1. 2x²-8=y² solve this given ​

To solve the equation (2x^2 - 8 = y^2), we can rewrite it and analyze it step-by-step:Given: [ 2x^2 - 8 = y^2 ]Step 1: Isolate terms. Rewrite as: [ y^2 = 2x^2 - 8 ]Step 2: The equation represents a relationship between (x) and (y). To find solutions, rearrange it as: [ y^2 = 2x^2 - 8 ]Step 3: For (y) to be real, the right side must be ≥ 0: [ 2x^2 - 8 \geq 0 ] [ 2x^2 \geq 8 ] [ x^2 \geq 4 ] [ x \leq -2 \quad \text{or} \quad x \geq 2 ]Step 4: Express (y) in terms of (x): [ y = \pm \sqrt{2x^2 - 8} ]Step 5: Example solutions:For (x=2): [ y = \pm \sqrt{2(2)^2 - 8} = \pm \sqrt{8 - 8} = 0 ]For (x=3): [ y = \pm \sqrt{2(9) - 8} = \pm \sqrt{18 - 8} = \pm \sqrt{10} ]Step 6: So all ((x, y)) pairs such that (x \leq -2) or (x \geq 2) and (y = \pm \sqrt{2x^2 - 8}) satisfy the equation.

Answered by romnickpallon | 2025-07-07