Question

1) x⁴ -18x² +32 =02) (x²+x) +6=0​

Answer 1

x^4-18x^2+32=0

По т. Виета: x1*x2=32 x1+x2=18 =>

x^2=16 => x=4 x=-4

x^2=2 => $x=\sqrt{2}$ $x=-\sqrt{2}$

Ответ: -4 ; $-\sqrt{2}$ ; $\sqrt{2}$ ; 4.

x^2+x+6=0

D=1^2-4*6*1<0 => решений нет.

Answer 2

Let's solve the two equations:

1. x⁴ - 18x² + 32 = 0

To solve this equation, we can make a substitution. Let y = x². Now the equation becomes:

y² - 18y + 32 = 0

This is a quadratic equation in terms of y. We can factor it:

(y - 16)(y - 2) = 0

Now, solve for y:

y - 16 = 0 or y - 2 = 0

y = 16 or y = 2

Now, substitute y back with x²:

x² = 16 or x² = 2

Now, take the square root of both sides:

x = ±4 or x = ±√2

So, the solutions to the equation are x = 4, x = -4, x = √2, and x = -√2.

2. (x² + x) + 6 = 0

This equation is a quadratic equation in terms of x. Let's solve it:

x² + x + 6 = 0

To solve this quadratic equation, you can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 1, b = 1, and c = 6. Plug these values into the quadratic formula:

x = (-1 ± √(1² - 4(1)(6))) / (2(1))

x = (-1 ± √(1 - 24)) / 2

x = (-1 ± √(-23)) / 2

Since the discriminant (the value inside the square root) is negative, there are no real solutions to this equation. The solutions are complex numbers:

x = (-1 ± √(-23))/2

So, the solutions to the equation are x = (-1 + √(-23))/2 and x = (-1 - √(-23))/2.