Question

solve for x.
2x = \sqrt{x^{2} + 3}
if there are two solutions, write one solution in each box. if there is one solution, leave the second box empty.
x = \boxed{} and x = \boxed{}

Explanation:

Step1: Square both sides to eliminate the square root.

Squaring both sides of the equation \(2x=\sqrt{x^{2}+3}\), we get \((2x)^{2}=(\sqrt{x^{2}+3})^{2}\), which simplifies to \(4x^{2}=x^{2}+3\).

Step2: Solve the resulting quadratic equation.

Subtract \(x^{2}\) and 3 from both sides: \(4x^{2}-x^{2}-3 = 0\), so \(3x^{2}-3 = 0\). Factor out 3: \(3(x^{2}-1)=0\), then \(x^{2}-1 = 0\). Factoring the difference of squares, we have \((x - 1)(x + 1)=0\), so \(x = 1\) or \(x=-1\).

Step3: Check for extraneous solutions.

Substitute \(x = 1\) into the original equation: Left side \(2(1)=2\), right side \(\sqrt{1^{2}+3}=\sqrt{4}=2\). So \(x = 1\) is valid.
Substitute \(x=-1\) into the original equation: Left side \(2(-1)=-2\), right side \(\sqrt{(-1)^{2}+3}=\sqrt{4}=2\). Since \(-2
eq2\), \(x=-1\) is extraneous.

Answer:

\(x = 1\) and \(x=\) (empty)