What are the solutions to the equation 5x^{27}x-53=0 ?

The solutions to the equation 5x^{27}x-53=0 are x=\sqrt[28]{53/5},x=-\sqrt[28]{53/5}

Find the zeros of 5x^{27}x-53=0

The zeros of 5x^{27}x-53=0 are x=\sqrt[28]{53/5},x=-\sqrt[28]{53/5}

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5 x^{27} x - 53 = 0

Decimal Notation

x = 1.08797 \dots, x = - 1.08797 \dots

Solution steps

5 x^{27} x - 53 = 0

Move

53

to the right side

5 x^{27} x = 53

Simplify

5 x^{28} = 53

Divide both sides by

5

\frac{5 x ^{28}}{5} = \frac{53}{5}

x^{28} = \frac{53}{5}

For

x^{n} = f (a)

, n is even, the solutions are

m^{4} + 2 m^{3} + 2 m^{2} = 0

x^{3} + x^{2} - x - 10 = 0

(x + 2) (2 x^{2} - 5 x) - x^{3} = 8

x, x^{3} + y^{3} + z^{3} = k^{1} - 100

4 x - 2 x^{4} - 2 = 0

What are the solutions to the equation 5x^{27}x-53=0 ?
The solutions to the equation 5x^{27}x-53=0 are x=\sqrt[28]{53/5},x=-\sqrt[28]{53/5}
Find the zeros of 5x^{27}x-53=0
The zeros of 5x^{27}x-53=0 are x=\sqrt[28]{53/5},x=-\sqrt[28]{53/5}