What is the answer to \sqrt[5]{2x^2+1}-2=0 ?

The answer to \sqrt[5]{2x^2+1}-2=0 is x=sqrt(31/2),x=-sqrt(31/2)

Solve for x: \sqrt[5]{2x^2+1}-2=0

The solution is x=sqrt(31/2),x=-sqrt(31/2)

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x = \frac{31}{2}, x = - \frac{31}{2}

Decimal Notation

x = 3.93700 \dots, x = - 3.93700 \dots

Solution steps

Move

2

to the right side

Take both sides of the equation to the power of

5 : 2 x^{2} + 1 = 32

2 x^{2} + 1 = 32

Solve

2 x^{2} + 1 = 32 : x = \frac{31}{2}, x = - \frac{31}{2}

x = \frac{31}{2}, x = - \frac{31}{2}

Verify Solutions:

x = \frac{31}{2}

True

, x = - \frac{31}{2}

True

The solutions are

x = \frac{31}{2}, x = - \frac{31}{2}

5 - x = x + 1

6 x + 1 = 5

x + x + 7 = 5

8 x - 7 = 5

5 x + 6 = 4

What is the answer to \sqrt[5]{2x^2+1}-2=0 ?
The answer to \sqrt[5]{2x^2+1}-2=0 is x=sqrt(31/2),x=-sqrt(31/2)
Solve for x: \sqrt[5]{2x^2+1}-2=0
The solution is x=sqrt(31/2),x=-sqrt(31/2)