what is the real # solution

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04-05-2017
Mathematics

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what is the real # solution

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jdoe0001 jdoe0001 · Answer 1 · 2017-05-04T01:09:12+02:00

[tex]\bf (x+5)(x+6)=0\implies \begin{cases} x+5=0\implies &x=-5\\ x+6=0\implies &x=-6 \end{cases} \\\\\\ (x-9)(x+8)=0\implies \begin{cases} x-9=0\implies &x=9\\ x+8=0\implies &x=-8 \end{cases} \\\\\\ (x+4)(x-4)=0\implies \begin{cases} x+4=0\implies &x=-4\\ x-4=0\implies &x=4 \end{cases} \\\\\\ x^2-14x+49\implies (x-7)^2\iff (x-7)(x-7) \\\\\\ (x-7)(x-7)=0\implies \begin{cases} x-7=0\implies &x=7\\ x-7=0\implies &x=7 \end{cases}[/tex]