Question
Solving a radical equation with a quadratic expression under the radical Solve for \( x \), where \( x \) is a real number. \( x=\sqrt{x^{2}+x+6} \) If there is more than one solution, separate them with commas. If there is no solution, click on "No solution".
Ask by Hardy Graham. in the United States
Feb 02,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
There is no real solution to the equation.
No solution
Solution
To solve the equation:
\[
x = \sqrt{x^{2} + x + 6}
\]
**Step 1: Square Both Sides**
To eliminate the square root, square both sides of the equation:
\[
x^2 = x^2 + x + 6
\]
**Step 2: Simplify the Equation**
Subtract \( x^2 \) from both sides:
\[
0 = x + 6
\]
**Step 3: Solve for \( x \)**
\[
x = -6
\]
**Step 4: Verify the Solution**
It's important to check whether this solution satisfies the original equation:
\[
-6 = \sqrt{(-6)^2 + (-6) + 6}
\]
\[
-6 = \sqrt{36 - 6 + 6}
\]
\[
-6 = \sqrt{36}
\]
\[
-6 = 6
\]
This is not true. Therefore, \( x = -6 \) is an **extraneous solution** and does not satisfy the original equation.
**Conclusion:**
There is **no real solution** to the equation.
**Answer:**
No solution
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Bonus Knowledge
To solve the equation \( x = \sqrt{x^{2} + x + 6} \), we start by squaring both sides: \[ x^2 = x^2 + x + 6 \] Then, we simplify this equation: \[ 0 = x + 6 \] This gives us: \[ x = -6 \] Next, we need to check if this is a valid solution by substituting \( x = -6 \) back into the original equation: \[ -6 = \sqrt{(-6)^{2} + (-6) + 6} \] Calculating the right side: \[ \sqrt{36 - 6 + 6} = \sqrt{36} = 6 \] This means: \[ -6 \neq 6 \] Therefore, \( x = -6 \) is not a valid solution. After checking the calculations, we find that the equation has no real solutions since the only candidate we found does not hold in the original equation. No solution.
