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

The answer to 1+sqrt(x)=sqrt(x^2+1) is x=0,x\approx 2.31459…

Solve for x: 1+sqrt(x)=sqrt(x^2+1)

The solution is x=0,x\approx 2.31459…

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1 + x = x^{2} + 1

x = 0, x \approx 2.31459 \dots

Solution steps

1 + x = x^{2} + 1

Remove square roots

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

Solve

4 x = x^{4} - 2 x^{3} + x^{2} : x = 0, x \approx 2.31459 \dots

x = 0, x \approx 2.31459 \dots

Verify Solutions:

x = 0

True

, x \approx 2.31459 \dots

True

The solutions are

x = 0, x \approx 2.31459 \dots

y = y

x, f (x - 3) = x + 3 + 2

x + 3 x + 4 = 9 x

n = 12 - n

What is the answer to 1+sqrt(x)=sqrt(x^2+1) ?
The answer to 1+sqrt(x)=sqrt(x^2+1) is x=0,x\approx 2.31459…
Solve for x: 1+sqrt(x)=sqrt(x^2+1)
The solution is x=0,x\approx 2.31459…