What are the solutions to the equation 9x^4=12x^2-4 ?

The solutions to the equation 9x^4=12x^2-4 are x=sqrt(2/3),x=-sqrt(2/3)

Find the zeros of 9x^4=12x^2-4

The zeros of 9x^4=12x^2-4 are x=sqrt(2/3),x=-sqrt(2/3)

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9 x^{4} = 12 x^{2} - 4

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

Decimal Notation

x = 0.81649 \dots, x = - 0.81649 \dots

Solution steps

9 x^{4} = 12 x^{2} - 4

Move

4

to the left side

9 x^{4} + 4 = 12 x^{2}

Move

12 x^{2}

to the left side

9 x^{4} + 4 - 12 x^{2} = 0

Write in the standard form

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

9 x^{4} - 12 x^{2} + 4 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

9 u^{2} - 12 u + 4 = 0

Solve

9 u^{2} - 12 u + 4 = 0 : u = \frac{2}{3}

u = \frac{2}{3}

Substitute back

u = x^{2},

solve for

x

Solve

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

The solutions are

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

(5 - 2 x)^{3} = - 1

x^{3} + 10 x^{2} + 16 x = 0

x^{4} + x^{3} - 8 x^{2} - 9 x - 9 = 0

(x - 3) (x - 1)^{2} (x^{2} + 4 x + 13) = 0

- 2 x^{3} + x^{2} + 13 x + 6 = 0

What are the solutions to the equation 9x^4=12x^2-4 ?
The solutions to the equation 9x^4=12x^2-4 are x=sqrt(2/3),x=-sqrt(2/3)
Find the zeros of 9x^4=12x^2-4
The zeros of 9x^4=12x^2-4 are x=sqrt(2/3),x=-sqrt(2/3)