We will solve the equation step by step, checking for extraneous solutions at each step.
First, we will remove the absolute value by considering two cases:
Case 1: (6-4x) is nonnegative:
6 - 4x ≥ 0
-4x ≥ -6
x ≤ 3/2
Case 2: (6-4x) is negative:
-(6-4x) ≥ 0
4x - 6 ≥ 0
4x ≥ 6
x ≥ 3/2
Combining both cases, we find that x ≤ 3/2 or x ≥ 3/2.
Next, we will square both sides of the equation to eliminate the absolute value:
[6 - 4x]^2 = (8x + 4)^2
Expanding both sides:
36 - 48x + 16x^2 = 64x^2 + 64x + 16
Simplifying:
16x^2 + 112x + 20 = 0
Dividing by 4 to simplify further:
4x^2 + 28x + 5 = 0
Now, we will solve this quadratic equation using factoring or the quadratic formula. The quadratic equation does not factor nicely, so we will use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Using the quadratic formula with a = 4, b = 28, and c = 5:
x = (-28 ± √(28^2 - 4(4)(5))) / (2(4))
x = (-28 ± √(784 - 80)) / 8
x = (-28 ± √704) / 8
x = (-28 ± 8√11) / 8
x = -7/2 ± √11
Therefore, the solutions to the equation are x = -7/2 + √11 and x = -7/2 - √11.
To check for extraneous solutions, we substitute these solutions back into the original equation:
When x = -7/2 + √11:
6|6 - 4(-7/2 + √11)| = 8(-7/2 + √11) + 4
6|20 + 4√11| = (-28 + 8√11) + 4
6(20 + 4√11) = -24 + 8√11
120 + 24√11 = -24 + 8√11
144√11 = -144
This equation is not true, so x = -7/2 + √11 is an extraneous solution.
When x = -7/2 - √11:
6|6 - 4(-7/2 - √11)| = 8(-7/2 - √11) + 4
6|-20 - 4√11| = (-28 - 8√11) + 4
6(20 + 4√11) = -32 - 8√11
120 + 24√11 = -32 - 8√11
32 + 32√11 = -120
This equation is not true, so x = -7/2 - √11 is also an extraneous solution.
Therefore, the equation has no valid solutions.
Solve the equation. Check for extraneous solutions.
6|6-4x|= 8x +4
1 answer