How do you solve #(8sqrt( 2x-1)) / 3 = x#?

Question

Eva Abramson · Accepted Answer

First, send the 3 to the other side in multiplication.

#8(sqrt(2x - 1)) = 3x# #sqrt(2x - 1) = 3/8x# Get rid of the square root by squaring both sides of the equation. #(sqrt(2x - 1))^2 = (3/8x)^2# #2x - 1 = 9/64x^2# #64(2x - 1) = 9x^2# #128x - 64 = 9x^2# Solving by completing the square: #-64 = 9(x^2 + 128/9 + m - m)# #m = (b/2)^2# #m = ((128/9)/2)^2# #m = 16384/324# #-64 = 9(x^2 + 128/9 + 16384/324 - 16384/324)# #-64 = 9(x^2 + 128/9 + 16384/324) - 147456/324# #(-64 + 147456/324)/9 = (x + 128/18)^2# #126720/2916 = (x + 128/18)^2# #+-sqrt(126720/2915) - 128/18 = x# #-0.52 ~= x and -13.70 ~= x# When you plug both these answers into the original equation, they both make the square root negative inside. This means there is no solution Hopefully this helps!

Julia Adams · Answer

undefined To solve the equation (8√(2x-1))/3 = x, you can follow these steps:

1. Multiply both sides of the equation by 3 to eliminate the fraction: 8√(2x-1) = 3x.
2. Square both sides of the equation to remove the square root: (8√(2x-1))^2 = (3x)^2.
3. Simplify the equation: 64(2x-1) = 9x^2.
4. Expand and rearrange the equation: 128x - 64 = 9x^2.
5. Move all terms to one side of the equation: 9x^2 - 128x + 64 = 0.
6. Factor the quadratic equation or use the quadratic formula to solve for x.
   - Factoring: (3x - 4)(3x - 16) = 0. Therefore, x = 4/3 or x = 16/3.
   - Quadratic formula: x = (-b ± √(b^2 - 4ac))/(2a), where a = 9, b = -128, and c = 64.
     Plugging in these values, you will get x = 4/3 or x = 16/3.

Therefore, the solutions to the equation are x = 4/3 or x = 16/3.