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

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

Solve for x

$$x = \frac{\sqrt{79} - 1}{3} \approx 2.629398139$$

$$x=\frac{-\sqrt{79}-1}{3}\approx -3.296064806$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+3\right), the least common multiple of x^{2}+x-6,x-2.

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

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

$$5=\left(x^{2}+x-6\right)\times 2+\left(x+3\right)\left(x-3\right)$$

Use the distributive property to multiply x^{2}+x-6 by 2.

$$5=2x^{2}+2x-12+\left(x+3\right)\left(x-3\right)$$

Consider \left(x+3\right)\left(x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$5=2x^{2}+2x-12+x^{2}-9$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$5=3x^{2}+2x-12-9$$

Subtract 9 from -12 to get -21.

$$5=3x^{2}+2x-21$$

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}+2x-21=5$$

Subtract 5 from both sides.

$$3x^{2}+2x-21-5=0$$

Subtract 5 from -21 to get -26.

$$3x^{2}+2x-26=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 2 for b, and -26 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 3\left(-26\right)}}{2\times 3}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 3\left(-26\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-2±\sqrt{4-12\left(-26\right)}}{2\times 3}$$

Multiply -12 times -26.

$$x=\frac{-2±\sqrt{4+312}}{2\times 3}$$

Add 4 to 312.

$$x=\frac{-2±\sqrt{316}}{2\times 3}$$

Take the square root of 316.

$$x=\frac{-2±2\sqrt{79}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-2±2\sqrt{79}}{6}$$

Now solve the equation x=\frac{-2±2\sqrt{79}}{6} when ± is plus. Add -2 to 2\sqrt{79}.

$$x=\frac{2\sqrt{79}-2}{6}$$

Divide -2+2\sqrt{79} by 6.

$$x=\frac{\sqrt{79}-1}{3}$$

Now solve the equation x=\frac{-2±2\sqrt{79}}{6} when ± is minus. Subtract 2\sqrt{79} from -2.

$$x=\frac{-2\sqrt{79}-2}{6}$$

Divide -2-2\sqrt{79} by 6.

$$x=\frac{-\sqrt{79}-1}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{79}-1}{3} x=\frac{-\sqrt{79}-1}{3}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+3\right), the least common multiple of x^{2}+x-6,x-2.

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

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

$$5=\left(x^{2}+x-6\right)\times 2+\left(x+3\right)\left(x-3\right)$$

Use the distributive property to multiply x^{2}+x-6 by 2.

$$5=2x^{2}+2x-12+\left(x+3\right)\left(x-3\right)$$

Consider \left(x+3\right)\left(x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$5=2x^{2}+2x-12+x^{2}-9$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$5=3x^{2}+2x-12-9$$

Subtract 9 from -12 to get -21.

$$5=3x^{2}+2x-21$$

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}+2x-21=5$$

Add 21 to both sides.

$$3x^{2}+2x=5+21$$

Add 5 and 21 to get 26.

$$3x^{2}+2x=26$$

Divide both sides by 3.

$$\frac{3x^{2}+2x}{3}=\frac{26}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{2}{3}x=\frac{26}{3}$$

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{26}{3}+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{26}{3}+\frac{1}{9}$$

Add \frac{26}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{79}{9}$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{79}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{79}{9}}$$

Simplify.

$$x+\frac{1}{3}=\frac{\sqrt{79}}{3} x+\frac{1}{3}=-\frac{\sqrt{79}}{3}$$

Subtract \frac{1}{3} from both sides of the equation.

$$x=\frac{\sqrt{79}-1}{3} x=\frac{-\sqrt{79}-1}{3}$$