Solve left(x^2-6xright)^2-2left(x-3right)^2-81=0

$${ \left( { x }^{ 2 } -6x \right) }^{ 2 } -2 { \left( x-3 \right) }^{ 2 } -81 = 0$$

Solve for x

$$x=2\sqrt{5}+3\approx 7.472135955$$

$$x=3-2\sqrt{5}\approx -1.472135955$$

$$x=3$$

Solution Steps

Simplify.

$$x^{4}-12x^{3}+34x^{2}+12x-99=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -99 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±99,±33,±11,±9,±3,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-12x^{3}+34x^{2}+12x-99 by x-3 to get x^{3}-9x^{2}+7x+33. Solve the equation where the result equals to 0.

$$x^{3}-9x^{2}+7x+33=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 33 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±33,±11,±3,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x^{2}+7x+33 by x-3 to get x^{2}-6x-11. Solve the equation where the result equals to 0.

$$x^{2}-6x-11=0$$

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 1 for a, -6 for b, and -11 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\left(-11\right)}}{2}$$

Do the calculations.

$$x=\frac{6±4\sqrt{5}}{2}$$

Solve the equation x^{2}-6x-11=0 when ± is plus and when ± is minus.

$$x=3-2\sqrt{5} x=2\sqrt{5}+3$$

List all found solutions.

$$x=3 x=3-2\sqrt{5} x=2\sqrt{5}+3$$