Find the roots of this quadratic equation
√3x^2+√3x+6√3
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/3x^2 + /3x + 6/3 = 0
By Formula method
By comparing to
ax^2 + bx + c =0
a=/3 ; b=/3 ; c=6/3
Discriminant = b^2 - 4ac
= (/3)^2 - 4(/3 × 6/3)
= 3 - 4×6×3
= 3-72
= - 69
x = - b +/- (/b^2-4ac) / 2a
x = - /3 +/- (/-69) / 2×/3
x = - /3 +/- (/-69) / 2/3
1) x = - /3 + /-69 / 2/3
2) x = - /3 - /-69 / 2/3
Hope it helps you!!! ☺☺
By Formula method
By comparing to
ax^2 + bx + c =0
a=/3 ; b=/3 ; c=6/3
Discriminant = b^2 - 4ac
= (/3)^2 - 4(/3 × 6/3)
= 3 - 4×6×3
= 3-72
= - 69
x = - b +/- (/b^2-4ac) / 2a
x = - /3 +/- (/-69) / 2×/3
x = - /3 +/- (/-69) / 2/3
1) x = - /3 + /-69 / 2/3
2) x = - /3 - /-69 / 2/3
Hope it helps you!!! ☺☺
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