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![\sqrt[4]{x^3+6 x^2}](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "\sqrt[4]{x^3+6 x^2}"){kind=icon}
02 Apr 2025, 09:53
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Tough: level 5
\(x=\sqrt[4]{x^3+6 x^2}\)
Quantity A: The sum of all possible roots of x
Quantity B: 1
\(x=\sqrt[4]{x^3+6 x^2}\)
Quantity A: The sum of all possible roots of x
Quantity B: 1
08 Apr 2025, 05:00
Kudos
Bookmarks
We are given the equation:
$$
\(x=\sqrt[4]{x^3+6 x^2}\)
$$
Step 1: Solve the Equation $\(x=\sqrt[4]{x^3+6 x^2}\)$
First, let's eliminate the fourth root by raising both sides to the fourth power:
$$
\(x^4=x^3+6 x^2\)
$$
Step 2: Rearrange the Equation
Bring all terms to one side to form a polynomial equation:
$$
\(x^4-x^3-6 x^2=0\)
$$
Step 3: Factor the Polynomial
Factor out $\(x^2\)$ :
$$
\(x^2\left(x^2-x-6\right)=0\)
$$
This gives us:
1. $\(x^2=0 \rightarrow x=0\)$
2. $\(x^2-x-6=0\)$
Now, solve the quadratic equation $\(x^2-x-6=0\)$ :
Using the quadratic formula:
$$
\(x=\frac{1 \pm \sqrt{1+24}}{2}=\frac{1 \pm 5}{2}\)
$$
So, the solutions are:
$$
\(\begin{gathered}\\
x=\frac{1+5}{2}=3 \\\\
x=\frac{1-5}{2}=-2\\
\end{gathered}\)
$$
Step 4: List All Possible Roots
From the factoring, the possible roots are:
1. $\(x=0\left(\right.\)$ from $\(\left.x^2=0\right)\)$
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 $\(x=0\)$ :
$$
\(0=\sqrt[4]{0^3+6 \cdot 0^2}=\sqrt[4]{0}=0\)
$$
This holds true.
2. For $\(x=3\)$ :
$$
\(3=\sqrt[4]{3^3+6 \cdot 3^2}=\sqrt[4]{27+54}=\sqrt[4]{81}=3\)
$$
This holds true.
3. For $\(x=-2\)$ :
$$
\(-2=\sqrt[4]{(-2)^3+6 \cdot(-2)^2}=\sqrt[4]{-8+24}=\sqrt[4]{16}=2\)
$$
Here, $\(-2 \neq 2\)$, 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:
$$
\(0+3=3\)
$$
In the end
A=3
B=1
A is greater and it is the final solution.
$$
\(x=\sqrt[4]{x^3+6 x^2}\)
$$
Step 1: Solve the Equation $\(x=\sqrt[4]{x^3+6 x^2}\)$
First, let's eliminate the fourth root by raising both sides to the fourth power:
$$
\(x^4=x^3+6 x^2\)
$$
Step 2: Rearrange the Equation
Bring all terms to one side to form a polynomial equation:
$$
\(x^4-x^3-6 x^2=0\)
$$
Step 3: Factor the Polynomial
Factor out $\(x^2\)$ :
$$
\(x^2\left(x^2-x-6\right)=0\)
$$
This gives us:
1. $\(x^2=0 \rightarrow x=0\)$
2. $\(x^2-x-6=0\)$
Now, solve the quadratic equation $\(x^2-x-6=0\)$ :
Using the quadratic formula:
$$
\(x=\frac{1 \pm \sqrt{1+24}}{2}=\frac{1 \pm 5}{2}\)
$$
So, the solutions are:
$$
\(\begin{gathered}\\
x=\frac{1+5}{2}=3 \\\\
x=\frac{1-5}{2}=-2\\
\end{gathered}\)
$$
Step 4: List All Possible Roots
From the factoring, the possible roots are:
1. $\(x=0\left(\right.\)$ from $\(\left.x^2=0\right)\)$
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 $\(x=0\)$ :
$$
\(0=\sqrt[4]{0^3+6 \cdot 0^2}=\sqrt[4]{0}=0\)
$$
This holds true.
2. For $\(x=3\)$ :
$$
\(3=\sqrt[4]{3^3+6 \cdot 3^2}=\sqrt[4]{27+54}=\sqrt[4]{81}=3\)
$$
This holds true.
3. For $\(x=-2\)$ :
$$
\(-2=\sqrt[4]{(-2)^3+6 \cdot(-2)^2}=\sqrt[4]{-8+24}=\sqrt[4]{16}=2\)
$$
Here, $\(-2 \neq 2\)$, 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 end
A=3
B=1
A is greater and it is the final solution.
