JanaeSpellman
JanaeSpellman
17-07-2017
Mathematics
contestada
How to solve and why the answer is (A)
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konrad509
konrad509
17-07-2017
[tex]\dfrac{1}{2}x^2+mx=5\\ \dfrac{1}{2}x^2+mx-5=0\\\\ \Delta=m^2-4\cdot\dfrac{1}{2}\cdot(-5)=m^2+10\\ \sqrt{\Delta}=\sqrt{m^2+10}\\\\ x=\dfrac{-m\pm\sqrt{m^2+10}}{2\cdot\dfrac{1}{2}}=-m\pm\sqrt{m^2+10}[/tex]
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