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Вопрос от Анонимного юзера 13 января 2025 17:15

-0,6x² - 12 = 0

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13 января 2025 17:15

To solve the quadratic equation (-0.6x^2 - 12 = 0), we can first isolate the (x^2) term.

  1. Add 12 to both sides: [ -0.6x^2 = 12 ]

  2. Now, divide both sides by -0.6: [ x^2 = \frac{12}{-0.6} ]

    Simplifying the right side: [ x^2 = -20 ]

  3. Since (x^2 = -20), we see that there are no real solutions because the square of a real number cannot be negative. Instead, we will look for complex solutions:

    Taking the square root of both sides: [ x = \pm \sqrt{-20} ]

  4. We can express (-20) as (20i^2) (since (i) is the imaginary unit): [ x = \pm \sqrt{20} \cdot i ]

  5. Simplifying (\sqrt{20}): [ \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} ]

  6. Therefore, the solutions are: [ x = \pm 2\sqrt{5}i ]

In conclusion, the solutions to the equation (-0.6x^2 - 12 = 0) are: [ x = 2\sqrt{5}i \quad \text{and} \quad x = -2\sqrt{5}i ]

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-0,6x² - 12 = 0
To solve the quadratic equation \(-0.6x^2 - 12 = 0\), we can first isolate the \(x^2\) term. 1. Add 12 to both sides: \[ -0.6x^2 = 12 \] 2. Now, divide both sides by -0.6: \[ x^2 = \frac{12}{-0.6} \] Simplifying the right side: \[ x^2 = -20 \] 3. Since \(x^2 = -20\), we see that there are no real solutions because the square of a real number cannot be negative. Instead, we will look for complex solutions: Taking the square root of both sides: \[ x = \pm \sqrt{-20} \] 4. We can express \(-20\) as \(20i^2\) (since \(i\) is the imaginary unit): \[ x = \pm \sqrt{20} \cdot i \] 5. Simplifying \(\sqrt{20}\): \[ \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \] 6. Therefore, the solutions are: \[ x = \pm 2\sqrt{5}i \] In conclusion, the solutions to the equation \(-0.6x^2 - 12 = 0\) are: \[ x = 2\sqrt{5}i \quad \text{and} \quad x = -2\sqrt{5}i \]