What is the answer to \sqrt[3]{2-x}+sqrt(x-1)=1 ?

The answer to \sqrt[3]{2-x}+sqrt(x-1)=1 is x=2,x=1,x=10

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\sqrt[3]{2-x}+sqrt(x-1)=1

x = 2, x = 1, x = 10

Solution steps

Rewrite the equation with

and

x = - u^{3} + 2

u + (- u^{3} + 2) - 1 = 1

Solve

u + (- u^{3} + 2) - 1 = 1 : u = 0, u = 1, u = - 2

u = 0, u = 1, u = - 2

Substitute back

solve for

x

Solve

x = 2, x = 1, x = 10

Verify Solutions:

x = 2

True

, x = 1

True

, x = 10

True

The solutions are

x = 2, x = 1, x = 10

x + 1 = x + 1

x^{2} - 3 x + 2 = 0

2 y + 6 + 3 = y + 2

x + 1 = x + 2

9 = n^{\frac{2}{3}}

What is the answer to \sqrt[3]{2-x}+sqrt(x-1)=1 ?
The answer to \sqrt[3]{2-x}+sqrt(x-1)=1 is x=2,x=1,x=10
Solve for x: \sqrt[3]{2-x}+sqrt(x-1)=1
The solution is x=2,x=1,x=10