Questions: Solve the radical equation. √(8-2x)=x

Solution Steps

To solve the radical equation \(\sqrt{8-2x} = x\), we need to eliminate the square root by squaring both sides of the equation. After that, we will solve the resulting quadratic equation. Finally, we will check the solutions to ensure they satisfy the original equation.

1. Square both sides of the equation to eliminate the square root.
2. Rearrange the resulting quadratic equation to standard form \(ax^2 + bx + c = 0\).
3. Solve the quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
4. Check the solutions in the original equation to verify their validity.

Starting with the equation: \[ \sqrt{8 - 2x} = x \] Square both sides to eliminate the square root: \[ (\sqrt{8 - 2x})^2 = x^2 \] This simplifies to: \[ 8 - 2x = x^2 \]

Rearrange the equation to the standard quadratic form \(ax^2 + bx + c = 0\): \[ x^2 + 2x - 8 = 0 \]

Solve the quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): \[ x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-8)}}{2(1)} \] \[ x = \frac{-2 \pm \sqrt{4 + 32}}{2} \] \[ x = \frac{-2 \pm \sqrt{36}}{2} \] \[ x = \frac{-2 \pm 6}{2} \] This gives us two solutions: \[ x = \frac{-2 + 6}{2} = 2 \] \[ x = \frac{-2 - 6}{2} = -4 \]

Check each solution in the original equation \(\sqrt{8 - 2x} = x\):

For \(x = 2\): \[ \sqrt{8 - 2(2)} = 2 \] \[ \sqrt{4} = 2 \] This is true.

For \(x = -4\): \[ \sqrt{8 - 2(-4)} = -4 \] \[ \sqrt{8 + 8} = -4 \] \[ \sqrt{16} = -4 \] This is false because \(\sqrt{16} = 4\), not \(-4\).

Final Answer