Question

1) х 4 -13х 2+36=0; 2) х 4 -х 2 -12=0; 3) (х 2 -3) 2 -4(х 2 -3)-12 =0;

Answer 1

Ответ:

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Объяснение:

ok

Answer 2

To solve these equations, let's go through each one step by step:

1. $x^4 - 13x^2 + 36 = 0$

We can notice that this equation is a quadratic equation in terms of $x^2$. Let's substitute $x^2$ with a variable, say $y$:

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

Now, we can solve this quadratic equation by factoring:

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

So, either $y - 4 = 0$ or $y - 9 = 0$.

For $y - 4 = 0$, we have $y = 4$. Substituting back $y = x^2$, we get $x^2 = 4$. Taking the square root of both sides, we have $x = \pm 2$.

For $y - 9 = 0$, we have $y = 9$. Substituting back $y = x^2$, we get $x^2 = 9$. Taking the square root of both sides, we have $x = \pm 3$.

Therefore, the solutions to the equation are $x = 2$, $x = -2$, $x = 3$, and $x = -3$.

2. $x^4 - x^2 - 12 = 0$

Similar to the previous equation, let's substitute $x^2$ with a variable, say $y$:

$y^2 - y - 12 = 0$

This quadratic equation can be factored:

$(y - 4)(y + 3) = 0$

So, either $y - 4 = 0$ or $y + 3 = 0$.

For $y - 4 = 0$, we have $y = 4$. Substituting back $y = x^2$, we get $x^2 = 4$. Taking the square root of both sides, we have $x = \pm 2$.

For $y + 3 = 0$, we have $y = -3$. Substituting back $y = x^2$, we get $x^2 = -3$. However, there are no real solutions for this case since the square root of a negative number is not real.

Therefore, the solutions to the equation are $x = 2$ and $x = -2$.

3. $(x^2 - 3)^2 - 4(x^2 - 3) - 12 = 0$

Let's simplify this equation step by step:

Expand the square terms:

$x^4 - 6x^2 + 9 - 4x^2 + 12 - 12 = 0$

Combine like terms:

$x^4 - 10x^2 + 9 = 0$

Now, let's substitute $x^2$ with a variable, say $y$:

$y^2 - 10y + 9 = 0$

This quadratic equation can be factored:

$(y - 1)(y - 9) = 0$

So, either $y - 1 = 0$ or $y - 9 = 0$.

For $y - 1 = 0$, we have $y = 1$. Substituting back $y = x^2$, we get (x