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# Help. ​

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Help.

Nov 3, 2017

#1
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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 .

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

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

#1
+7354
+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 .

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

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