Question

1. 4 х⁴ - 37 х² + 9=0

Answer 1

Пусть x^2=t, тогда получим
4t^2-37t+9
D=37^2-4*4*9=1225
t1=(37+35)/8=9, t2=(37-35)/8=1/4
Если t1=9, то x1=-3 или x2=3;
Если t2=1/4, то x3=-1/2 или x4=1/2.

Answer 2

We can solve this quadratic equation by using substitution. Let's make a substitution:

Let u = x²

Then the equation becomes:

4u² - 37u + 9 = 0

We can factor this equation to get:

(4u - 1)(u - 9) = 0

Substituting back u = x², we have:

(4x² - 1)(x² - 9) = 0

Now we can solve for x:

4x² - 1 = 0 or x² - 9 = 0

Solving for 4x² - 1 = 0, we get:

4x² = 1

x² = 1/4

x = ± 1/2

Solving for x² - 9 = 0, we get:

x² = 9

x = ± 3

Therefore, the solutions to the equation 4x⁴ - 37x² + 9 = 0 are:

x = -3, -1/2, 1/2, 3