Solution 2.3:5a

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In this exercise we can use the technique for writing equations in factorized form. Consider in our case the equation
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<center> [[Image:2_3_5a.gif]] </center>
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<math>\left( x+7 \right)\left( x+7 \right)=0</math>
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This equation has only
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<math>x=-\text{7}</math>
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as a root because both factors become zero only when
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<math>x=-\text{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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<math>\left( x+7 \right)\left( x+7 \right)=x^{2}+14x+49</math>
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Thus, one answer is the equation
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<math>x^{2}+14x+49=0</math>.
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NOTE: All second-degree equations which have
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<math>x=-\text{7}</math>
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as a root can be written as
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<math>ax^{2}+14ax+49a=0</math>
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where
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<math>a</math>
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is a non-zero constant.

Revision as of 09:44, 21 September 2008

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


\displaystyle \left( x+7 \right)\left( x+7 \right)=0


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


\displaystyle \left( x+7 \right)\left( x+7 \right)=x^{2}+14x+49


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

NOTE: All second-degree equations which have \displaystyle x=-\text{7} as a root can be written as


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


where \displaystyle a is a non-zero constant.

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