Question

[tex]x^{4} -13 x^{2} +36=0

Answer 1

$x^{4}-13x^{2}+36=0\\ x^{2}=t\\ t^{2}-13t+36=0\\ t_{1}=9\\ t_{2}=4\\ x^{2}=9\\ x=3;x=-3;\\ x^{2}=4\\ x=2;x=-2;$

Ответ: -2, 2, -3, 3.

Answer 2

To solve the quadratic equation $x^4 - 13x^2 + 36 = 0$, we can use a substitution to simplify it. Let $y = x^2$, so the equation becomes:

$y^2 - 13y + 36 = 0$

Now, we have a quadratic equation in terms of $y$. We can factor this equation or use the quadratic formula to find the values of $y$ and then solve for $x$. Let's factor the equation:

$(y - 9)(y - 4) = 0$

Now, we have two possible values for $y$:

1. $y - 9 = 0$, which gives $y = 9$.
2. $y - 4 = 0$, which gives $y = 4$.

Now, we can solve for $x$ by taking the square root of both sides for each value of $y$:

1. For $y = 9$, we have $x^2 = 9$, which gives $x = \pm 3$.
2. For $y = 4$, we have $x^2 = 4$, which gives $x = \pm 2$.

So, the solutions to the original equation $x^4 - 13x^2 + 36 = 0$ are $x = \pm 3$ and $x = \pm 2$.