x^2-(3√2+2i)x+6√2i=0 find
b^2-4ac,
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x²-(3√2+2i)x+6√2i=0
here , a = 1 ,b= - ( 3√2+2i) ,c= (6+√2i)
substitute these values in equation b²-4ac
we get,
- (3√2+2i)²-4×(1)×(6+√2i)
[ -(3√2)²- (2i)²+2× (-3√2)×(-2i) ] - [4 ( 6+√2i) ]
[18+4i²+12√2i] - [24+4√2i)
18 + 4i² + 12√2i - 24 - 4√2i
4i²+ 12√2i - 4√2i - 24 + 18
4i² + 8√2i - 6
so value of b²- 4ac from equation
x²- (3√2+2i)x+6√2=0 is 4i² + 8√2 - 6
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