What is the answer to \sqrt[5]{2x^2+9}-2=0 ?

The answer to \sqrt[5]{2x^2+9}-2=0 is x=sqrt(23/2),x=-sqrt(23/2)

Solve for x: \sqrt[5]{2x^2+9}-2=0

The solution is x=sqrt(23/2),x=-sqrt(23/2)

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x = \frac{23}{2}, x = - \frac{23}{2}

Decimal Notation

x = 3.39116 \dots, x = - 3.39116 \dots

Solution steps

Move

2

to the right side

Take both sides of the equation to the power of

5 : 2 x^{2} + 9 = 32

2 x^{2} + 9 = 32

Solve

2 x^{2} + 9 = 32 : x = \frac{23}{2}, x = - \frac{23}{2}

x = \frac{23}{2}, x = - \frac{23}{2}

Verify Solutions:

x = \frac{23}{2}

True

, x = - \frac{23}{2}

True

The solutions are

x = \frac{23}{2}, x = - \frac{23}{2}

- 243 = - (- 3 - 12 x)^{\frac{5}{4}}

(c - 1)^{\frac{1}{2}} - 1.33 = 0

25 - 4 x = x + 5

a x = 2 x + 3

a, 2 3 a + b = c

What is the answer to \sqrt[5]{2x^2+9}-2=0 ?
The answer to \sqrt[5]{2x^2+9}-2=0 is x=sqrt(23/2),x=-sqrt(23/2)
Solve for x: \sqrt[5]{2x^2+9}-2=0
The solution is x=sqrt(23/2),x=-sqrt(23/2)