How do you solve the rational equation #(4-8x)/(1-x+4)=8/(x+1)#?

Question

Savannah Anderson · Accepted Answer

#[4-8x]/[1-x+4]=8/[x+1]#

#[4-8x]/[5-x]=8/[x+1]#

Cross multiply to remove the fractions

#(4-8x)(x+1)=8(5-x)#

Expand the brackets

#4x+4-8x^2-8x=40-8x#

#4-4x-8x^2=40-8x#

Add #8x^2# to both sides

#4-4x=8x^2-8x+40#

add #4x# to both sides

4=8x^2-4x+40#

subtract 4 from both sides

#8x^2-4x+36=0#

Divide both sides by 4

#2x^2-x+9=0#

Put into the quadratic formula

#x=[1\pmsqrt[1-4xx2xx9]]/[2xx2]#

#x=[1\pmsqrt[-71]]/4#

No solutions as you have a negative number in the square root sign

Abigail Allen · Answer

undefined To solve the rational equation (4-8x)/(1-x+4)=8/(x+1), we can start by cross-multiplying to eliminate the fractions. This gives us (4-8x)(x+1) = 8(1-x+4). Expanding both sides of the equation, we get 4x + 4 - 8x^2 - 8x = 8 - 8x + 32. Simplifying further, we have -8x^2 - 4x - 8x + 4 = 8 - 8x + 32. Combining like terms, we obtain -8x^2 - 20x + 4 = 40 - 8x. Rearranging the equation, we have -8x^2 - 20x + 8x + 4 - 40 = 0. Simplifying, we get -8x^2 - 12x - 36 = 0. Factoring out -4, we have -4(2x^2 + 3x + 9) = 0. Setting each factor equal to zero, we find that 2x^2 + 3x + 9 = 0. However, this quadratic equation does not factor easily, so we can use the quadratic formula to find the solutions. The quadratic formula states that x = (-b ± √(b^2 - 4ac))/(2a), where a, b, and c are the coefficients of the quadratic equation. In this case, a = 2, b = 3, and c = 9. Plugging these values into the quadratic formula, we get x = (-3 ± √(3^2 - 4(2)(9)))/(2(2)). Simplifying further, we have x = (-3 ± √(9 - 72))/4, which becomes x = (-3 ± √(-63))/4. Since the square root of a negative number is not a real number, this equation has no real solutions. Therefore, the rational equation (4-8x)/(1-x+4)=8/(x+1) has no solution.