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Solve -2x² + 2 = 0

The solutions of -2x² + 2 = 0 are x = -1 or x = 1.

x² +x += 0
−2x2+2=0⇒x=−1, 1-2x^2 + 2 = 0 \Rightarrow x = -1,\ 1
Discriminant D = 16 · x ≈ -1, 1

Change a, b or c (whole numbers, a not 0) and the exact solutions follow.

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Graph of y = -2x² + 2-3-2-1123-20-15-10-55x = -1x = 1vertexy = -2x² + 2
Graph of y = -2x² + 2. The curve crosses the x-axis at x = -1 and x = 1; the vertex is at (0, 2).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = -2, b = 0, c = 2.

    −2x2+2=0-2x^2 + 2 = 0
  2. 2

    Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.

    D=02−4⋅(−2)⋅2=0+16=16D = 0^2 - 4\cdot(-2)\cdot2 = 0 + 16 = 16
  3. 3

    D is positive, so there are two different real solutions.

  4. 4

    Method 1, factoring. The roots are rational, so the left side factors:

    −2(x+1)(x−1)=0-2\left(x + 1\right)\left(x - 1\right) = 0
  5. 5

    A product is zero when one of its factors is zero, so set each factor equal to 0 and solve:

    x+1=0⇒x=−1x−1=0⇒x=1x + 1 = 0 \Rightarrow x = -1\qquad x - 1 = 0 \Rightarrow x = 1
  6. 6

    Method 2, The quadratic formula works for every quadratic:

    x=−b±D2a=0±16−4x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{16}}{-4}
  7. 7

    Simplify to get the solutions:

    x=−1orx=1x = -1\quad\text{or}\quad x = 1
  8. 8

    Check by putting x = -1 back in: a·x² + b·x + c = -2·(1) + 0·(-1) + 2 = 0 ✓

LaTeX source of the formulas on this page
\[ -2x^2 + 2 = 0\ \Rightarrow\ x = -1,\ 1 \]
\[ -2x^2 + 2 = 0 \]
\[ D = 0^2 - 4\cdot(-2)\cdot2 = 0 + 16 = 16 \]
\[ -2\left(x + 1\right)\left(x - 1\right) = 0 \]
\[ x + 1 = 0 \Rightarrow x = -1\qquad x - 1 = 0 \Rightarrow x = 1 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{16}}{-4} \]
\[ x = -1\quad\text{or}\quad x = 1 \]

The parabola y = -2x² + 2

xy
-3-16
-2-6
-10
02
10
2-6
3-16

Questions

What are the solutions of -2x² + 2 = 0?

The solutions of -2x² + 2 = 0 are x = -1 or x = 1.

Does -2x² + 2 = 0 factor?

Yes: -2(x+1)(x-1) = 0.

What is the vertex of y = -2x² + 2?

(0, 2). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 0, product = c/a = -1.

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