Question

(X^2+4x^2)+(x+2)^2=4

Answer 1

x²+4x²+x²+4x+4-4=0

6x²+4x=0

3x²+2x=0

x(3x+2)=0

x1= 0, x2= -2/3

Answer 2

To solve the equation (x^2+4x^2)+(x+2)^2=4, we can begin by simplifying the left-hand side of the equation:

(x^2+4x^2)+(x+2)^2=4 5x^2 + (x^2 + 4x + 4) = 4 6x^2 + 4x = 0

Now we can factor out 2x from the expression on the right-hand side:

2x(3x+2) = 0

This equation is true if either 2x = 0 or 3x+2 = 0. Solving for x in each of these equations, we get:

2x = 0, x = 0 3x+2 = 0, x = -2/3

Therefore, the solutions to the equation (x^2+4x^2)+(x+2)^2=4 are x = 0 and x = -2/3.