Solve for k so that the rational equation has exactly one extraneous solution and one real solution.

Question

2/x^2-4x+k = 2x/x-1 + 1/x-3

Answer 1

2/(x² - 4x + k) = 2x/(x - 1) + 1/(x – 3) – corrected with proper parentheses

Notice that the product of the denominators on the right hand side is:

(x – 1)(x – 3) = x² - 4x +1

Which is the same expression as denominator on the left hand side if k = 3

2/(x² - 4x + 3) = 2x/(x - 1) + 1/(x – 3)

2/[(x - 1)(x - 3)] = 2x/(x - 1) + 1/(x – 3)

then multiply the whole equation by (x – 1)(x – 3) to get

2 = 2x(x - 3) + (x – 1)

2x² - 6x + x – 3 = 0

2x² - 5x - 3 = 0

Factoring

(2x + 1)(x - 3) = 0

First factor could be zero

2x + 1 = 0

2x = -1

x = -1/2

Second factor could be zero

x – 3 = 0

x = 3

Finally, notice that x cannot equal 3 since that would create a denominator = 0 in the original problem

Therefore, x = -1/2 is the real solution and x = 3 is an extraneous solution if k = 3

Answer 2

(x-3)(x-1) = x^2 -4x +3 = x^2-4x +k

where k=3

then

2/(x-3)(x-1) = 2x/(x-1) + 1/(x-3)

multiply by a common denominator

to get

2 = 2x(x-3) + x-1 = 2x^2-6x+x -1 = 2x^2-5x -1

2x^2 -5x -3 = 0

(2x+1)(x-3)=0

x=3 or -1/2

if by "extraneous, you mean a solution not in the domain

the domain of the original equation included all real numbers except 1 and 3

which make denominators = 0

but 1 and 3 are real numbers

extraneous means a solution that's not in the original equation

whenever you multiply by a variable term, you risk introducing a solution that's not in the original equation

a solution that's not in the domain for x

simple example is 1/x =x

when you multiply by x to eliminate the fraction you get 1=x^2 which has 2 solutions, x=1 and x=-1

in this case only x=0 would be extraneous

another example

x=i

x^2=-1

sqrx^2 =+/-sqr-1 = +i or -i, but -i is not a solution to the original equation, as -i does not = +i

yet you've distinguished extraneous as not a real solution, as if you want an imaginary solution

if you mean an imaginary solution as the "extraneous" solution, a non-real solution

as opposed to a real number. the domain does exclude imaginary numbers

keep k in the problem and multiply by a common denominator

eliminate the fractions

combine like terms

get a 4th degree polynomial

use descartes rule of signs

and get maximum one positive real solution

and max 2 imaginary solutions, 1 zero solution, if k>6

find k so that the terms alternate minimum times to minimize the real solution down to one

Answer 3

In order for a solution to be extraneous, it must make a denominator equal to zero. For the example given, if we use x=1, that will make the denominator x-1 equal to zero. If we use x=3, that will make the denominator x-3 equal to zero. This means that our two options for the extraneous solution could be either x=1 or x=3.

Looking at the third denominator, x-4x+k, we want this to be the product of the two factors already given, x-1 and x-3. If we multiply (x-1)(x-3), we get x²-4x+3. So let’s try to plug 3 in for k and solve the rational expression.

2/x^2-4x+3 = 2x/x-1 + 1/x-3

We want to multiply each term by all the factors that appear in the denominators.

2*(x-1)(x-3)/x^2-4x+3 = 2x*(x-1)(x-3)/x-1 + 1*(x-1)(x-3)/x-3

This will simplify to

2 = 2x(x-3) + 1(x-1)

2 = 2x²- 6x + x - 1

2 = 2x² - 5x - 1

0 = 2x² - 5x - 3

Now we can solve this quadratic by factoring using the AC Method.

(Find two numbers that multiply to equal A*C and add to equal B)

___*___ = 2*-3 = -6

___+___=-5

-6 and -1 will work to satisfy these conditions so replace the -5x with -6x and 1x

0 = 2x² - 6x + x - 3 Factor out the greatest common factor from the first two terms and the last two terms

0 = 2x(x-3) + 1(x-3) Combine the gcf’s in their own parentheses, and rewrite the factor that was leftover

0 = (2x+1)(x-3) Now set each factor equal to zero and solve

2x + 1 = 0 x - 3 = 0

2x = -1 x = 3

x = -1/2

In the beginning, we figured out that x = 3 is extraneous. This leaves us with one extraneous solution and one real solution.