Evaluate and simplify in a+bi form:
(6-�ã-18))+(2+�ã-50)
5 answers
ã- radical
(6-�ã-18))+(2+�ã-50)
= (6 - √-18) + (2 + √-50)
=(6 - 3√2√-1) + (2 + 5√2√-1)
=(6 - 3√2 i) + (2 + 5√2 i)
= 8 + 2√2 i
= (6 - √-18) + (2 + √-50)
=(6 - 3√2√-1) + (2 + 5√2√-1)
=(6 - 3√2 i) + (2 + 5√2 i)
= 8 + 2√2 i
To show the square-root sign (√), you can type:
"& r a d i c ;"
but omit the intervening spaces and the double quotes.
I assume you are asking to simplify:
(6-√(-18))+(2+&raidc;(-50))
We first remove the outer parentheses and convert the radicals to i to get:
6-√(-18)+2+√(-50)
8 - √(18)i + √(50)i
= 8 - 3√2 i + 5√2 i
= 8 +2√2 i
"& r a d i c ;"
but omit the intervening spaces and the double quotes.
I assume you are asking to simplify:
(6-√(-18))+(2+&raidc;(-50))
We first remove the outer parentheses and convert the radicals to i to get:
6-√(-18)+2+√(-50)
8 - √(18)i + √(50)i
= 8 - 3√2 i + 5√2 i
= 8 +2√2 i
Isn't it same as 8+2i&radic2
The question asked for:
"Evaluate and simplify in a+bi form:"
In any case the "i" is usually put after the coefficient. √2 is considered part of the coefficient.
"Evaluate and simplify in a+bi form:"
In any case the "i" is usually put after the coefficient. √2 is considered part of the coefficient.