What is the answer to sqrt(x^2-\sqrt{2x+1)}=2 ?

The answer to sqrt(x^2-\sqrt{2x+1)}=2 is x\approx 2.54302…

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sqrt(x^2-\sqrt{2x+1)}=2

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

x \approx 2.54302 \dots

Solution steps

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

Remove square roots

2 x + 1 = 16 - 8 x^{2} + x^{4}

Solve

2 x + 1 = 16 - 8 x^{2} + x^{4} : x \approx 2.54302 \dots, x \approx 1.42715 \dots

x \approx 2.54302 \dots, x \approx 1.42715 \dots

Verify Solutions:

x \approx 2.54302 \dots

True

, x \approx 1.42715 \dots

False

The solution is

x \approx 2.54302 \dots

7 x = 13 x

y \cdot y = 15

x + 3 + 1 = 3 x

y^{2} - 9 = 9 - y

\frac{2 x}{5 + x} = 3 x

What is the answer to sqrt(x^2-\sqrt{2x+1)}=2 ?
The answer to sqrt(x^2-\sqrt{2x+1)}=2 is x\approx 2.54302…
Solve for x: sqrt(x^2-\sqrt{2x+1)}=2
The solution is x\approx 2.54302…