# Irrational Equation A3
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/exercises/irrational-equation-a-3/
Fetched from algebrica.org test 7199; source modified 2025-04-25T17:06:25.
Before you begin, review the key concepts behind [irrational equations](../irrational-equations.md) to better follow the solution process.
Solve the irrational equation:
\\[
\\sqrt{4 -x} = 3- \\sqrt{5 +x}
\\]
* * *
Determine the set of admissible solutions. The equation is of the form:
\\[
\\sqrt\[n\]{f(x)} = g(x)
\\]
where \\(n\\), the index of the root is even. The set of admissible solutions is determined by solving the system of inequalities:
\\[
\\begin{cases} f(x) \\geq 0 \\\\\[2ex\] g(x) \\geq 0 \\\\ \\end{cases}
\\]
The system becomes:
\\[
\\begin{cases} 4-x \\geq 0 \\\\\[2ex\] 5+x \\geq 0 \\\\ \\end{cases}
\\]
* * *
For \\(4-x \\geq 0\\) we have \\(x \\leq 4\\) and for \\(5+x \\geq 0\\) we have \\(x \\geq -5\\). The admissible set of solutions is given by the intersection of the intervals found for \\(f(x)\\) and \\(g(x)\\). Use the graphical method to determine it:
\\[
-5
\\]
\\[
4
\\]
The admissible set of solutions is within the interval \\(\[-5, 4\]\\).
* * *
Solve the initial equation by squaring both members:
\\[
\\begin{align} &(\\sqrt{4 -x})^2 = (3- \\sqrt{5+x})^2 \\\\\[2ex\] &4 -x = (3- \\sqrt{5+x})^2 \\end{align}
\\]
The member to the right of the equal sign is the [square of a binomial](https://algebrica.org/square-of-a-binomial/) \\((a+b)^2 = a^2+2ab +b^2\\).
The equation becomes:
\\[
\\begin{align} &4-x = 9- 6\\sqrt{5 +x} + (5 +x) \\\\\[2ex\] &- 6\\sqrt{5 +x} = 4 -x -9 -5 -x\\\\\[2ex\] &6\\sqrt{5 +x} = 2x +10 \\\\\[2ex\] &3\\sqrt{5 +x} = x +5 \\end{align}
\\]
* * *
We have obtained an equation similar to the starting one. Now, square both sides again:
\\[
\\begin{align} &(3\\sqrt{5+x})^2 = (x+5)^2\\\\\[1ex\] &9\\cdot(5+x) = x^2+10x + 25\\\\\[1ex\] &45 +9x- x^2 -10x -25= 0 \\\\\[1ex\] &x^2 +10x -9x +25 -45= 0 \\\\\[1ex\] &x^2 +x -20= 0 \\end{align}
\\]
* * *
The [second-degree equation](../quadratic-equations.md) can be solved using the [quadratic formula](../quadratic-formula.md):
\\[
\\begin{align\*} x\_{1,2} &= \\frac{-1 \\pm \\sqrt{1^2 -4(1)(-20)}}{2(1)} \\\\\[2ex\] &= \\frac{-1 \\pm \\sqrt{1 +80}}{2}\\\\\[2ex\] &= \\frac{-1 \\pm 9}{2}\\\\\[2ex\] \\end{align\*}
\\]
We have:
\\[
x = -\\frac{10}{2} \\to x = -5
\\]
\\[
x = \\frac{8}{2} \\to x = 4
\\]
The solution to the equation is:
\\[
x = -5 \\quad x = 4
\\]
* * *
Both values are acceptable since they fall within the set of admissible solutions of the equation. Verify if they satisfy the initial equation. Remember that this step is crucial because solutions may occur in the set of acceptable solutions but do not satisfy the identity of the equation.
For \\(x = -5 \\) we have:
\\[
\\begin{align} &\\sqrt{4-(-5)} = 3- \\sqrt{5+(-5)}\\\\\[1ex\] &\\sqrt{9} = 3 - \\sqrt{0}\\\\\[1ex\] &3 = 3 \\end{align}
\\]
The equality is verified, so \\(x = -5\\;\\) is a solution to the equation.
For \\(x =4 \\) we have:
\\[
\\begin{align} &\\sqrt{4-4} = 3- \\sqrt{5+4}\\\\\[1ex\] &0 = 3 - \\sqrt{9}\\\\\[1ex\] &0 = 3-3\\\\\[1ex\] &0 = 0 \\end{align}
\\]
The equality is verified, so \\(x = 4\\;\\) is a solution to the equation.
The solution to the equation is:
\\[
x= -5 \\quad \\quad x = 4
\\]
