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

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

Solve for x

$$x=-2$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3.

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

Subtract 6 from both sides.

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

Add x to both sides.

$$2x^{2}-6+x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$2x^{2}+x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=1 ab=2\left(-6\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

Rewrite 2x^{2}+x-6 as \left(2x^{2}-3x\right)+\left(4x-6\right).

$$\left(2x^{2}-3x\right)+\left(4x-6\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(2x-3\right)+2\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(x+2\right)$$

To find equation solutions, solve 2x-3=0 and x+2=0.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3.

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

Subtract 6 from both sides.

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

Add x to both sides.

$$2x^{2}-6+x=0$$

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.

$$2x^{2}+x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 1 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-1±\sqrt{1-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$x=\frac{-1±\sqrt{1+48}}{2\times 2}$$

Add 1 to 48.

$$x=\frac{-1±\sqrt{49}}{2\times 2}$$

Take the square root of 49.

$$x=\frac{-1±7}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±7}{4}$$

Now solve the equation x=\frac{-1±7}{4} when ± is plus. Add -1 to 7.

$$x=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-1±7}{4} when ± is minus. Subtract 7 from -1.

$$x=-\frac{8}{4}$$

Divide -8 by 4.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3.

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

Add x to both sides.

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

Divide both sides by 2.

$$\frac{2x^{2}+x}{2}=\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

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

Divide 6 by 2.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=3+\frac{1}{16}$$

Add 3 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{49}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{7}{4} x+\frac{1}{4}=-\frac{7}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

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