+0  
 
0
247
1
avatar+4688 

Help.

 Nov 3, 2017

Best Answer 

 #1
avatar+7347 
+1

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 .

 

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

.
 Nov 3, 2017
edited by hectictar  Nov 3, 2017
 #1
avatar+7347 
+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

35 Online Users

avatar
avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.