Question

2(4x-1) ^{4} +3(4x-1) ^{2} -5=0 Решите уравнение методом замены переменной

Answer 1

Решение этого задания на фото

Answer 2

$2(4x-1)^4+3(4x-1)^2-5=0$
заменим переменную $y=(4x-1)^2$
$2y^2+3y-5=0 \\ D=b^2-4ac=49 \\ y_1_,_2= \frac{-b^+_- \sqrt{D} }{2a} \\ y_1=-2.5 \\ y_2=1 \\ \\ (4x-1)^2=-2.5 \\ 16x^2-8x+1=-2.5 \\ 16x^2-8x+3.5=0 \\ D=-160\ \textless \ 0$
действительных корней нет

$16x^2-8x+1=1 \\ 16x^2-8x=0 \\ x(16x-8)=0 \\ x_1=0 \\ 16x-8=0 \\ 16x=8 \\ x_2=0.5$

Answer 3

Solving the Equation Using Variable Substitution

To solve the equation 2(4x-1)^4 + 3(4x-1)^2 - 5 = 0 using the method of variable substitution, we can let u = (4x-1). This will allow us to transform the given equation into a quadratic equation in terms of u.

Substituting the Variable

Let's substitute u = (4x-1) into the equation:

2u^4 + 3u^2 - 5 = 0

Now, we have a quadratic equation in terms of u.

Solving the Quadratic Equation

We can solve this quadratic equation using the quadratic formula:

u = (-b ± √(b^2 - 4ac)) / (2a)

Where a = 2, b = 3, and c = -5.

Calculating the Discriminant

First, let's calculate the discriminant D = b^2 - 4ac:

D = 3^2 - 4*2*(-5) D = 9 + 40 D = 49

Since the discriminant is positive, the quadratic equation has two distinct real roots.

Finding the Roots

Now, we can use the quadratic formula to find the roots of the equation:

u = (-3 ± √49) / (2*2)

This simplifies to:

u = (-3 ± 7) / 4

So, the roots are:

u1 = (-3 + 7) / 4 = 1 u2 = (-3 - 7) / 4 = -2.5

Reverting to the Original Variable

Now that we have found the values of u, we can revert back to the original variable x using the substitution u = (4x-1):

For u = 1: 4x - 1 = 1 4x = 2 x = 0.5

For u = -2.5: 4x - 1 = -2.5 4x = -1.5 x = -0.375

Therefore, the solutions to the original equation are: x = 0.5 and x = -0.375.