Solve 4-8/3x+1=3x^2+5/3x+1

$$4 - \frac { 8 } { 3 x + 1 } = \frac { 3 x ^ { 2 } + 5 } { 3 x + 1 }$$

Solve for x

$$x=1$$

$$x=3$$

Steps Using Factoring By Grouping

Variable x cannot be equal to -\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by 3x+1.

$$\left(3x+1\right)\times 4-8=3x^{2}+5$$

Use the distributive property to multiply 3x+1 by 4.

$$12x+4-8=3x^{2}+5$$

Subtract 8 from 4 to get -4.

$$12x-4=3x^{2}+5$$

Subtract 3x^{2} from both sides.

$$12x-4-3x^{2}=5$$

Subtract 5 from both sides.

$$12x-4-3x^{2}-5=0$$

Subtract 5 from -4 to get -9.

$$12x-9-3x^{2}=0$$

Divide both sides by 3.

$$4x-3-x^{2}=0$$

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

$$-x^{2}+4x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=4 ab=-\left(-3\right)=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=3 b=1$$

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

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

Factor out -x in -x^{2}+3x.

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

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

$$\left(x-3\right)\left(-x+1\right)$$

To find equation solutions, solve x-3=0 and -x+1=0.

$$x=3 x=1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by 3x+1.

$$\left(3x+1\right)\times 4-8=3x^{2}+5$$

Use the distributive property to multiply 3x+1 by 4.

$$12x+4-8=3x^{2}+5$$

Subtract 8 from 4 to get -4.

$$12x-4=3x^{2}+5$$

Subtract 3x^{2} from both sides.

$$12x-4-3x^{2}=5$$

Subtract 5 from both sides.

$$12x-4-3x^{2}-5=0$$

Subtract 5 from -4 to get -9.

$$12x-9-3x^{2}=0$$

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.

$$-3x^{2}+12x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 12 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-3\right)\left(-9\right)}}{2\left(-3\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-3\right)\left(-9\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-12±\sqrt{144+12\left(-9\right)}}{2\left(-3\right)}$$

Multiply 12 times -9.

$$x=\frac{-12±\sqrt{144-108}}{2\left(-3\right)}$$

Add 144 to -108.

$$x=\frac{-12±\sqrt{36}}{2\left(-3\right)}$$

Take the square root of 36.

$$x=\frac{-12±6}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-12±6}{-6}$$

Now solve the equation x=\frac{-12±6}{-6} when ± is plus. Add -12 to 6.

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

Divide -6 by -6.

$$x=1$$

Now solve the equation x=\frac{-12±6}{-6} when ± is minus. Subtract 6 from -12.

$$x=-\frac{18}{-6}$$

Divide -18 by -6.

$$x=3$$

The equation is now solved.

$$x=1 x=3$$

Steps for Completing the Square

Variable x cannot be equal to -\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by 3x+1.

$$\left(3x+1\right)\times 4-8=3x^{2}+5$$

Use the distributive property to multiply 3x+1 by 4.

$$12x+4-8=3x^{2}+5$$

Subtract 8 from 4 to get -4.

$$12x-4=3x^{2}+5$$

Subtract 3x^{2} from both sides.

$$12x-4-3x^{2}=5$$

Add 4 to both sides.

$$12x-3x^{2}=5+4$$

Add 5 and 4 to get 9.

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

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.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 12 by -3.

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

Divide 9 by -3.

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

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

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

Square -2.

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

Add -3 to 4.

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

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=1 x-2=-1$$

Add 2 to both sides of the equation.

$$x=3 x=1$$