What is the answer to \sqrt[4]{1-x}+\sqrt[4]{1+x}=\sqrt[4]{2} ?

The answer to \sqrt[4]{1-x}+\sqrt[4]{1+x}=\sqrt[4]{2} is x=1,x=-1

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\sqrt[4]{1-x}+\sqrt[4]{1+x}=\sqrt[4]{2}

4 1 - x + 4 1 + x = 42

x = 1, x = - 1

Solution steps

4 1 - x + 4 1 + x = 42

Rewrite the equation with

4 1 - x = u

and

x = - u^{4} + 1

u + 4 1 + (- u^{4} + 1) = 42

Solve

u + 4 1 + (- u^{4} + 1) = 42 : u = 0, u \approx 1.18920 \dots

u = 0, u \approx 1.18920 \dots

Substitute back

u = 4 1 - x,

solve for

x

Solve

4 1 - x = 0 : x = 1

Solve

4 1 - x = 1.18920 \dots : x = - 1

x = 1, x = - 1

Verify Solutions:

x = 1

True

, x = - 1

True

The solutions are

x = 1, x = - 1

x - x - 3 = 3

3 (2 - x)^{\frac{3}{2}} = - 24

2 - 3 x = 7

0.2 x + \frac{0.5}{x - 0.05} = 0

x - 8 = x + 4

What is the answer to \sqrt[4]{1-x}+\sqrt[4]{1+x}=\sqrt[4]{2} ?
The answer to \sqrt[4]{1-x}+\sqrt[4]{1+x}=\sqrt[4]{2} is x=1,x=-1
Solve for x: \sqrt[4]{1-x}+\sqrt[4]{1+x}=\sqrt[4]{2}
The solution is x=1,x=-1