Answers

2014-06-30T18:28:20+08:00
m^2+7m- \frac{51}{4}=0
m^2+7m= \frac{51}{4}
m^2+7m+( \frac{7}{2})^2= \frac{51}{4}+ ( \frac{7}{2} )^2
\sqrt{(m+ \frac{7}{2})^2 }= +or-\sqrt{ \frac{100}{4} }
m+ \frac{7}{2}= +or- \frac{10}{2}
m= \frac{10}{2}- \frac{7}{2}
m= \frac{3}{2}
m=- \frac{10}{2}- \frac{7}{2}
m=- \frac{17}{2}

Check:
m= \frac{3}{2}
( \frac{3}{2})^2+7( \frac{3}{2})- \frac{51}{4}=0
 \frac{9}{4}+ \frac{21}{2}-  \frac{51}{4}  =0
 \frac{51}{4} - \frac{51}{4}=0
0=0

m= -\frac{17}{2}
( -\frac{17}{2})^2+7( -\frac{17}{2})- \frac{51}{4}=0
 \frac{289}{4}- \frac{119}{2}- \frac{51}{4}=0
 \frac{51}{4}-  \frac{51}{4} =0
0=0

w^2+6w-11=0
w^2+6w=11
w^2+6w+(3)^2=11+(3)^2
 \sqrt{(w+3)^2}= \sqrt{20}
w+3=+or- \sqrt{20}
w=+or- \sqrt{20}-3
 \left \{ {{w= \sqrt{20}-3 } \atop {w= -\sqrt{20}-3 }} \right.

(In checking you can just use a scientific calculator... Just substitute the variables with the given value...) ^_^ \/

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