What are the solutions to the equation 4(x^4+1)=5x^2,x>0 ?

The solutions to the equation 4(x^4+1)=5x^2,x>0 are

Find the zeros of 4(x^4+1)=5x^2,x>0

The zeros of 4(x^4+1)=5x^2,x>0 are

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4 (x^{4} + 1) = 5 x^{2}, x > 0

x = \frac{13}{4} + \frac{3}{4} i, x = - \frac{13}{4} - \frac{3}{4} i, x = \frac{13}{4} - \frac{3}{4} i, x = - \frac{13}{4} + \frac{3}{4} i

Solution steps

4 (x^{4} + 1) = 5 x^{2}

Expand

4 (x^{4} + 1) : 4 x^{4} + 4

4 x^{4} + 4 = 5 x^{2}

Move

5 x^{2}

to the left side

4 x^{4} + 4 - 5 x^{2} = 0

Write in the standard form

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

4 x^{4} - 5 x^{2} + 4 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

4 u^{2} - 5 u + 4 = 0

Solve

4 u^{2} - 5 u + 4 = 0 : u = \frac{5}{8} + i \frac{39}{8}, u = \frac{5}{8} - i \frac{39}{8}

u = \frac{5}{8} + i \frac{39}{8}, u = \frac{5}{8} - i \frac{39}{8}

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = \frac{5}{8} + i \frac{39}{8} : x = \frac{13}{4} + \frac{3}{4} i, x = - \frac{13}{4} - \frac{3}{4} i

Solve

x^{2} = \frac{5}{8} - i \frac{39}{8} : x = \frac{13}{4} - \frac{3}{4} i, x = - \frac{13}{4} + \frac{3}{4} i

The solutions are

x = \frac{13}{4} + \frac{3}{4} i, x = - \frac{13}{4} - \frac{3}{4} i, x = \frac{13}{4} - \frac{3}{4} i, x = - \frac{13}{4} + \frac{3}{4} i

4 x^{4} - 221 x^{2} + 1225 = 0

x = x^{3} - x

2 x^{3} - 128 = 0

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

9 x^{3} + x^{2} - 81 x - 9 = 0

What are the solutions to the equation 4(x^4+1)=5x^2,x>0 ?
The solutions to the equation 4(x^4+1)=5x^2,x>0 are
Find the zeros of 4(x^4+1)=5x^2,x>0
The zeros of 4(x^4+1)=5x^2,x>0 are