We are given the equation:
\(x=4√x3+6x2\)
Step 1: Solve the Equation $
x=4√x3+6x2$
First, let's eliminate the fourth root by raising both sides to the fourth power:
\(x4=x3+6x2\)
Step 2: Rearrange the Equation
Bring all terms to one side to form a polynomial equation:
\(x4−x3−6x2=0\)
Step 3: Factor the Polynomial
Factor out $
x2$ :
\(x2(x2−x−6)=0\)
This gives us:
1. $
x2=0→x=0$
2. $
x2−x−6=0$
Now, solve the quadratic equation $
x2−x−6=0$ :
Using the quadratic formula:
\(x=1±√1+242=1±52\)
So, the solutions are:
\(x=1+52=3x=1−52=−2\)
Step 4: List All Possible Roots
From the factoring, the possible roots are:
1. $
x=0($ from $
x2=0)$
2. $
x=3$ (from the quadratic)
3. $
x=−2$ (from the quadratic)
Step 5: Verify the Roots in the Original Equation
It's crucial to verify each potential root in the original equation because raising both sides to a power can introduce extraneous solutions.
1. For $
x=0$ :
\(0=4√03+6⋅02=4√0=0\)
This holds true.
2. For $
x=3$ :
\(3=4√33+6⋅32=4√27+54=4√81=3\)
This holds true.
3. For $
x=−2$ :
\(−2=4√(−2)3+6⋅(−2)2=4√−8+24=4√16=2\)
Here, $
−2≠2$, so $
x=−2$ is not a valid solution.
Step 6: Identify Valid Real Roots
From the verification:
- Valid roots: $
x=0,x=3$
- Invalid root: $
x=−2$
Step 7: Calculate the Sum of All Possible Real Roots
Sum of valid roots:
\(0+3=3\)
In the endA=3
B=1
A is greater and it is the final solution.
PS: notice how in the stem we do not have 4 before the root but is the same root of 4. I fixed it