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NotSoSmart  Nov 3, 2017

Best Answer 

 #1
avatar+5256 
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5.     \(\frac{-6+i}{-5+i}\)

 

Multiply the numerator and denominator by  -5 - i .

 

\(=\frac{(-6+i)(-5-i)}{(-5+i)(-5-i)}=\frac{30+6i-5i-i^2}{25+5i-5i-i^2}=\frac{30+i-i^2}{25-i^2}\)

 

Replace the  i2  's  with  -1  since   i2  =  -1

 

\(=\frac{30+i-(-1)}{25-(-1)} =\frac{30+i+1}{25+1} =\frac{31+i}{26}\)

 

 

 

6.      \(\frac12x^2-x+5=0\)

 

We can use the quadratic formula to solve this for  x, with  a = 1/2 ,  b = -1 ,  and  c = 5 .

 

\(x = {-(-1) \pm \sqrt{(-1)^2-4(\frac12)(5)} \over 2(\frac12)} \\~\\ x = {1 \pm \sqrt{1-10} \over 1} \\~\\ x=1\pm\sqrt{-9} \\~\\ x=1\pm i\sqrt9\)

hectictar  Nov 3, 2017
edited by hectictar  Nov 3, 2017
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1+0 Answers

 #1
avatar+5256 
+1
Best Answer

5.     \(\frac{-6+i}{-5+i}\)

 

Multiply the numerator and denominator by  -5 - i .

 

\(=\frac{(-6+i)(-5-i)}{(-5+i)(-5-i)}=\frac{30+6i-5i-i^2}{25+5i-5i-i^2}=\frac{30+i-i^2}{25-i^2}\)

 

Replace the  i2  's  with  -1  since   i2  =  -1

 

\(=\frac{30+i-(-1)}{25-(-1)} =\frac{30+i+1}{25+1} =\frac{31+i}{26}\)

 

 

 

6.      \(\frac12x^2-x+5=0\)

 

We can use the quadratic formula to solve this for  x, with  a = 1/2 ,  b = -1 ,  and  c = 5 .

 

\(x = {-(-1) \pm \sqrt{(-1)^2-4(\frac12)(5)} \over 2(\frac12)} \\~\\ x = {1 \pm \sqrt{1-10} \over 1} \\~\\ x=1\pm\sqrt{-9} \\~\\ x=1\pm i\sqrt9\)

hectictar  Nov 3, 2017
edited by hectictar  Nov 3, 2017

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