What are the solutions to the equation (x+2)(x+2)=(x+2)2 ?

The solutions to the equation (x+2)(x+2)=(x+2)2 are x=0,x=-2

Find the zeros of (x+2)(x+2)=(x+2)2

The zeros of (x+2)(x+2)=(x+2)2 are x=0,x=-2

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(x + 2) (x + 2) = (x + 2) 2

x = 0, x = - 2

Solution steps

(x + 2) (x + 2) = (x + 2) \cdot 2

Move

(x + 2) 2

to the left side

(x + 2) (x + 2) - (x + 2) \cdot 2 = 0

Factor

(x + 2) (x + 2) - (x + 2) \cdot 2 : x (x + 2)

x (x + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + 2 = 0

Solve

x + 2 = 0 : x = - 2

The solutions to the quadratic equation are:

x = 0, x = - 2

4 b^{2} - 4 b - 35 = 0

x, x^{2} + y^{2} - z^{2} = 0

6 = - \frac{1}{2} t^{2} + 6 t - 9

(x - 7)^{2} - 18 = 0

k^{2} + 3 k - 18 = 0

What are the solutions to the equation (x+2)(x+2)=(x+2)2 ?
The solutions to the equation (x+2)(x+2)=(x+2)2 are x=0,x=-2
Find the zeros of (x+2)(x+2)=(x+2)2
The zeros of (x+2)(x+2)=(x+2)2 are x=0,x=-2