- #1
danny12345
- 22
- 0
1)y^3+6y^2+11y+6=0
2)y^3+6y^2+12y+8=0
find it's root and tell me how you obtained it.
2)y^3+6y^2+12y+8=0
find it's root and tell me how you obtained it.
dansingh said:1)y^3+6y^2+11y+6=0
2)y^3+6y^2+12y+8=0
find it's root and tell me how you obtained it.
Wilmer said:y^3+6y^2+11y+6=0
y^3+6y^2+12y+8=0
Well, if I got that on a timed test,
I'd simply subtract the equations
to get y+2 = 0, so y = -2
Would you give me a pass mark, Mark :)
YA! On a timed test, that's all I'd want.MarkFL said:I don't think the two equations are simultaneous (even though the OP said "find its root" as if the two equations were one entity)...so if my interpretation is correct and unless you used this in lieu of the rational roots theorem to find a root as a starting point for both, I would have to deduct some points for not finding ALL roots of both equations. :)
The roots of this equation are -1, -2, and -3.
The roots can be solved by factoring the equation or by using the quadratic formula.
No, as the highest exponent of the variable y is 3, there can only be a maximum of three roots.
The coefficients represent the values used to create the equation and can affect the location and nature of the roots.
This equation can be graphed by plotting points with various values for y and solving for the corresponding x values. The roots will be where the graph intersects the x-axis.