Solve 3x^2-8x+4/x-2=5x+8

$$\frac { 3 x ^ { 2 } - 8 x + 4 } { x - 2 } = 5 x + 8$$

Solve for x

$$x=-5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$3x^{2}-8x+4=5x\left(x-2\right)+\left(x-2\right)\times 8$$

Use the distributive property to multiply 5x by x-2.

$$3x^{2}-8x+4=5x^{2}-10x+\left(x-2\right)\times 8$$

Use the distributive property to multiply x-2 by 8.

$$3x^{2}-8x+4=5x^{2}-10x+8x-16$$

Combine -10x and 8x to get -2x.

$$3x^{2}-8x+4=5x^{2}-2x-16$$

Subtract 5x^{2} from both sides.

$$3x^{2}-8x+4-5x^{2}=-2x-16$$

Combine 3x^{2} and -5x^{2} to get -2x^{2}.

$$-2x^{2}-8x+4=-2x-16$$

Add 2x to both sides.

$$-2x^{2}-8x+4+2x=-16$$

Combine -8x and 2x to get -6x.

$$-2x^{2}-6x+4=-16$$

Add 16 to both sides.

$$-2x^{2}-6x+4+16=0$$

Add 4 and 16 to get 20.

$$-2x^{2}-6x+20=0$$

Divide both sides by 2.

$$-x^{2}-3x+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-10=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=2 b=-5$$

Rewrite -x^{2}-3x+10 as \left(-x^{2}+2x\right)+\left(-5x+10\right).

$$\left(-x^{2}+2x\right)+\left(-5x+10\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(-x+2\right)+5\left(-x+2\right)$$

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

$$\left(-x+2\right)\left(x+5\right)$$

To find equation solutions, solve -x+2=0 and x+5=0.

$$x=2 x=-5$$

Variable x cannot be equal to 2.

$$x=-5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$3x^{2}-8x+4=5x\left(x-2\right)+\left(x-2\right)\times 8$$

Use the distributive property to multiply 5x by x-2.

$$3x^{2}-8x+4=5x^{2}-10x+\left(x-2\right)\times 8$$

Use the distributive property to multiply x-2 by 8.

$$3x^{2}-8x+4=5x^{2}-10x+8x-16$$

Combine -10x and 8x to get -2x.

$$3x^{2}-8x+4=5x^{2}-2x-16$$

Subtract 5x^{2} from both sides.

$$3x^{2}-8x+4-5x^{2}=-2x-16$$

Combine 3x^{2} and -5x^{2} to get -2x^{2}.

$$-2x^{2}-8x+4=-2x-16$$

Add 2x to both sides.

$$-2x^{2}-8x+4+2x=-16$$

Combine -8x and 2x to get -6x.

$$-2x^{2}-6x+4=-16$$

Add 16 to both sides.

$$-2x^{2}-6x+4+16=0$$

Add 4 and 16 to get 20.

$$-2x^{2}-6x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -6 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-2\right)\times 20}}{2\left(-2\right)}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-2\right)\times 20}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-6\right)±\sqrt{36+8\times 20}}{2\left(-2\right)}$$

Multiply 8 times 20.

$$x=\frac{-\left(-6\right)±\sqrt{36+160}}{2\left(-2\right)}$$

Add 36 to 160.

$$x=\frac{-\left(-6\right)±\sqrt{196}}{2\left(-2\right)}$$

Take the square root of 196.

$$x=\frac{-\left(-6\right)±14}{2\left(-2\right)}$$

The opposite of -6 is 6.

$$x=\frac{6±14}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{6±14}{-4}$$

Now solve the equation x=\frac{6±14}{-4} when ± is plus. Add 6 to 14.

$$x=\frac{20}{-4}$$

Divide 20 by -4.

$$x=-5$$

Now solve the equation x=\frac{6±14}{-4} when ± is minus. Subtract 14 from 6.

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

Divide -8 by -4.

$$x=2$$

The equation is now solved.

$$x=-5 x=2$$

Variable x cannot be equal to 2.

$$x=-5$$

Steps for Completing the Square

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

$$3x^{2}-8x+4=5x\left(x-2\right)+\left(x-2\right)\times 8$$

Use the distributive property to multiply 5x by x-2.

$$3x^{2}-8x+4=5x^{2}-10x+\left(x-2\right)\times 8$$

Use the distributive property to multiply x-2 by 8.

$$3x^{2}-8x+4=5x^{2}-10x+8x-16$$

Combine -10x and 8x to get -2x.

$$3x^{2}-8x+4=5x^{2}-2x-16$$

Subtract 5x^{2} from both sides.

$$3x^{2}-8x+4-5x^{2}=-2x-16$$

Combine 3x^{2} and -5x^{2} to get -2x^{2}.

$$-2x^{2}-8x+4=-2x-16$$

Add 2x to both sides.

$$-2x^{2}-8x+4+2x=-16$$

Combine -8x and 2x to get -6x.

$$-2x^{2}-6x+4=-16$$

Subtract 4 from both sides.

$$-2x^{2}-6x=-16-4$$

Subtract 4 from -16 to get -20.

$$-2x^{2}-6x=-20$$

Divide both sides by -2.

$$\frac{-2x^{2}-6x}{-2}=-\frac{20}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{6}{-2}\right)x=-\frac{20}{-2}$$

Divide -6 by -2.

$$x^{2}+3x=-\frac{20}{-2}$$

Divide -20 by -2.

$$x^{2}+3x=10$$

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

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=10+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=10+\frac{9}{4}$$

Add 10 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{49}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{7}{2} x+\frac{3}{2}=-\frac{7}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=2 x=-5$$

Variable x cannot be equal to 2.

$$x=-5$$