What are the solutions to the equation 25x^4+16=41x^2 ?

The solutions to the equation 25x^4+16=41x^2 are x=1,x=-1,x= 4/5 ,x=-4/5

Find the zeros of 25x^4+16=41x^2

The zeros of 25x^4+16=41x^2 are x=1,x=-1,x= 4/5 ,x=-4/5

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25 x^{4} + 16 = 41 x^{2}

x = 1, x = - 1, x = \frac{4}{5}, x = - \frac{4}{5}

Decimal Notation

x = 1, x = - 1, x = 0.8, x = - 0.8

Solution steps

25 x^{4} + 16 = 41 x^{2}

Move

41 x^{2}

to the left side

25 x^{4} + 16 - 41 x^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

25 x^{4} - 41 x^{2} + 16 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

25 u^{2} - 41 u + 16 = 0

Solve

25 u^{2} - 41 u + 16 = 0 : u = 1, u = \frac{16}{25}

u = 1, u = \frac{16}{25}

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = 1 : x = 1, x = - 1

Solve

x^{2} = \frac{16}{25} : x = \frac{4}{5}, x = - \frac{4}{5}

The solutions are

x = 1, x = - 1, x = \frac{4}{5}, x = - \frac{4}{5}

3 x^{3} + 15 x^{2} - 2 x = 10

(x + 4) \cdot (- 4 x^{2} + 4 x - 1) = 0

n^{4} (n + 5) - 2 (5 + 7 n) = - 70

z^{4} = 6 z^{2} + 7

7 z^{3} + 30 z^{2} = 25 z

What are the solutions to the equation 25x^4+16=41x^2 ?
The solutions to the equation 25x^4+16=41x^2 are x=1,x=-1,x= 4/5 ,x=-4/5
Find the zeros of 25x^4+16=41x^2
The zeros of 25x^4+16=41x^2 are x=1,x=-1,x= 4/5 ,x=-4/5