What is the answer to sqrt(x^4+x^2+1)=x^2-3x+1 ?

The answer to sqrt(x^4+x^2+1)=x^2-3x+1 is x=0

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sqrt(x^4+x^2+1)=x^2-3x+1

x^{4} + x^{2} + 1 = x^{2} - 3 x + 1

x = 0

Solution steps

x^{4} + x^{2} + 1 = x^{2} - 3 x + 1

Square both sides:

x^{4} + x^{2} + 1 = x^{4} - 6 x^{3} + 11 x^{2} - 6 x + 1

x^{4} + x^{2} + 1 = x^{4} - 6 x^{3} + 11 x^{2} - 6 x + 1

Solve

x^{4} + x^{2} + 1 = x^{4} - 6 x^{3} + 11 x^{2} - 6 x + 1 : x = 0

x = 0

Verify Solutions:

x = 0

True

The solution is

x = 0

x + 40 + x = 10

b^{\frac{4}{3}} = 625

x^{2} - 6 x + 2 = 0

7 2 x + 3 = 5

3 \cdot x^{2} + 2 2 x - 23 = 0

What is the answer to sqrt(x^4+x^2+1)=x^2-3x+1 ?
The answer to sqrt(x^4+x^2+1)=x^2-3x+1 is x=0
Solve for x: sqrt(x^4+x^2+1)=x^2-3x+1
The solution is x=0