What is the answer to \sqrt[3]{x^2-3x-5}=0 ?

The answer to \sqrt[3]{x^2-3x-5}=0 is x=(3+sqrt(29))/2 ,x=(3-sqrt(29))/2

Solve for x: \sqrt[3]{x^2-3x-5}=0

The solution is x=(3+sqrt(29))/2 ,x=(3-sqrt(29))/2

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x = \frac{3 + 29}{2}, x = \frac{3 - 29}{2}

Decimal Notation

x = 4.19258 \dots, x = - 1.19258 \dots

Solution steps

Take both sides of the equation to the power of

3 : x^{2} - 3 x - 5 = 0

x^{2} - 3 x - 5 = 0

Solve

x^{2} - 3 x - 5 = 0 : x = \frac{3 + 29}{2}, x = \frac{3 - 29}{2}

x = \frac{3 + 29}{2}, x = \frac{3 - 29}{2}

Verify Solutions:

x = \frac{3 + 29}{2}

True

, x = \frac{3 - 29}{2}

True

The solutions are

x = \frac{3 + 29}{2}, x = \frac{3 - 29}{2}

2 x - 1 = x + 1

3 x + 2 - x - 2 = 0

q, p = q r - 1

3 - 1 + 4 x = x - 2

What is the answer to \sqrt[3]{x^2-3x-5}=0 ?
The answer to \sqrt[3]{x^2-3x-5}=0 is x=(3+sqrt(29))/2 ,x=(3-sqrt(29))/2
Solve for x: \sqrt[3]{x^2-3x-5}=0
The solution is x=(3+sqrt(29))/2 ,x=(3-sqrt(29))/2