Solve left(x^2right)^2-6x^3+6x^2+9x=4

$${ \left( { x }^{ 2 } \right) }^{ 2 } -6 { x }^{ 3 } +6 { x }^{ 2 } +9x=4$$

Solve for x

$$x=\frac{3-\sqrt{5}}{2}\approx 0.381966011$$

$$x=4$$

$$x = \frac{\sqrt{5} + 3}{2} \approx 2.618033989$$

$$x=-1$$

Solution Steps

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

$$x^{4}-6x^{3}+6x^{2}+9x=4$$

Subtract 4 from both sides.

$$x^{4}-6x^{3}+6x^{2}+9x-4=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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 x^{4}-6x^{3}+6x^{2}+9x-4 by x+1 to get x^{3}-7x^{2}+13x-4. Solve the equation where the result equals to 0.

$$x^{3}-7x^{2}+13x-4=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-7x^{2}+13x-4 by x-4 to get x^{2}-3x+1. Solve the equation where the result equals to 0.

$$x^{2}-3x+1=0$$

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, -3 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$x=\frac{3±\sqrt{5}}{2}$$

Solve the equation x^{2}-3x+1=0 when ± is plus and when ± is minus.

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

List all found solutions.

$$x=-1 x=4 x=\frac{3-\sqrt{5}}{2} x=\frac{\sqrt{5}+3}{2}$$