# How do you solve using the completing the square method 4x^2 + 4x + 1 = 49 ?

May 15, 2018

$x = - 4$ or $3$.

#### Explanation:

$4 {x}^{2} + 4 x + 1 = 49$ can be written as

${\left(2 x\right)}^{2} + 2 \times \left(2 x\right) \times 1 + {1}^{2} - 49 = 0$

or ${\left(2 x + 1\right)}^{2} - {7}^{2} = 0$

and using ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$ this becomes

$\left(2 x + 1 + 7\right) \left(2 x + 1 - 7\right) = 0$

or $\left(2 x + 8\right) \left(2 x - 6\right) = 0$

and either $2 x + 8 = 0$ i.e. $2 x = - 8$ and $x = - \frac{8}{2} = - 4$

or $2 x - 6 = 0$ i.e. $2 x = 6$ and $x = \frac{6}{2} = 3$

May 15, 2018

3, and - 4

#### Explanation:

1. Transforming Method
$y = 4 {x}^{2} + 4 x - 48 = 0$
$y = 4 \left({x}^{2} + x - 12\right) = 0$
Solve the quadratic equation ${x}^{2} + x - 12 = 0$, by the new Transforming Method - case a = 1 (Socratic, Google Search)
Find 2 real roots, that have opposite signs (ac < 0), knowing the sum (-b = -1) and the product (c = -12).
They are 3 and - 4
2. Completing the square method.
$4 {x}^{2} + 4 x = 48$
${x}^{2} + x = 12$
${x}^{2} + x + \frac{1}{4} = 12 + \frac{1}{4} = \frac{49}{4}$
${\left(x + \frac{1}{2}\right)}^{2} = \frac{49}{4}$
$x + \frac{1}{2} = \pm \frac{7}{2}$
$x 1 = \frac{7}{2} - \frac{1}{2} = \frac{6}{2} = 3$
$x 2 = - \frac{7}{2} - \frac{1}{2} = - \frac{8}{2} = - 4$
May 15, 2018

$x = - 4 \mathmr{and} x = 3$

$4 {\left(x + \frac{1}{2}\right)}^{2} - 49 = 0$

#### Explanation:

Given: $4 {x}^{2} + 4 x + 1 = 49 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(1\right)$

Write as:
$4 \left({x}^{2} + \frac{4}{4} x\right) + 1 - 49 = 0 \leftarrow$ As yet no values have changed.

The next step will change things in as much as it introduces a value that is not in the original equation. It is removed by introducing an as yet unknown constant of $k$. It will turn the introduced value into 0.

$4 \left({x}^{\textcolor{red}{2}} + \underbrace{\frac{4}{4}} \textcolor{w h i t e}{\text{d}} x\right) + 1 - 49 = 0$
$\textcolor{w h i t e}{\text{ddddddd}} \downarrow$
color(white)("dddd")"Halve this."
$\textcolor{w h i t e}{\text{ddddddd}} \downarrow$
4(xcolor(white)("d")+color(white)("d")1/2color(white)("d"))^(color(red)(2))+k-48=0" "..................Equation(2)

Notice the exponent $\left(\textcolor{red}{2}\right)$ has been moved outside the bracket, the $x$ from $\frac{4}{4} x$ has been removed and that $k$ has been introduced. Also as indicated, the $\frac{4}{4}$ has been halved as well.

$\textcolor{b r o w n}{\text{The above set of steps form the basis for completing the square}}$

Now we do the correction bit.

Set $4 {\left(\frac{1}{2}\right)}^{2} + k = 0 \textcolor{w h i t e}{\text{ddd")=>color(white)("ddd") 1+k=0color(white)("d}} \implies k = - 1$

Substitute into $E q u a t i o n \left(2\right)$ giving:

$4 {\left(x + \frac{1}{2}\right)}^{2} - 1 - 48 = 0$

$4 {\left(x + \frac{1}{2}\right)}^{2} - 49 = 0 \leftarrow$Completed square form
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$4 {\left(x + \frac{1}{2}\right)}^{2} = 49$

Divide both sides by 4

${\left(x + \frac{1}{2}\right)}^{2} = \frac{49}{4}$

Square root both sides

$x + \frac{1}{2} = \pm \frac{7}{2}$

Subtract $\frac{1}{2}$ from both sides

$x = - \frac{1}{2} \pm \frac{7}{2}$

$x = - 4 \mathmr{and} x = 3$