How do you solve #x^2 / (x+3) - 5 / (x+3) = 0# and find any extraneous solutions?

Question

Kennedy Adams · Accepted Answer

#x = +-sqrt5#

We can cross-multiply since there are only two fractions, so let's move one to the other side.

#color(blue)(x^2/(x+3) = 5/(x+3)" "larr (x+3 !=0))#

#x^2(x+3) = 5(x+3)#

#x^3+3x^2= 5x+15" "larr# no fractions! Make equal to 0

#x^3+3x^2 -5x-15 =0" "larr# find factors by grouping

#x^2(x+3) -5(x+3)=0" "larr# common bracket

#(x+3)(x^2-5) =0#

If #x+3 = 0 " "rarr x =-3# (extraneous solution #x-3 !=0) #

If #x^2 -5 = 0 " "rarr x = +-sqrt5 " "larr# the only solutions.

But we could have found the solution much more quickly if we had taken the fraction into consideration from the start!

#color(blue)(x^2/(x+3) = 5/(x+3)" "larr (x+3 !=0))#

Since the denominators and numerators are equal, they must also be.

#x^2 = 5#

#x = +-sqrt5#

#x+3 !=0, " so " x !=-3#

Julia Adams · Answer

undefined To solve the equation x^2 / (x+3) - 5 / (x+3) = 0 and find any extraneous solutions, we can start by combining the fractions on the left side of the equation with a common denominator of (x+3). This gives us (x^2 - 5) / (x+3) = 0.

Next, we can set the numerator equal to zero and solve for x. So, x^2 - 5 = 0.

Using the difference of squares formula, we can factor this equation as (x - √5)(x + √5) = 0.

Setting each factor equal to zero, we have x - √5 = 0 and x + √5 = 0.

Solving for x in each equation, we find x = √5 and x = -√5.

Now, we need to check if these solutions are extraneous by substituting them back into the original equation.

When we substitute √5 into the equation, we get (√5)^2 / (√5 + 3) - 5 / (√5 + 3) = 0. Simplifying this, we get 5 / (√5 + 3) - 5 / (√5 + 3) = 0.

Since the denominators are the same, the fractions cancel out, resulting in 0 = 0.

Similarly, when we substitute -√5 into the equation, we get (-√5)^2 / (-√5 + 3) - 5 / (-√5 + 3) = 0. Simplifying this, we get 5 / (-√5 + 3) - 5 / (-√5 + 3) = 0.

Again, the fractions cancel out, resulting in 0 = 0.

Since both solutions satisfy the original equation, there are no extraneous solutions.

Therefore, the solutions to the equation x^2 / (x+3) - 5 / (x+3) = 0 are x = √5 and x = -√5.