Question

1) x+3√x-18=0 2) (x+1)^4-2(x+1)^2-3=0

Answer 1

Ответ: x=36

1) x+3√x-18=0

3√x-18= - x

9(x-18)=x^2

9x-162=x^2

9x-162-x^2=0

-x^2+9x-162=0

x= -(-9)±$\sqrt{(-9)^{2}-4*1*162}$ \ 2*1

x1= $\frac{9+\sqrt{81-648} }{2}$

x2= $\frac{9+\sqrt{81+648} }{2}$

x1= $\frac{9+\sqrt{-567} }{2}$=6

x2=$\frac{9-\sqrt{-567} }{2}$=-3

$\sqrt{x} =6\\x=36$

$\sqrt{x} =-3$=∉

Answer 2

It looks like you have provided two equations. Let's solve them one by one:

Equation 1: x + 3√x - 18 = 0

To solve this equation, we can isolate the square root term and then solve for x:

1. Move the terms around: x + 3√x = 18

2. Square both sides of the equation to eliminate the square root: (x + 3√x)^2 = 18^2 x^2 + 6x√x + 9x = 324

3. Combine like terms: x^2 + 9x + 6x√x - 324 = 0

4. Factor out the common factor of √x from the last two terms: x^2 + 9x + 6√x(x) - 324 = 0

5. Substitute y = √x: x^2 + 9x + 6xy - 324 = 0

6. Now this is a quadratic equation in terms of x and y. Solve it for y: y = (-6x ± √(36x^2 - 4(x^2 - 324))) / 2 y = (-6x ± √(4*324)) / 2 y = (-6x ± 36) / 2 y = -3x ± 18

7. Substitute back y = √x: √x = -3x ± 18

8. Square both sides to solve for x: x = (-3x ± 18)^2 x = 9x^2 ± 108x + 324 8x^2 ± 108x + 324 = 0

Now you have two quadratic equations to solve for x. To continue, you can use the quadratic formula or factorization methods to find the solutions.

Equation 2: (x + 1)^4 - 2(x + 1)^2 - 3 = 0

Let's solve this equation:

1. Expand the terms: x^4 + 4x^3 + 6x^2 + 4x + 1 - 2(x^2 + 2x + 1) - 3 = 0

2. Distribute the -2 in the second term: x^4 + 4x^3 + 6x^2 + 4x + 1 - 2x^2 - 4x - 2 - 3 = 0

3. Combine like terms: x^4 + 4x^3 + 4x^2 = 0

4. Factor out x^2: x^2(x^2 + 4x + 4) = 0

5. Factor the quadratic term: x^2(x + 2)^2 = 0

6. Set each factor equal to zero: x^2 = 0 --> x = 0 x + 2 = 0 --> x = -2

So, the solutions for this equation are x = 0 and x = -2.