Question

solve the radical equation.
\sqrt{19 - 2x} = x + 8

the solution set is \\{\\ \\}.

Explanation:

Step1: Square both sides to eliminate the square root.

Squaring both sides of the equation \(\sqrt{19 - 2x}=x + 8\), we get \((\sqrt{19 - 2x})^2=(x + 8)^2\), which simplifies to \(19-2x=x^{2}+16x + 64\).

Step2: Rearrange into standard quadratic form.

Bring all terms to one side: \(x^{2}+16x + 64-19 + 2x = 0\), combining like terms gives \(x^{2}+18x + 45 = 0\).

Step3: Factor the quadratic equation.

Factor \(x^{2}+18x + 45\) as \((x + 3)(x + 15)=0\).

Step4: Solve for x.

Setting each factor equal to zero: \(x+3 = 0\) or \(x + 15=0\), so \(x=-3\) or \(x=-15\).

Step5: Check for extraneous solutions.

• For \(x=-3\): Left side \(\sqrt{19-2(-3)}=\sqrt{19 + 6}=\sqrt{25}=5\), Right side \(-3 + 8 = 5\). So \(x=-3\) is valid.
• For \(x=-15\): Left side \(\sqrt{19-2(-15)}=\sqrt{19 + 30}=\sqrt{49}=7\), Right side \(-15 + 8=-7\). Since \(7

eq - 7\), \(x=-15\) is extraneous.

Answer:

\(\{-3\}\)