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How do you solve #(2x)/(4-x)=x^2/(x-4)#?

Answer 1

#x=0#
#x=-2#

#(2x)/(4-x)=x^2/(x-4)# or #-(2x)/(x-4)=x^2/(x-4)# or #x^2=-2x# or #x^2+2x=0# or #x(x+2)=0# or #x=0#------------------Ans #1# or #x+2=0# or #x=-2#----------------------------Ans #2#

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Answer 2

Jace Anderson

To solve the equation (2x)/(4-x) = x^2/(x-4), we can start by cross-multiplying to eliminate the fractions. This gives us 2x(x-4) = x^2(4-x). Expanding both sides of the equation, we get 2x^2 - 8x = 4x^2 - x^3. Rearranging the terms, we have x^3 - 2x^2 - 8x = 0. Factoring out an x, we get x(x^2 - 2x - 8) = 0. Setting each factor equal to zero, we have x = 0 and x^2 - 2x - 8 = 0. Solving the quadratic equation, we can factor it as (x - 4)(x + 2) = 0, giving us x = 4 and x = -2 as the solutions. Therefore, the equation (2x)/(4-x) = x^2/(x-4) is satisfied when x equals 0, 4, or -2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.