Solve -1/2x+5/2=2/x

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

Solve for x

$$x=1$$

$$x=4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x, the least common multiple of 2,x.

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

Cancel out 2 and 2.

$$-xx+2x\times \frac{5}{2}=2\times 2$$

Multiply x and x to get x^{2}.

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

Cancel out 2 and 2.

$$-x^{2}+5x=2\times 2$$

Multiply 2 and 2 to get 4.

$$-x^{2}+5x=4$$

Subtract 4 from both sides.

$$-x^{2}+5x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4\left(-4\right)}}{2\left(-1\right)}$$

Multiply 4 times -4.

$$x=\frac{-5±\sqrt{25-16}}{2\left(-1\right)}$$

Add 25 to -16.

$$x=\frac{-5±\sqrt{9}}{2\left(-1\right)}$$

Take the square root of 9.

$$x=\frac{-5±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±3}{-2}$$

Now solve the equation x=\frac{-5±3}{-2} when ± is plus. Add -5 to 3.

$$x=-\frac{2}{-2}$$

Divide -2 by -2.

$$x=1$$

Now solve the equation x=\frac{-5±3}{-2} when ± is minus. Subtract 3 from -5.

$$x=-\frac{8}{-2}$$

Divide -8 by -2.

$$x=4$$

The equation is now solved.

$$x=1 x=4$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x, the least common multiple of 2,x.

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

Cancel out 2 and 2.

$$-xx+2x\times \frac{5}{2}=2\times 2$$

Multiply x and x to get x^{2}.

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

Cancel out 2 and 2.

$$-x^{2}+5x=2\times 2$$

Multiply 2 and 2 to get 4.

$$-x^{2}+5x=4$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 5 by -1.

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

Divide 4 by -1.

$$x^{2}-5x=-4$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-4+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-4+\frac{25}{4}$$

Add -4 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{9}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

Add \frac{5}{2} to both sides of the equation.

$$x=4 x=1$$