How do you solve # 1/(4 -x) = x/8#?

Question

Isaiah Anthony · Accepted Answer

Below

#1/(4-x)=x/8# #1/(4-x)times8/8 = x/8times(4-x)/(4-x)# #8/(8(4-x)) = (x(4-x))/(8(4-x))# #8 = x(4-x)# #8 = 4x-x^2# #x^2-4x+8=0# Using the quadratic formula: #x = (-b+-sqrt(b^2-4ac))/(2a)# #x = (4+-sqrt(16-4times1times8))/(2times1)# #x = (4+-sqrt(-16))/2# If you haven't learnt complex numbers yet, then you can ignore the following lines and just state that since the discriminant is less than 0, then there is no solution to the equation. #x = (4+-sqrt(16times-1))/2# #x = (4+-4i)/2# #x = 2+-2i# #x^2-4x+8=0# #(x-(2-2i))(x-(2+2i))=0# #(x-2+2i)(x-2-2i)=0#

Isaiah Anthony · Answer

undefined To solve the equation 1/(4 - x) = x/8, you can start by cross-multiplying to eliminate the fractions. Multiply both sides of the equation by (4 - x) and 8 to get rid of the denominators. This will give you 8 = x(4 - x). Simplify the equation by distributing x, which results in 8 = 4x - x^2. Rearrange the equation to form a quadratic equation: x^2 - 4x + 8 = 0. To solve this quadratic equation, you can use factoring, completing the square, or the quadratic formula. In this case, the quadratic equation does not factor easily, so you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a). Plugging in the values from the quadratic equation, you will get x = (4 ± √(-16)) / 2. Since the discriminant (b^2 - 4ac) is negative, the equation has no real solutions. Therefore, there are no real values of x that satisfy the original equation 1/(4 - x) = x/8.