Solve {c}{1/x-3=1/x+2-1/2}{text{Whatis}}{text{itsdegne}}

$$\left. \begin{array} { c } { \frac { 1 } { x - 3 } = \frac { 1 } { x + 2 } - \frac { 1 } { 2 } } \\ { \text { What is } } \\ { \text { its degne } } \end{array} \right.$$

Solve for x

$$x=\frac{1+\sqrt{15}i}{2}\approx 0.5+1.936491673i$$

$$x=\frac{-\sqrt{15}i+1}{2}\approx 0.5-1.936491673i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,3 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-3\right)\left(x+2\right), the least common multiple of x-3,x+2,2.

$$2x+4=2x-6+2\left(x-3\right)\left(x+2\right)\left(-\frac{1}{2}\right)$$

Multiply 2 and -\frac{1}{2} to get -1.

$$2x+4=2x-6-\left(x-3\right)\left(x+2\right)$$

Use the distributive property to multiply -1 by x-3.

$$2x+4=2x-6+\left(-x+3\right)\left(x+2\right)$$

Use the distributive property to multiply -x+3 by x+2 and combine like terms.

$$2x+4=2x-6-x^{2}+x+6$$

Combine 2x and x to get 3x.

$$2x+4=3x-6-x^{2}+6$$

Add -6 and 6 to get 0.

$$2x+4=3x-x^{2}$$

Subtract 3x from both sides.

$$2x+4-3x=-x^{2}$$

Combine 2x and -3x to get -x.

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

Add x^{2} to both sides.

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

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.

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

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

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

Multiply -4 times 4.

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

Add 1 to -16.

$$x=\frac{-\left(-1\right)±\sqrt{-15}}{2}$$

Take the square root of -15.

$$x=\frac{-\left(-1\right)±\sqrt{15}i}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{15}i}{2}$$

Now solve the equation x=\frac{1±\sqrt{15}i}{2} when ± is plus. Add 1 to i\sqrt{15}.

$$x=\frac{1+\sqrt{15}i}{2}$$

Now solve the equation x=\frac{1±\sqrt{15}i}{2} when ± is minus. Subtract i\sqrt{15} from 1.

$$x=\frac{-\sqrt{15}i+1}{2}$$

The equation is now solved.

$$x=\frac{1+\sqrt{15}i}{2} x=\frac{-\sqrt{15}i+1}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,3 since division by zero is not defined. Multiply both sides of the equation by 2\left(x-3\right)\left(x+2\right), the least common multiple of x-3,x+2,2.

$$2x+4=2x-6+2\left(x-3\right)\left(x+2\right)\left(-\frac{1}{2}\right)$$

Multiply 2 and -\frac{1}{2} to get -1.

$$2x+4=2x-6-\left(x-3\right)\left(x+2\right)$$

Use the distributive property to multiply -1 by x-3.

$$2x+4=2x-6+\left(-x+3\right)\left(x+2\right)$$

Use the distributive property to multiply -x+3 by x+2 and combine like terms.

$$2x+4=2x-6-x^{2}+x+6$$

Combine 2x and x to get 3x.

$$2x+4=3x-6-x^{2}+6$$

Add -6 and 6 to get 0.

$$2x+4=3x-x^{2}$$

Subtract 3x from both sides.

$$2x+4-3x=-x^{2}$$

Combine 2x and -3x to get -x.

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

Add x^{2} to both sides.

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

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -4 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=-\frac{15}{4}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{15}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{15}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{15}i}{2} x-\frac{1}{2}=-\frac{\sqrt{15}i}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{1+\sqrt{15}i}{2} x=\frac{-\sqrt{15}i+1}{2}$$