What are the solutions to the equation 3x^4+26=19x^2 ?

The solutions to the equation 3x^4+26=19x^2 are x=sqrt(13/3),x=-sqrt(13/3),x=sqrt(2),x=-sqrt(2)

Find the zeros of 3x^4+26=19x^2

The zeros of 3x^4+26=19x^2 are x=sqrt(13/3),x=-sqrt(13/3),x=sqrt(2),x=-sqrt(2)

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3 x^{4} + 26 = 19 x^{2}

x = \frac{13}{3}, x = - \frac{13}{3}, x = 2, x = - 2

Solution steps

3 x^{4} + 26 = 19 x^{2}

Move

19 x^{2}

to the left side

3 x^{4} + 26 - 19 x^{2} = 0

Write in the standard form

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

3 x^{4} - 19 x^{2} + 26 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

3 u^{2} - 19 u + 26 = 0

Solve

3 u^{2} - 19 u + 26 = 0 : u = \frac{13}{3}, u = 2

u = \frac{13}{3}, u = 2

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = \frac{13}{3} : x = \frac{13}{3}, x = - \frac{13}{3}

Solve

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

The solutions are

x = \frac{13}{3}, x = - \frac{13}{3}, x = 2, x = - 2

0 = x^{4} + 4 x + 3

x (x^{2} + 5 x + 8) = 0

x^{3} - 9 x^{2} + 14 x = 0

- 12 x^{2} (x^{3} + 2) = 0

x^{3} + 2 x - 42 = 0

What are the solutions to the equation 3x^4+26=19x^2 ?
The solutions to the equation 3x^4+26=19x^2 are x=sqrt(13/3),x=-sqrt(13/3),x=sqrt(2),x=-sqrt(2)
Find the zeros of 3x^4+26=19x^2
The zeros of 3x^4+26=19x^2 are x=sqrt(13/3),x=-sqrt(13/3),x=sqrt(2),x=-sqrt(2)