Question

44 solve the radical equation. check for extraneous solutions.
\sqrt4{2x + 3} = \sqrt4{x + 9}

Explanation:

Step1: Raise both sides to the 4th power

To eliminate the fourth - root, we raise both sides of the equation \(\sqrt[4]{2x + 3}=\sqrt[4]{x + 9}\) to the power of 4. According to the property \((\sqrt[n]{a})^n=a\) (for \(a\geq0\) when \(n\) is even), we have \((\sqrt[4]{2x + 3})^4=(\sqrt[4]{x + 9})^4\), which simplifies to \(2x+3=x + 9\).

Step2: Solve the linear equation

Subtract \(x\) from both sides of the equation \(2x+3=x + 9\): \(2x - x+3=x - x+9\), which gives \(x+3 = 9\). Then subtract 3 from both sides: \(x+3-3=9 - 3\), so \(x = 6\).

Step3: Check for extraneous solutions

We substitute \(x = 6\) back into the original equation.
For the left - hand side: \(\sqrt[4]{2(6)+3}=\sqrt[4]{12 + 3}=\sqrt[4]{15}\)
For the right - hand side: \(\sqrt[4]{6 + 9}=\sqrt[4]{15}\)
Since the left - hand side is equal to the right - hand side when \(x = 6\), \(x = 6\) is a valid solution.

Answer:

\(x = 6\)