Answer:
C) x = 2 is extraneous
D) x = -1
Step-by-step explanation:
Given equation:
[tex]\sqrt{x+2}+1=\sqrt{3-x}[/tex]
Subtract 1 from both sides of the equation:
[tex]\sqrt{x+2} = \sqrt{3 - x} - 1[/tex]
Square both sides:
[tex](\sqrt{x+2})^2 = (\sqrt{3 - x} - 1)^2[/tex]
Simplify both sides, applying the Perfect Squares formula, (a - b)² = a² - 2ab + b², to the right side:
[tex]x+2=(\sqrt{3-x})^2-2\sqrt{3-x}+1[/tex]
[tex]x+2=3-x-2\sqrt{3-x}+1[/tex]
[tex]x+2=4-x-2\sqrt{3-x}[/tex]
Isolate the square root term:
[tex]2\sqrt{3-x}=4-x-x-2[/tex]
[tex]2\sqrt{3-x}=2-2x[/tex]
Divide both sides by 2:
[tex]\sqrt{3-x}=1-x[/tex]
Square both sides again:
[tex](\sqrt{3-x})^2=(1-x)^2[/tex]
[tex]3-x=1-2x+x^2[/tex]
Rearrange into the form ax² + bx + c = 0:
[tex]x^2-x-2=0[/tex]
Factor the quadratic:
[tex]x^2-2x+x-2=0[/tex]
[tex]x(x-2)+1(x-2)=0[/tex]
[tex](x+1)(x-2)=0[/tex]
Apply the zero product property to find the solutions:
[tex]x+1=0 \implies x=-1[/tex]
[tex]x-2=0 \implies x=2[/tex]
So, the solutions are x = -1 and x = 2. However, we need to check if these solutions satisfy the original equation, as sometimes extraneous solutions may occur when squaring both sides of an equation.
Substitute x = -1 into the original equation:
[tex]\begin{aligned}\sqrt{(-1)+2}+1&\overset{?}=\sqrt{3-(-1)}\\\sqrt{1}+1&\overset{?}=\sqrt{4}\\1+1&\overset{?}=2\\2&=2\end{aligned}[/tex]
Substitute x = 2 into the original equation:
[tex]\begin{aligned}\sqrt{2+2}+1&\overset{?}=\sqrt{3-2}\\\sqrt{4}+1&\overset{?}=\sqrt{1}\\2+1&\overset{?}=1\\3&\neq1\end{aligned}[/tex]
By testing both x = -1 and x = 2 in the original equation, we find that only x = -1 is a valid solution and x = 2 is extraneous.
