Solve 2/x+3-7/x=5

$$\frac { 2 } { x + 3 } - \frac { 7 } { x } = 5$$

Solve for x (complex solution)

$$x=\frac{\sqrt{5}i}{5}-2\approx -2+0.447213595i$$

$$x=-\frac{\sqrt{5}i}{5}-2\approx -2-0.447213595i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+3\right), the least common multiple of x+3,x.

$$x\times 2-\left(x+3\right)\times 7=5x\left(x+3\right)$$

Use the distributive property to multiply x+3 by 7.

$$x\times 2-\left(7x+21\right)=5x\left(x+3\right)$$

To find the opposite of 7x+21, find the opposite of each term.

$$x\times 2-7x-21=5x\left(x+3\right)$$

Combine x\times 2 and -7x to get -5x.

$$-5x-21=5x\left(x+3\right)$$

Use the distributive property to multiply 5x by x+3.

$$-5x-21=5x^{2}+15x$$

Subtract 5x^{2} from both sides.

$$-5x-21-5x^{2}=15x$$

Subtract 15x from both sides.

$$-5x-21-5x^{2}-15x=0$$

Combine -5x and -15x to get -20x.

$$-20x-21-5x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-5x^{2}-20x-21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -20 for b, and -21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-5\right)\left(-21\right)}}{2\left(-5\right)}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\left(-5\right)\left(-21\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-20\right)±\sqrt{400+20\left(-21\right)}}{2\left(-5\right)}$$

Multiply 20 times -21.

$$x=\frac{-\left(-20\right)±\sqrt{400-420}}{2\left(-5\right)}$$

Add 400 to -420.

$$x=\frac{-\left(-20\right)±\sqrt{-20}}{2\left(-5\right)}$$

Take the square root of -20.

$$x=\frac{-\left(-20\right)±2\sqrt{5}i}{2\left(-5\right)}$$

The opposite of -20 is 20.

$$x=\frac{20±2\sqrt{5}i}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{20±2\sqrt{5}i}{-10}$$

Now solve the equation x=\frac{20±2\sqrt{5}i}{-10} when ± is plus. Add 20 to 2i\sqrt{5}.

$$x=\frac{20+2\sqrt{5}i}{-10}$$

Divide 20+2i\sqrt{5} by -10.

$$x=-\frac{\sqrt{5}i}{5}-2$$

Now solve the equation x=\frac{20±2\sqrt{5}i}{-10} when ± is minus. Subtract 2i\sqrt{5} from 20.

$$x=\frac{-2\sqrt{5}i+20}{-10}$$

Divide 20-2i\sqrt{5} by -10.

$$x=\frac{\sqrt{5}i}{5}-2$$

The equation is now solved.

$$x=-\frac{\sqrt{5}i}{5}-2 x=\frac{\sqrt{5}i}{5}-2$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+3\right), the least common multiple of x+3,x.

$$x\times 2-\left(x+3\right)\times 7=5x\left(x+3\right)$$

Use the distributive property to multiply x+3 by 7.

$$x\times 2-\left(7x+21\right)=5x\left(x+3\right)$$

To find the opposite of 7x+21, find the opposite of each term.

$$x\times 2-7x-21=5x\left(x+3\right)$$

Combine x\times 2 and -7x to get -5x.

$$-5x-21=5x\left(x+3\right)$$

Use the distributive property to multiply 5x by x+3.

$$-5x-21=5x^{2}+15x$$

Subtract 5x^{2} from both sides.

$$-5x-21-5x^{2}=15x$$

Subtract 15x from both sides.

$$-5x-21-5x^{2}-15x=0$$

Combine -5x and -15x to get -20x.

$$-20x-21-5x^{2}=0$$

Add 21 to both sides. Anything plus zero gives itself.

$$-20x-5x^{2}=21$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-5x^{2}-20x=21$$

Divide both sides by -5.

$$\frac{-5x^{2}-20x}{-5}=\frac{21}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{20}{-5}\right)x=\frac{21}{-5}$$

Divide -20 by -5.

$$x^{2}+4x=\frac{21}{-5}$$

Divide 21 by -5.

$$x^{2}+4x=-\frac{21}{5}$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=-\frac{21}{5}+2^{2}$$

Square 2.

$$x^{2}+4x+4=-\frac{21}{5}+4$$

Add -\frac{21}{5} to 4.

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

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+2\right)^{2}=-\frac{1}{5}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{-\frac{1}{5}}$$

Simplify.

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

Subtract 2 from both sides of the equation.

$$x=\frac{\sqrt{5}i}{5}-2 x=-\frac{\sqrt{5}i}{5}-2$$