Respuesta :

konrad509
[tex](x+a)(x+a)=x^2+2ax+a^2\\\\ 2a=10 \wedge a^2=24\\ a=5 \wedge (a=\sqrt{24} \vee a=-\sqrt{24})\\ a=5 \wedge (a=2\sqrt{6} \vee a=-2\sqrt{6})[/tex]

The above is contradiction, so there are no such values of 'a'.

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