What is the answer to sqrt(x^2+2x+1)=sqrt(x+1) ?

The answer to sqrt(x^2+2x+1)=sqrt(x+1) is x=0,x=-1

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sqrt(x^2+2x+1)=sqrt(x+1)

x^{2} + 2 x + 1 = x + 1

x = 0, x = - 1

Solution steps

x^{2} + 2 x + 1 = x + 1

Square both sides:

x^{2} + 2 x + 1 = x + 1

x^{2} + 2 x + 1 = x + 1

Solve

x^{2} + 2 x + 1 = x + 1 : x = 0, x = - 1

x = 0, x = - 1

Verify Solutions:

x = 0

True

, x = - 1

True

The solutions are

x = 0, x = - 1

x - 2 x - 6 = 0

(x - 6)^{\frac{1}{2}} = 4

120 = \frac{t}{0.005}

k, f = k^{\frac{1}{3}} \cdot l^{\frac{1}{2}}

289 + x = \frac{( 1 )}{( 5 )}

What is the answer to sqrt(x^2+2x+1)=sqrt(x+1) ?
The answer to sqrt(x^2+2x+1)=sqrt(x+1) is x=0,x=-1
Solve for x: sqrt(x^2+2x+1)=sqrt(x+1)
The solution is x=0,x=-1