Solve 4 - 8/(3x + 1) = (3x ^ 2 + 5)/(3x + 1) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

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

Solve for x

$x=1$
$x=3$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

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.

$$x=3$$ $$x=1$$