How do you find all the real and complex roots of $3 x^{2} - x + 2 = 0$ $3x^2 - x + 2 = 0$ ?

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Answer 1 · 2018-08-08T09:53:47

$x = \frac{1 + \sqrt{23} \cdot i}{6} or x = \frac{1 - \sqrt{23} \cdot i}{6}$ $x=(1+sqrt23*i)/6 or x=(1-sqrt23*i)/6$

Here ,

$3 x^{2} - x + 2 = 0$ $3x^2-x+2=0$

Comparing with $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

$a = 3, b = - 1 and c = 2$ $a=3,b=-1 and c=2$

$:.Delta=b^2-4ac=(-1)^2-4(3)(2)$

$:.color(red)(Delta=1-24=-23 < 0=>ul(x !inRR and x inCC)$

$:.Delta=23(-1)$

$:.Delta=23i^2to[because i^2=-1]$

$:.sqrtDelta=sqrt23*i$

So,

$x=(-b+-sqrtDelta)/(2a)=(1+-sqrt23*i)/(2xx3)$

$:.x=(1+-sqrt23*i)/6$

Hence ,

$x=(1+sqrt23*i)/6 or x=(1-sqrt23*i)/6$

How do you find all the real and complex roots of 3x^2 - x + 2 = 03x2−x+2=03x^2 - x + 2 = 0?