What are the solutions to the equation 16x^4-52x^2+36=0 ?

The solutions to the equation 16x^4-52x^2+36=0 are x= 3/2 ,x=-3/2 ,x=1,x=-1

Find the zeros of 16x^4-52x^2+36=0

The zeros of 16x^4-52x^2+36=0 are x= 3/2 ,x=-3/2 ,x=1,x=-1

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16 x^{4} - 52 x^{2} + 36 = 0

x = \frac{3}{2}, x = - \frac{3}{2}, x = 1, x = - 1

Decimal Notation

x = 1.5, x = - 1.5, x = 1, x = - 1

Solution steps

16 x^{4} - 52 x^{2} + 36 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

16 u^{2} - 52 u + 36 = 0

Solve

16 u^{2} - 52 u + 36 = 0 : u = \frac{9}{4}, u = 1

u = \frac{9}{4}, u = 1

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = \frac{9}{4} : x = \frac{3}{2}, x = - \frac{3}{2}

Solve

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

The solutions are

x = \frac{3}{2}, x = - \frac{3}{2}, x = 1, x = - 1

x^{3} + x^{2} + 4 x + 4

x^{5} = 18 x^{3} - 81 x

m^{5} + 5 m^{4} - 2 m^{3} - 10 m^{2} + m + 5 = 0

(x - 1)^{4} - (x - 1)^{2} - 12 = 0

x, (x^{2} + y^{2})^{3} = y^{4}

What are the solutions to the equation 16x^4-52x^2+36=0 ?
The solutions to the equation 16x^4-52x^2+36=0 are x= 3/2 ,x=-3/2 ,x=1,x=-1
Find the zeros of 16x^4-52x^2+36=0
The zeros of 16x^4-52x^2+36=0 are x= 3/2 ,x=-3/2 ,x=1,x=-1