Question

Solve the equation 6(2x + 4)2 = (2x + 4) + 2.

Answer 1

ANSWER
$x = - \frac{5}{3}$

or

$x = - \frac{9}{4}$

EXPLANATION

We want to solve the equation,

$6 {(2x + 4)}^{2} = (2x + 4) + 2$

One interesting way to solve this equation is to equate everything to zero and solve as quadratic equation by factoring.

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

The above is a quadratic equation in
$(2x + 4)$

where

$a=6,b=1,c=-2$

and
$a \times c = - 2 \times 6 = - 12$

We use the factors,
$-4,3$
to split the middle term to get,

$6 {(2x + 4)}^{2} + 3 (2x + 4) - 4(2x + 4) - 2 = 0$

We factor to obtain,

$3 (2x + 4)[2(2x + 4) + 1] - 2 [ 2 (2x + 4) + 1]= 0$

We factor further to obtain,

$(3(2x + 4) - 2)(2(2x + 4) + 1) = 0$

$(6x + 12 - 2)(4x + 8 + 1) = 0$

$(6x + 10)(4x + 9) = 0$

$x = - \frac{10}{6} = - \frac{5}{3}$

or

$x = - \frac{9}{4}$

Answer 2

The solution of the equation $6{\left( {2x + 4} \right)^2} = \left( {2x + 4} \right) + 2$ is $\boxed{x=- \frac{5}{3}}{\text{ and }}\boxed{x =- \frac{9}{4}}.$

Further explanation:

The Fundamental Theorem of Algebra states that the polynomial has n roots if the degree of the polynomial is $n$.

$f\left( x \right) = a{x^n} + b{x^{n - 1}} +\ldots+ cx + d$

The polynomial function has n roots or zeroes.

Given:

The equation is $6{\left( {2x + 4} \right)^2} = \left( {2x + 4} \right) + 2.$

Explanation:

The equation $6{\left( {2x + 4} \right)^2} = \left( {2x + 4} \right) + 2$ is a quadratic equation and has 2 solutions or zeros.

Solve the equation $6{\left( {2x + 4} \right)^2} = \left( {2x + 4} \right) + 2.$

$\begin{aligned}6{\left( {2x + 4} \right)^2} &= \left( {2x + 4} \right) + 2\\6{\left( {2x + 4} \right)^2} - \left( {2x + 4} \right) - 2 &= 0\\\end{aligned}$

Further solve the above equation using middle term splitting.

$\begin{aligned}6{\left( {2x + 4} \right)^2} - \left( {2x + 4} \right) - 2 &= 0 \hfill\\6{\left( {2x + 4} \right)^2} - 4\left( {2x + 4} \right) + 3\left( {2x + 4} \right) - 2 &= 0 \hfill\\3\left( {2x + 4} \right)\left[ {2\left( {2x + 4} \right) + 1} \right] - 2\left[ {2\left( {2x + 4} \right) + 1} \right] &= 0 \hfill\\\left[ {2\left( {2x + 4} \right) + 1} \right]\left[ {3\left( {2x + 4} \right) - 2} \right] &= 0 \hfill\\\end{aligned}$

Again solve the above equation.

$\begin{aligned}\left( {4x + 8 + 1} \right)\left( {6x + 12 - 2} \right) &= 0\\\left( {4x + 9} \right)\left( {6x + 10} \right) &= 0\\4x + 9 &= 0\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,6x + 10 = 0 \\ 4x - 9&=0\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6x = - 10\\x &= \frac{{ - 9}}{4}\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{{ - 10}}{6} \\\end{aligned}$

The solution of the equation $6{\left( {2x + 4}\right)^2} = \left( {2x + 4}\right) + 2$ is $\boxed{x= - \frac{5}{3}}{\text{ and }}\boxed{x = - \frac{9}{4}}.$

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Quadratic equation

Keywords: roots, linear equation, quadratic equation, zeros, function, polynomial, solution, cubic function, degree of the function.