Question

9х^4+23х^2-12=0 решите

Answer 1

Смотри....................

Answer 2

To solve the quadratic equation $9x^4 + 23x^2 - 12 = 0$, we can use substitution to convert it into a standard quadratic equation. Let's make a substitution $y = x^2$ and rewrite the equation:

$9y^2 + 23y - 12 = 0$

Now, we can solve this quadratic equation for $y$. To do that, we can factor it or use the quadratic formula:

The factored form of the equation is:

$9y^2 + 23y - 12 = (3y - 4)(3y + 3) = 0$

Now, set each factor to zero and solve for $y$:

1. $3y - 4 = 0$ $y = \frac{4}{3}$

2. $3y + 3 = 0$ $y = -1$

Now that we have the values of $y$, we can revert to the substitution $y = x^2$ to find the corresponding values of $x$:

1. $y = \frac{4}{3}$ $x^2 = \frac{4}{3}$ $x = \pm \sqrt{\frac{4}{3}}$ $x = \pm \frac{2}{\sqrt{3}}$

2. $y = -1$ $x^2 = -1$

Since we cannot take the square root of a negative number in the real number system, there are no real solutions for $x$ in this case.

Therefore, the solutions to the original equation $9x^4 + 23x^2 - 12 = 0$ are:

$x = \pm \frac{2}{\sqrt{3}}$