What are the solutions to the equation 16x^4-41x^2+25=0 ?

The solutions to the equation 16x^4-41x^2+25=0 are x= 5/4 ,x=-5/4 ,x=1,x=-1

Find the zeros of 16x^4-41x^2+25=0

The zeros of 16x^4-41x^2+25=0 are x= 5/4 ,x=-5/4 ,x=1,x=-1

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

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

Decimal Notation

x = 1.25, x = - 1.25, x = 1, x = - 1

Solution steps

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

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

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

Solve

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

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

Substitute back

u = x^{2},

solve for

x

Solve

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

Solve

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

The solutions are

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

x^{6} - 124 x^{3} - 125 = 0

48 x + 4 x^{2} - 2 x^{3} = 0

216 x = 6 x^{3}

x^{4} + x^{2} - 20 = 0

x^{3} + x^{2} - 25 x - 25 = 0

What are the solutions to the equation 16x^4-41x^2+25=0 ?
The solutions to the equation 16x^4-41x^2+25=0 are x= 5/4 ,x=-5/4 ,x=1,x=-1
Find the zeros of 16x^4-41x^2+25=0
The zeros of 16x^4-41x^2+25=0 are x= 5/4 ,x=-5/4 ,x=1,x=-1