produces division by zero which is undefined!

One could argue that x=-1 is not a valid answer since it doesn't satisfy the original equation.

What do you think about this dilemma?

Answer 2

There are two solutions to the quadratic equation as all have stated...  x= -1 or x=3

 

However, there is a hole in the graph at x= 1 and at x=-1

 

Since x cannot be -1, there is only one answer.

x=3

Answer 3

2/(x+1) + 4/(x²-1) = 1

 

      2(x²-1)/(x+1) + 4 = x² - 1

      2(x+1)(x-1)/(x+1) + 4 = x² - 1

      2x - 2 + 4 = x² - 1

      0 = x² - 2x - 3

 

Factors to:

 

      0 = (x-3)(x+1)

      x = 3 and -1

 

There are two solutions.

 

You could also look at the discriminant, b²-4ac, where a=1, b=-2, and c=-3:

 

      b2 - 4ac

      (-2)² - 4(1)(-3) = 4 + 12 = 16

 

Since the discriminant > 0, there will be two solutions.  If the discriminant = 0, there will be one solution.  If it is less than zero, there are no real solutions.

 

Answer 4

We have a rational equation.  First, we need to find the LCD of this equation because we have different denominators.  The LCD is (x + 1)(x - 1).

 

 

(2(x - 1) + 4) / (x + 1)(x - 1) = ((x + 1)(x - 1)) / (x + 1)(x - 1)

 

 

Next, we equate numerators.

 

2(x - 1) + 4 = (x + 1)(x - 1)

 

2x - 2 + 4 = x² - 1

 

2x + 2 = x² - 1

 

 

Subtract 2x and subtract 2 on both sides of equation.

 

0 = x² - 2x - 3

 

 

We have a quadratic equation.  In these types of equations, we will always have 2 solutions.