Solve x+3(2-3x^2)^2=2

$$x + 3 ( 2 - 3 x ^ { 2 } ) ^ { 2 } = 2$$

Solve for x

$$x=\frac{1-\sqrt{21}}{6}\approx -0.597095949$$

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=\frac{\sqrt{21}+1}{6}\approx 0.930429282$$

$$x=-1$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-3x^{2}\right)^{2}.

$$x+3\left(4-12x^{2}+9\left(x^{2}\right)^{2}\right)=2$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x+3\left(4-12x^{2}+9x^{4}\right)=2$$

Use the distributive property to multiply 3 by 4-12x^{2}+9x^{4}.

$$x+12-36x^{2}+27x^{4}=2$$

Subtract 2 from both sides.

$$x+12-36x^{2}+27x^{4}-2=0$$

Subtract 2 from 12 to get 10.

$$x+10-36x^{2}+27x^{4}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$27x^{4}-36x^{2}+x+10=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 10 and q divides the leading coefficient 27. List all candidates \frac{p}{q}.

$$±\frac{10}{27},±\frac{10}{9},±\frac{10}{3},±10,±\frac{5}{27},±\frac{5}{9},±\frac{5}{3},±5,±\frac{2}{27},±\frac{2}{9},±\frac{2}{3},±2,±\frac{1}{27},±\frac{1}{9},±\frac{1}{3},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 27x^{4}-36x^{2}+x+10 by x+1 to get 27x^{3}-27x^{2}-9x+10. Solve the equation where the result equals to 0.

$$27x^{3}-27x^{2}-9x+10=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 10 and q divides the leading coefficient 27. List all candidates \frac{p}{q}.

$$±\frac{10}{27},±\frac{10}{9},±\frac{10}{3},±10,±\frac{5}{27},±\frac{5}{9},±\frac{5}{3},±5,±\frac{2}{27},±\frac{2}{9},±\frac{2}{3},±2,±\frac{1}{27},±\frac{1}{9},±\frac{1}{3},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=\frac{2}{3}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 27x^{3}-27x^{2}-9x+10 by 3\left(x-\frac{2}{3}\right)=3x-2 to get 9x^{2}-3x-5. Solve the equation where the result equals to 0.

$$9x^{2}-3x-5=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 9 for a, -3 for b, and -5 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 9\left(-5\right)}}{2\times 9}$$

Do the calculations.

$$x=\frac{3±3\sqrt{21}}{18}$$

Solve the equation 9x^{2}-3x-5=0 when ± is plus and when ± is minus.

$$x=\frac{1-\sqrt{21}}{6} x=\frac{\sqrt{21}+1}{6}$$

List all found solutions.

$$x=-1 x=\frac{2}{3} x=\frac{1-\sqrt{21}}{6} x=\frac{\sqrt{21}+1}{6}$$