Solve the following equation:
(x−1)2x3+1=x+4
To solve this equation, we follow these steps:
- Step 1: Multiply both sides by (x−1)2 to eliminate the fraction.
- Step 2: Expand and simplify both sides of the equation.
- Step 3: Rearrange the equation to form a polynomial equal to zero.
- Step 4: Solve the resulting polynomial using factorization or the quadratic formula.
Now, let's execute these steps:
Step 1: Multiply both sides by (x−1)2:
(x3+1)=(x+4)(x−1)2
Step 2: Expand the right side:
(x+4)(x2−2x+1)=x(x2−2x+1)+4(x2−2x+1)
Calculating each part yields:
x(x2−2x+1)=x3−2x2+x
4(x2−2x+1)=4x2−8x+4
Add these together:
x3−2x2+x+4x2−8x+4=x3+2x2−7x+4
Step 3: Combine terms and rearrange:
x3+1=x3+2x2−7x+4
Simplify by cancelling x3 from both sides:
1=2x2−7x+4
Move 1 to the right side:
0=2x2−7x+3
Step 4: Solve the quadratic equation 2x2−7x+3=0.
Using the quadratic formula, x=2a−b±b2−4ac, where a=2, b=−7, and c=3.
Calculate the discriminant:
b2−4ac=(−7)2−4⋅2⋅3=49−24=25
Now plug into the quadratic formula:
x=47±25
Simplify:
x=47±5
Two solutions arise:
x=412=3 and x=42=21
Since x=1 would make the denominator zero, it is not a valid solution for the original equation.
Therefore, the solution to the problem is x=3 or x=21.
x=3,21