x2-x+3=0 quadratic equation
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apsaditya:
It has imaginary roots
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Answered by
5
Using quadratic formula:
x = (-(-1) +- √(1^2 - 4(1)(3)))/(2(1))
= (1 +- √(-11))/(2)
= (1 +- (3.31)(i))/(2)
Therefore, roots are:
[0.5 + 1.65i] and [0.5 - 1.65i]
x = (-(-1) +- √(1^2 - 4(1)(3)))/(2(1))
= (1 +- √(-11))/(2)
= (1 +- (3.31)(i))/(2)
Therefore, roots are:
[0.5 + 1.65i] and [0.5 - 1.65i]
Answered by
2
ANSWER: x²-x+3=0
ax²+bx+c=0 with
a= 1,b= -1 and c=
x= -b=root b²-4ac/2.
1=root -11/
1=root 1
=1/2 ,=root 11/2
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