Question

$\sqrt{5-x}=(5-x^2)$

Answer 1

The solutions are $x = \frac{-1 + \sqrt{21}}{2}$ and $x = \frac{1 - \sqrt{17}}{2}$.

Answer 2

The problem asks to solve the equation $\sqrt{5-x}=(5-x^2)$. The given steps correctly derive the equation $10=(2x^2+1) \pm (2x-1)$ from the original equation $\sqrt{a-x}=a-x^2$ by setting $a=5$.

Solution:

The original equation is $\sqrt{5-x}=5-x^2$. For the square root to be defined, $5-x \ge 0 \implies x \le 5$. For the right side to be non-negative (since it equals a square root), $5-x^2 \ge 0 \implies x^2 \le 5 \implies -\sqrt{5} \le x \le \sqrt{5}$. Combining these conditions, the valid range for $x$ is $[-\sqrt{5}, \sqrt{5}]$, which is approximately $[-2.236, 2.236]$.

The provided steps lead to the equation: $10 = (2x^2+1) \pm (2x-1)$

We consider two cases based on the $\pm$ sign, which arises from $\sqrt{(2x-1)^2} = |2x-1|$.

Case 1: $10 = (2x^2+1) + (2x-1)$ $10 = 2x^2 + 2x$ Dividing by 2: $5 = x^2 + x$ $x^2 + x - 5 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-5)}}{2(1)}$ $x = \frac{-1 \pm \sqrt{1 + 20}}{2}$ $x = \frac{-1 \pm \sqrt{21}}{2}$

Let's check these values against the valid range $[-\sqrt{5}, \sqrt{5}]$: $x_1 = \frac{-1 + \sqrt{21}}{2} \approx \frac{-1 + 4.58}{2} = \frac{3.58}{2} = 1.79$. This value is within the range. For this solution, $2x-1 = \sqrt{21}-2 \approx 2.58 > 0$, so this corresponds to the '+' sign in $10 = (2x^2+1) + (2x-1)$. This is a valid solution.

$x_2 = \frac{-1 - \sqrt{21}}{2} \approx \frac{-1 - 4.58}{2} = \frac{-5.58}{2} = -2.79$. This value is outside the range. Thus, $x_2$ is an extraneous solution.

Case 2: $10 = (2x^2+1) - (2x-1)$ $10 = 2x^2 + 1 - 2x + 1$ $10 = 2x^2 - 2x + 2$ Dividing by 2: $5 = x^2 - x + 1$ $x^2 - x - 4 = 0$ Using the quadratic formula: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-4)}}{2(1)}$ $x = \frac{1 \pm \sqrt{1 + 16}}{2}$ $x = \frac{1 \pm \sqrt{17}}{2}$

Let's check these values against the valid range $[-\sqrt{5}, \sqrt{5}]$: $x_3 = \frac{1 + \sqrt{17}}{2} \approx \frac{1 + 4.12}{2} = \frac{5.12}{2} = 2.56$. This value is outside the range. Thus, $x_3$ is an extraneous solution.

$x_4 = \frac{1 - \sqrt{17}}{2} \approx \frac{1 - 4.12}{2} = \frac{-3.12}{2} = -1.56$. This value is within the range. For this solution, $2x-1 = 1-\sqrt{17}-1 = -\sqrt{17} \approx -4.12 < 0$, so this corresponds to the '-' sign in $10 = (2x^2+1) - (2x-1)$. This is a valid solution.

Both valid solutions must satisfy the original equation. We have verified both solutions in the thought process.

The solutions to the equation are $x = \frac{-1 + \sqrt{21}}{2}$ and $x = \frac{1 - \sqrt{17}}{2}$.

Explanation of the solution:

1. Establish domain for $x$: $5-x \ge 0 \implies x \le 5$ and $5-x^2 \ge 0 \implies -\sqrt{5} \le x \le \sqrt{5}$. Combined, $x \in [-\sqrt{5}, \sqrt{5}]$.
2. Square both sides of $\sqrt{5-x}=5-x^2$ to get $5-x = (5-x^2)^2$.
3. Rearrange the squared equation to form a quadratic in $x^2$ or $x$, or solve for $a=5$ in the general form $a^2-(2x^2+1)a+x^4+x=0$.
4. The given steps correctly lead to $10 = (2x^2+1) \pm (2x-1)$.
5. Solve two separate quadratic equations for $x$:
   • $10 = 2x^2+1 + 2x-1 \implies x^2+x-5=0 \implies x = \frac{-1 \pm \sqrt{21}}{2}$.
   • $10 = 2x^2+1 - (2x-1) \implies x^2-x-4=0 \implies x = \frac{1 \pm \sqrt{17}}{2}$.
6. Filter solutions based on the domain $x \in [-\sqrt{5}, \sqrt{5}]$.
   • $\frac{-1 + \sqrt{21}}{2} \approx 1.79$ is valid.
   • $\frac{-1 - \sqrt{21}}{2} \approx -2.79$ is extraneous.
   • $\frac{1 + \sqrt{17}}{2} \approx 2.56$ is extraneous.
   • $\frac{1 - \sqrt{17}}{2} \approx -1.56$ is valid.
7. The valid solutions are $\frac{-1 + \sqrt{21}}{2}$ and $\frac{1 - \sqrt{17}}{2}$.