Solve a^2+4/a=2

$$a ^ { 2 } + \frac { 4 } { a } = 2$$

Solve for a (complex solution)

$$a=1+i$$

$$a=1-i$$

$$a=-2$$

Solution Steps

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by a.

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

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$a^{3}+4=2a$$

Subtract 2a from both sides.

$$a^{3}+4-2a=0$$

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

$$a^{3}-2a+4=0$$

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.

$$a=-2$$

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

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

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 2 for c in the quadratic formula.

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

Do the calculations.

$$a=\frac{2±\sqrt{-4}}{2}$$

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

$$a=1-i a=1+i$$

List all found solutions.

$$a=-2 a=1-i a=1+i$$

Solve for a

$$a=-2$$

Solution Steps

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by a.

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

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$a^{3}+4=2a$$

Subtract 2a from both sides.

$$a^{3}+4-2a=0$$

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

$$a^{3}-2a+4=0$$

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.

$$a=-2$$

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

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

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 2 for c in the quadratic formula.

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

Do the calculations.

$$a=\frac{2±\sqrt{-4}}{2}$$

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.

$$a=-2$$