Solve 3/x-3+x+1/x^2-5x+6=1

$$\frac { 3 } { x - 3 } + \frac { x + 1 } { x ^ { 2 } - 5 x + 6 } = 1$$

Solve for x

$$x = \frac{\sqrt{37} + 9}{2} \approx 7.541381265$$

$$x = \frac{9 - \sqrt{37}}{2} \approx 1.458618735$$

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 \left(x-3\right)\left(x-2\right), the least common multiple of x-3,x^{2}-5x+6.

$$\left(x-2\right)\times 3+x+1=\left(x-3\right)\left(x-2\right)$$

Use the distributive property to multiply x-2 by 3.

$$3x-6+x+1=\left(x-3\right)\left(x-2\right)$$

Combine 3x and x to get 4x.

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

Add -6 and 1 to get -5.

$$4x-5=\left(x-3\right)\left(x-2\right)$$

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

$$4x-5=x^{2}-5x+6$$

Subtract x^{2} from both sides.

$$4x-5-x^{2}=-5x+6$$

Add 5x to both sides.

$$4x-5-x^{2}+5x=6$$

Combine 4x and 5x to get 9x.

$$9x-5-x^{2}=6$$

Subtract 6 from both sides.

$$9x-5-x^{2}-6=0$$

Subtract 6 from -5 to get -11.

$$9x-11-x^{2}=0$$

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.

$$-x^{2}+9x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 9 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-1\right)\left(-11\right)}}{2\left(-1\right)}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-1\right)\left(-11\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-9±\sqrt{81+4\left(-11\right)}}{2\left(-1\right)}$$

Multiply 4 times -11.

$$x=\frac{-9±\sqrt{81-44}}{2\left(-1\right)}$$

Add 81 to -44.

$$x=\frac{-9±\sqrt{37}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-9±\sqrt{37}}{-2}$$

Now solve the equation x=\frac{-9±\sqrt{37}}{-2} when ± is plus. Add -9 to \sqrt{37}.

$$x=\frac{\sqrt{37}-9}{-2}$$

Divide -9+\sqrt{37} by -2.

$$x=\frac{9-\sqrt{37}}{2}$$

Now solve the equation x=\frac{-9±\sqrt{37}}{-2} when ± is minus. Subtract \sqrt{37} from -9.

$$x=\frac{-\sqrt{37}-9}{-2}$$

Divide -9-\sqrt{37} by -2.

$$x=\frac{\sqrt{37}+9}{2}$$

The equation is now solved.

$$x=\frac{9-\sqrt{37}}{2} x=\frac{\sqrt{37}+9}{2}$$

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 \left(x-3\right)\left(x-2\right), the least common multiple of x-3,x^{2}-5x+6.

$$\left(x-2\right)\times 3+x+1=\left(x-3\right)\left(x-2\right)$$

Use the distributive property to multiply x-2 by 3.

$$3x-6+x+1=\left(x-3\right)\left(x-2\right)$$

Combine 3x and x to get 4x.

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

Add -6 and 1 to get -5.

$$4x-5=\left(x-3\right)\left(x-2\right)$$

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

$$4x-5=x^{2}-5x+6$$

Subtract x^{2} from both sides.

$$4x-5-x^{2}=-5x+6$$

Add 5x to both sides.

$$4x-5-x^{2}+5x=6$$

Combine 4x and 5x to get 9x.

$$9x-5-x^{2}=6$$

Add 5 to both sides.

$$9x-x^{2}=6+5$$

Add 6 and 5 to get 11.

$$9x-x^{2}=11$$

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.

$$-x^{2}+9x=11$$

Divide both sides by -1.

$$\frac{-x^{2}+9x}{-1}=\frac{11}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{9}{-1}x=\frac{11}{-1}$$

Divide 9 by -1.

$$x^{2}-9x=\frac{11}{-1}$$

Divide 11 by -1.

$$x^{2}-9x=-11$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-11+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-11+\frac{81}{4}$$

Add -11 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{37}{4}$$

Factor x^{2}-9x+\frac{81}{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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{37}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{37}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{\sqrt{37}}{2} x-\frac{9}{2}=-\frac{\sqrt{37}}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=\frac{\sqrt{37}+9}{2} x=\frac{9-\sqrt{37}}{2}$$