Solve (2x+6)/(4x+2)=(4x+12)/(2) | 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

$$\frac{ 2x+6 }{ 4x+2 } = \frac{ 4x+12 }{ 2 }$$

Solve for x

$x=-3$
$x=-\frac{1}{4}=-0.25$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $-\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $2\left(2x+1\right)$, the least common multiple of $4x+2,2$.

$$2x+6=\left(2x+1\right)\left(4x+12\right)$$

Use the distributive property to multiply $2x+1$ by $4x+12$ and combine like terms.

$$2x+6=8x^{2}+28x+12$$

Subtract $8x^{2}$ from both sides.

$$2x+6-8x^{2}=28x+12$$

Subtract $28x$ from both sides.

$$2x+6-8x^{2}-28x=12$$

Combine $2x$ and $-28x$ to get $-26x$.

$$-26x+6-8x^{2}=12$$

Subtract $12$ from both sides.

$$-26x+6-8x^{2}-12=0$$

Subtract $12$ from $6$ to get $-6$.

$$-26x-6-8x^{2}=0$$

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.

$$-8x^{2}-26x-6=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-8$ for $a$, $-26$ for $b$, and $-6$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\left(-8\right)\left(-6\right)}}{2\left(-8\right)}$$

Square $-26$.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\left(-8\right)\left(-6\right)}}{2\left(-8\right)}$$

Multiply $-4$ times $-8$.

$$x=\frac{-\left(-26\right)±\sqrt{676+32\left(-6\right)}}{2\left(-8\right)}$$

Multiply $32$ times $-6$.

$$x=\frac{-\left(-26\right)±\sqrt{676-192}}{2\left(-8\right)}$$

Add $676$ to $-192$.

$$x=\frac{-\left(-26\right)±\sqrt{484}}{2\left(-8\right)}$$

Take the square root of $484$.

$$x=\frac{-\left(-26\right)±22}{2\left(-8\right)}$$

The opposite of $-26$ is $26$.

$$x=\frac{26±22}{2\left(-8\right)}$$

Multiply $2$ times $-8$.

$$x=\frac{26±22}{-16}$$

Now solve the equation $x=\frac{26±22}{-16}$ when $±$ is plus. Add $26$ to $22$.

$$x=\frac{48}{-16}$$

Divide $48$ by $-16$.

$$x=-3$$

Now solve the equation $x=\frac{26±22}{-16}$ when $±$ is minus. Subtract $22$ from $26$.

$$x=\frac{4}{-16}$$

Reduce the fraction $\frac{4}{-16}$ to lowest terms by extracting and canceling out $4$.

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

The equation is now solved.

$$x=-3$$ $$x=-\frac{1}{4}$$

Steps for Completing the Square

Variable $x$ cannot be equal to $-\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $2\left(2x+1\right)$, the least common multiple of $4x+2,2$.

$$2x+6=\left(2x+1\right)\left(4x+12\right)$$

Use the distributive property to multiply $2x+1$ by $4x+12$ and combine like terms.

$$2x+6=8x^{2}+28x+12$$

Subtract $8x^{2}$ from both sides.

$$2x+6-8x^{2}=28x+12$$

Subtract $28x$ from both sides.

$$2x+6-8x^{2}-28x=12$$

Combine $2x$ and $-28x$ to get $-26x$.

$$-26x+6-8x^{2}=12$$

Subtract $6$ from both sides.

$$-26x-8x^{2}=12-6$$

Subtract $6$ from $12$ to get $6$.

$$-26x-8x^{2}=6$$

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 $x^{2}+bx=c$.

$$-8x^{2}-26x=6$$

Divide both sides by $-8$.

$$\frac{-8x^{2}-26x}{-8}=\frac{6}{-8}$$

Dividing by $-8$ undoes the multiplication by $-8$.

$$x^{2}+\left(-\frac{26}{-8}\right)x=\frac{6}{-8}$$

Reduce the fraction $\frac{-26}{-8}$ to lowest terms by extracting and canceling out $2$.

$$x^{2}+\frac{13}{4}x=\frac{6}{-8}$$

Reduce the fraction $\frac{6}{-8}$ to lowest terms by extracting and canceling out $2$.

$$x^{2}+\frac{13}{4}x=-\frac{3}{4}$$

Divide $\frac{13}{4}$, the coefficient of the $x$ term, by $2$ to get $\frac{13}{8}$. Then add the square of $\frac{13}{8}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{13}{4}x+\left(\frac{13}{8}\right)^{2}=-\frac{3}{4}+\left(\frac{13}{8}\right)^{2}$$

Square $\frac{13}{8}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{13}{4}x+\frac{169}{64}=-\frac{3}{4}+\frac{169}{64}$$

Add $-\frac{3}{4}$ to $\frac{169}{64}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{13}{4}x+\frac{169}{64}=\frac{121}{64}$$

Factor $x^{2}+\frac{13}{4}x+\frac{169}{64}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{13}{8}\right)^{2}=\frac{121}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{8}\right)^{2}}=\sqrt{\frac{121}{64}}$$

Simplify.

$$x+\frac{13}{8}=\frac{11}{8}$$ $$x+\frac{13}{8}=-\frac{11}{8}$$

Subtract $\frac{13}{8}$ from both sides of the equation.

$$x=-\frac{1}{4}$$ $$x=-3$$