An interesting Math problem

📘 Solve the equation:

\[ \sqrt{5 - x} = 5 - x^2 \]

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✏️ Step 1: Eliminate the square root

Square both sides:

\[ (\sqrt{5 - x})^2 = (5 - x^2)^2 \] \[ 5 - x = 25 + x^4 - 10x^2 \]

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✏️ Step 2: Rearrange into a standard polynomial form

Move all terms to one side:

\[ 0 = x^4 - 5 - 10x^2 + x + 25 \]

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✏️ Step 3: Let’s use the quadratic formula

We write this as a quadratic in terms of \(5\):
Let:

• \(a = 1\)
• \(b = -(2x^2 + 1)\)
• \(c = x^4 + x\)

Use the quadratic formula:

\[ 5 = \frac{(2x^2 + 1) \pm \sqrt{(2x^2 + 1)^2 - 4(x^4 + x)}}{2} \]

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✏️ Step 4: Apply the Quadratic Formula and Simplify

We are solving:

\[ 5 = \frac{2x^2 + 1 \pm \sqrt{(-2x^2 - 1)^2 - 4(x^4 + x)}}{2} \]

Now simplify the discriminant:

\[ = \frac{2x^2 + 1 \pm \sqrt{4x^2 - 4x + 1}}{2} \]

Factor the square root:

\[ = \frac{2x^2 + 1 \pm (2x - 1)}{2} \]

Now evaluate the two cases from here, which can be solve using standard quard formula.

✏️ Step 5: Solve the two quadratic equations

Using the quadratic formula:

Equation 1:

\[ x^2 + x - 5 = 0 \Rightarrow x = \frac{-1 \pm \sqrt{1^2 + 4 \cdot 5}}{2} = \frac{-1 \pm \sqrt{21}}{2} \]

Equation 2:

\[ x^2 - x - 4 = 0 \Rightarrow x = \frac{1 \pm \sqrt{1^2 + 4 \cdot 4}}{2} = \frac{1 \pm \sqrt{17}}{2} \]

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✅ Final Answer:

Check all four roots against the original equation. Only two satisfy it:

\[ \boxed{\frac{1}{2}(1 - \sqrt{17}) \quad \text{and} \quad \frac{1}{2}(\sqrt{21} - 1)} \]