Solve 6(3+x^2)-12x^2/9+6x^2+x^4=0

$$\frac { 6 ( 3 + x ^ { 2 } ) - 12 x ^ { 2 } } { 9 + 6 x ^ { 2 } + x ^ { 4 } } = 0$$

Solve for x

$$x=\sqrt{3}\approx 1.732050808$$

$$x=-\sqrt{3}\approx -1.732050808$$

Steps by Finding Square Root

Multiply both sides of the equation by \left(x^{2}+3\right)^{2}.

$$6\left(3+x^{2}\right)-12x^{2}=0$$

Use the distributive property to multiply 6 by 3+x^{2}.

$$18+6x^{2}-12x^{2}=0$$

Combine 6x^{2} and -12x^{2} to get -6x^{2}.

$$18-6x^{2}=0$$

Subtract 18 from both sides. Anything subtracted from zero gives its negation.

$$-6x^{2}=-18$$

Divide both sides by -6.

$$x^{2}=\frac{-18}{-6}$$

Divide -18 by -6 to get 3.

$$x^{2}=3$$

Take the square root of both sides of the equation.

$$x=\sqrt{3} x=-\sqrt{3}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by \left(x^{2}+3\right)^{2}.

$$6\left(3+x^{2}\right)-12x^{2}=0$$

Use the distributive property to multiply 6 by 3+x^{2}.

$$18+6x^{2}-12x^{2}=0$$

Combine 6x^{2} and -12x^{2} to get -6x^{2}.

$$18-6x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-6x^{2}+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 0 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-6\right)\times 18}}{2\left(-6\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-6\right)\times 18}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{0±\sqrt{24\times 18}}{2\left(-6\right)}$$

Multiply 24 times 18.

$$x=\frac{0±\sqrt{432}}{2\left(-6\right)}$$

Take the square root of 432.

$$x=\frac{0±12\sqrt{3}}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{0±12\sqrt{3}}{-12}$$

Now solve the equation x=\frac{0±12\sqrt{3}}{-12} when ± is plus.

$$x=-\sqrt{3}$$

Now solve the equation x=\frac{0±12\sqrt{3}}{-12} when ± is minus.

$$x=\sqrt{3}$$

The equation is now solved.

$$x=-\sqrt{3} x=\sqrt{3}$$