Solve (x^2)^4*x^-5=8

$$( x ^ { 2 } ) ^ { 4 } \cdot x ^ { - 5 } = 8$$

Solve for x (complex solution)

$$x=-\sqrt{3}i-1\approx -1-1.732050808i$$

$$x=2$$

$$x=-1+\sqrt{3}i\approx -1+1.732050808i$$

Solution Steps

To raise a power to another power, multiply the exponents. Multiply 2 and 4 to get 8.

$$x^{8}x^{-5}=8$$

To multiply powers of the same base, add their exponents. Add 8 and -5 to get 3.

$$x^{3}=8$$

Subtract 8 from both sides.

$$x^{3}-8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -8 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±8,±4,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-8 by x-2 to get x^{2}+2x+4. Solve the equation where the result equals to 0.

$$x^{2}+2x+4=0$$

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, 2 for b, and 4 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$x=\frac{-2±\sqrt{-12}}{2}$$

Solve the equation x^{2}+2x+4=0 when ± is plus and when ± is minus.

$$x=-\sqrt{3}i-1 x=-1+\sqrt{3}i$$

List all found solutions.

$$x=2 x=-\sqrt{3}i-1 x=-1+\sqrt{3}i$$

Solve for x

$$x=2$$

Solution Steps

To raise a power to another power, multiply the exponents. Multiply 2 and 4 to get 8.

$$x^{8}x^{-5}=8$$

To multiply powers of the same base, add their exponents. Add 8 and -5 to get 3.

$$x^{3}=8$$

Subtract 8 from both sides.

$$x^{3}-8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -8 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±8,±4,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-8 by x-2 to get x^{2}+2x+4. Solve the equation where the result equals to 0.

$$x^{2}+2x+4=0$$

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, 2 for b, and 4 for c in the quadratic formula.

$$x=\frac{-2±\sqrt{2^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$x=\frac{-2±\sqrt{-12}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=2$$