Algebra · real student question

Solve the inequality -2x^2 + 8 <= 0.

Question

Solve

2x2+80-2x^{2}+8\le0

Step-by-step solution

  1. Isolate the squared term. Subtract 88 from both sides:

    2x28-2x^{2}\le-8

  2. Divide by 2-2 and reverse the inequality. Dividing (or multiplying) an inequality by a negative number flips its direction — the single rule that governs this problem:

    x24x^{2}\ge4

    Keeping \le here would give the exactly wrong answer 2x2-2\le x\le2, so the flip is not optional.

  3. Solve x24x^{2}\ge4 as a distance statement. Since x2=x2x^{2}=|x|^{2}, the condition is x2|x|\ge2: the point xx is at least 22 away from the origin. That is satisfied on the two rays outside [2,2][-2,2], not between them:

    x2orx2x\le-2\qquad\text{or}\qquad x\ge2

  4. Write the solution set.

    (,2][2,)(-\infty,-2]\cup[2,\infty)

    Both endpoints are included: at x=±2x=\pm2, 2(4)+8=0-2(4)+8=0, and 000\le0 is true ✓.

  5. Confirm with the graph and by testing. y=2x2+8y=-2x^{2}+8 is a downward parabola with roots at ±2\pm2 and vertex (0,8)(0,8); it is at or below zero exactly outside the roots. Testing: x=0x=0 gives 808\le0 ✗ (correctly excluded), x=3x=3 gives 18+8=100-18+8=-10\le0 ✓, x=3x=-3 gives 100-10\le0 ✓. The raw inequality and the claimed set agree at 5151 exact rational points ✓.

Answer

x2orx2x\le-2\quad\text{or}\quad x\ge2

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