From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	In this exercise we can use the technique for writing equations in factorized form. Consider in our case the equation
-	<center> [[Image:2_3_5a.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>(x+7)(x+7)=0\,\textrm{.}</math>}}
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		+	This equation has only <math>x=-7</math> as a root because both factors become zero only when <math>x=-7</math>. In addition, it is an second-degree equation, which we can clearly see if the left-hand side is expanded,
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		+	{{Displayed math||<math>(x+7)(x+7) = x^{2}+14x+49\,\textrm{.}</math>}}
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		+	Thus, one answer is the equation <math>x^{2}+14x+49=0\,</math>.
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		+	Note: All second-degree equations which have <math>x=-7</math> as its sole root can be written as
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		+	{{Displayed math||<math>ax^{2}+14ax+49a=0\,,</math>}}
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		+	where ''a'' is a non-zero constant.

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Current revision

In this exercise we can use the technique for writing equations in factorized form. Consider in our case the equation

\displaystyle (x+7)(x+7)=0\,\textrm{.}

This equation has only \displaystyle x=-7 as a root because both factors become zero only when \displaystyle x=-7. In addition, it is an second-degree equation, which we can clearly see if the left-hand side is expanded,

\displaystyle (x+7)(x+7) = x^{2}+14x+49\,\textrm{.}

Thus, one answer is the equation \displaystyle x^{2}+14x+49=0\,.

Note: All second-degree equations which have \displaystyle x=-7 as its sole root can be written as

\displaystyle ax^{2}+14ax+49a=0\,,

where a is a non-zero constant.