What are the solutions to the equation (x+1)^2-3x(x+1)^2=0 ?

The solutions to the equation (x+1)^2-3x(x+1)^2=0 are x=-1,x= 1/3

Find the zeros of (x+1)^2-3x(x+1)^2=0

The zeros of (x+1)^2-3x(x+1)^2=0 are x=-1,x= 1/3

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(x + 1)^{2} - 3 x (x + 1)^{2} = 0

x = - 1, x = \frac{1}{3}

Decimal Notation

x = - 1, x = 0.33333 \dots

Solution steps

(x + 1)^{2} - 3 x (x + 1)^{2} = 0

Factor

(x + 1)^{2} - 3 x (x + 1)^{2} : - (x + 1)^{2} (3 x - 1)

- (x + 1)^{2} (3 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or 3 x - 1 = 0

Solve

x + 1 = 0 : x = - 1

Solve

3 x - 1 = 0 : x = \frac{1}{3}

The solutions are

x = - 1, x = \frac{1}{3}

3 a^{4} + 2 a - 46 = 54

- 2 x^{4} + 9 = 2 x - 1

- 2 x^{4} + 9 = 2 x - 3

x^{4} + x^{3} - 4 x^{2} + 2 x - 12 = 0

3 x^{3} - 3 x^{2} - 270 x = 0

What are the solutions to the equation (x+1)^2-3x(x+1)^2=0 ?
The solutions to the equation (x+1)^2-3x(x+1)^2=0 are x=-1,x= 1/3
Find the zeros of (x+1)^2-3x(x+1)^2=0
The zeros of (x+1)^2-3x(x+1)^2=0 are x=-1,x= 1/3