# How do you solve using the completing the square method x^2 + 4x = 21?

Mar 9, 2016

$\textcolor{b l u e}{\text{Vertex} \to \left(x , y\right) = \left(- 2 , - 25\right)}$

#### Explanation:

Write as:$\text{ } {x}^{2} + 4 x - 21 = 0$...................(1)

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$\textcolor{b l u e}{\text{Shortcut method}}$

By sight:

${\left(x + 2\right)}^{2} - 4 - 21 = 0$

$\textcolor{b l u e}{{\left(x + 2\right)}^{2} - 25 = 0}$

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$\textcolor{b l u e}{\text{The logic behind the shortcut method}}$

$\textcolor{g r e e n}{\text{Consider only the left hand side}}$

$\textcolor{g r e e n}{\text{Place the two left most terms in brackets}}$

$\text{ } \left({x}^{\textcolor{m a \ge n t a}{2}} + 4 x\right) - 21$

$\textcolor{g r e e n}{\text{Move the index of "color(magenta)(2)" to the outside of the brackets}}$

$\text{ } {\left(x + 4 \textcolor{red}{x}\right)}^{\textcolor{m a \ge n t a}{2}} - 21$

$\textcolor{g r e e n}{\text{Remove the "color(red)(x)" from the "4x" inside the brackets}}$

$\text{ } {\left(x + \textcolor{b l u e}{4}\right)}^{2} - 21$

$\textcolor{g r e e n}{\text{Apply } \left(\frac{1}{2}\right) \times \textcolor{b l u e}{4} = \textcolor{red}{2}}$

$\text{ } {\left(x + \textcolor{red}{2}\right)}^{2} - 21$

$\textcolor{g r e e n}{\text{Add the constant } \textcolor{red}{k}}$

$\text{ } {\left(x + 2\right)}^{2} + \textcolor{red}{k} - 21$........................(2)

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$\textcolor{b r o w n}{\text{Determine the value of } k}$

Set equation (1) = equation (2)

$\text{ } {x}^{2} + 4 x - 21 = {\left(x + 2\right)}^{2} + k - 21$

$\text{ } \cancel{{x}^{2}} + \cancel{4 x} - \cancel{21} = \cancel{{x}^{2}} + \cancel{4 x} + 4 + k - \cancel{21}$

$\text{ } k = - 4$

So for ${\left(x + b\right)}^{2} + k + c \text{ } \to k = - {b}^{2}$

$\text{ "(x+2)^2-25=0" } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({2}_{a}\right)$

So $\textcolor{b l u e}{\text{Vertex} \to \left(x , y\right) = \left(\left(- 1\right) \times 2 , - 25\right) = \left(- 2 , - 25\right)}$

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