Question

Решить уравнения.x⁴-7x²-18=0​

Answer 1

Ответ:

-3; 3.

Объяснение:

Решить уравнение $x^{4} -7x^{2} -18=0$

Данное уравнение биквадратное. Решим заменой.

Пусть x² = t, t ≥ 0 . Тогда уравнение принимает вид: t²-7t - 18 =0.

$D =(-7) ^{2} -4\cdot 1 \cdot( -18) = 49+ 72 =121 = 11^{2} ;\\\\t{_1}= \dfrac{7-11}{2} =-\dfrac{4}{2} =-2;\\\\t{_2}= \dfrac{7+11}{2} =-\dfrac{18}{2} =9.$

Так как  t ≥ 0, то  t = 9.

Значит, х² = 9

            х= - 3 или х= 3.

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Answer 2

To solve the quadratic equation x⁴ - 7x² - 18 = 0, we can use a substitution method to convert it into a quadratic equation in terms of another variable. Let's introduce a new variable y = x². Then the equation becomes:

y² - 7y - 18 = 0

Now we have a simple quadratic equation in terms of y. Let's solve for y:

To solve y² - 7y - 18 = 0, we can factor it:

(y - 9)(y + 2) = 0

Now set each factor to zero and solve for y:

y - 9 = 0 or y + 2 = 0

If y - 9 = 0, then y = 9 If y + 2 = 0, then y = -2

Now, we recall that y = x², so we substitute the values of y back into the equation:

For y = 9: x² = 9 x = ±√9 x = ±3

For y = -2: x² = -2

However, the square root of a negative number is not a real number. Thus, there are no real solutions for this case.

Therefore, the solutions to the original equation x⁴ - 7x² - 18 = 0 are x = 3 and x = -3.