Solve 1/4*a^3=2

$$\frac { 1 } { 4 } \times a ^ { 3 } = 2$$

Solve for a (complex solution)

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

$$a=2$$

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

Solution Steps

Multiply both sides by 4, the reciprocal of \frac{1}{4}.

$$a^{3}=2\times 4$$

Multiply 2 and 4 to get 8.

$$a^{3}=8$$

Subtract 8 from both sides.

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

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.

$$a=2$$

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

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

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.

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

Do the calculations.

$$a=\frac{-2±\sqrt{-12}}{2}$$

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

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

List all found solutions.

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

Solve for a

$$a=2$$

Solution Steps

Multiply both sides by 4, the reciprocal of \frac{1}{4}.

$$a^{3}=2\times 4$$

Multiply 2 and 4 to get 8.

$$a^{3}=8$$

Subtract 8 from both sides.

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

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.

$$a=2$$

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

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

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.

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

Do the calculations.

$$a=\frac{-2±\sqrt{-12}}{2}$$

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

$$a\in \emptyset$$

List all found solutions.

$$a=2$$