Question

1)X^4-13x^2+36=0 2)x^4+15x^2-16=0

Answer 1

Ответ:

на фото....

________________________________

Answer 2

To solve these equations, we can use a substitution method to transform them into quadratic equations in terms of a new variable, say $y$. Let's solve each equation separately:

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

Let's make a substitution: $y = x^2$. The equation becomes:

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

This quadratic equation can be factored as $(y - 9)(y - 4) = 0$, which gives two possible values for $y$:

1. $y - 9 = 0 \implies y = 9$
2. $y - 4 = 0 \implies y = 4$

Now, we substitute back $x^2$ for $y$:

For $y = 9$: $x^2 = 9 \implies x = \pm 3$

For $y = 4$: $x^2 = 4 \implies x = \pm 2$

So, the solutions for the equation $x^4 - 13x^2 + 36 = 0$ are $x = -3, -2, 2, 3$.

2) $x^4 + 15x^2 - 16 = 0$

Again, let's make a substitution: $y = x^2$. The equation becomes:

$y^2 + 15y - 16 = 0$

This quadratic equation can be factored as $(y + 16)(y - 1) = 0$, which gives two possible values for $y$:

1. $y + 16 = 0 \implies y = -16$
2. $y - 1 = 0 \implies y = 1$

Now, we substitute back $x^2$ for $y$:

For $y = -16$: $x^2 = -16$ This equation has no real solutions because the square of a real number cannot be negative.

For $y = 1$: $x^2 = 1 \implies x = \pm 1$

So, the solutions for the equation $x^4 + 15x^2 - 16 = 0$ are $x = -1, 1$.