How to solve 4 + 8 = 3

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 4 + 8 = 3 | Equatix

Question

$$\frac { x } { 4 } + \frac { 8 } { x } = 3$$

Solve for x

$x=4$
$x=8$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $4x$, the least common multiple of $4,x$.

$$xx+4\times 8=12x$$

Multiply $x$ and $x$ to get $x^{2}$.

$$x^{2}+4\times 8=12x$$

Multiply $4$ and $8$ to get $32$.

$$x^{2}+32=12x$$

Subtract $12x$ from both sides.

$$x^{2}+32-12x=0$$

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

$$x^{2}-12x+32=0$$

To solve the equation, factor $x^{2}-12x+32$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-12$$ $$ab=32$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $32$.

$$-1,-32$$ $$-2,-16$$ $$-4,-8$$

Calculate the sum for each pair.

$$-1-32=-33$$ $$-2-16=-18$$ $$-4-8=-12$$

The solution is the pair that gives sum $-12$.

$$a=-8$$ $$b=-4$$

Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.

$$\left(x-8\right)\left(x-4\right)$$

To find equation solutions, solve $x-8=0$ and $x-4=0$.

$$x=8$$ $$x=4$$

Steps Using Factoring By Grouping

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $4x$, the least common multiple of $4,x$.

$$xx+4\times 8=12x$$

Multiply $x$ and $x$ to get $x^{2}$.

$$x^{2}+4\times 8=12x$$

Multiply $4$ and $8$ to get $32$.

$$x^{2}+32=12x$$

Subtract $12x$ from both sides.

$$x^{2}+32-12x=0$$

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

$$x^{2}-12x+32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+32$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-12$$ $$ab=1\times 32=32$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $32$.

$$-1,-32$$ $$-2,-16$$ $$-4,-8$$

Calculate the sum for each pair.

$$-1-32=-33$$ $$-2-16=-18$$ $$-4-8=-12$$

The solution is the pair that gives sum $-12$.

$$a=-8$$ $$b=-4$$

Rewrite $x^{2}-12x+32$ as $\left(x^{2}-8x\right)+\left(-4x+32\right)$.

$$\left(x^{2}-8x\right)+\left(-4x+32\right)$$

Factor out $x$ in the first and $-4$ in the second group.

$$x\left(x-8\right)-4\left(x-8\right)$$

Factor out common term $x-8$ by using distributive property.

$$\left(x-8\right)\left(x-4\right)$$

To find equation solutions, solve $x-8=0$ and $x-4=0$.

$$x=8$$ $$x=4$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $4x$, the least common multiple of $4,x$.

$$xx+4\times 8=12x$$

Multiply $x$ and $x$ to get $x^{2}$.

$$x^{2}+4\times 8=12x$$

Multiply $4$ and $8$ to get $32$.

$$x^{2}+32=12x$$

Subtract $12x$ from both sides.

$$x^{2}+32-12x=0$$

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.

$$x^{2}-12x+32=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-12$ for $b$, and $32$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 32}}{2}$$

Square $-12$.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 32}}{2}$$

Multiply $-4$ times $32$.

$$x=\frac{-\left(-12\right)±\sqrt{144-128}}{2}$$

Add $144$ to $-128$.

$$x=\frac{-\left(-12\right)±\sqrt{16}}{2}$$

Take the square root of $16$.

$$x=\frac{-\left(-12\right)±4}{2}$$

The opposite of $-12$ is $12$.

$$x=\frac{12±4}{2}$$

Now solve the equation $x=\frac{12±4}{2}$ when $±$ is plus. Add $12$ to $4$.

$$x=\frac{16}{2}$$

Divide $16$ by $2$.

$$x=8$$

Now solve the equation $x=\frac{12±4}{2}$ when $±$ is minus. Subtract $4$ from $12$.

$$x=\frac{8}{2}$$

Divide $8$ by $2$.

$$x=4$$

The equation is now solved.

$$x=8$$ $$x=4$$

Steps for Completing the Square

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $4x$, the least common multiple of $4,x$.

$$xx+4\times 8=12x$$

Multiply $x$ and $x$ to get $x^{2}$.

$$x^{2}+4\times 8=12x$$

Multiply $4$ and $8$ to get $32$.

$$x^{2}+32=12x$$

Subtract $12x$ from both sides.

$$x^{2}+32-12x=0$$

Subtract $32$ from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-12x=-32$$

Divide $-12$, the coefficient of the $x$ term, by $2$ to get $-6$. Then add the square of $-6$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=-32+\left(-6\right)^{2}$$

Square $-6$.

$$x^{2}-12x+36=-32+36$$

Add $-32$ to $36$.

$$x^{2}-12x+36=4$$

Factor $x^{2}-12x+36$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-6=2$$ $$x-6=-2$$

Add $6$ to both sides of the equation.

$$x=8$$ $$x=4$$

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Example

Question

Solve for x

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square