EgmatQuantExpert wrote:
Hey
levloans,
We know that both the roots of a quadratic equation can be found out by applying \(\frac{-b +- \sqrt{b^2 -4ac}}{2a}\).
So, if one root is of the form \(\frac{-b + \sqrt{b^2 -4ac}}{2a}\) then other root will be of the form \(\frac{-b - \sqrt{b^2 -4ac}}{2a}\).
Notice the difference in sign.
Therefore, if one root is \(3- 2\sqrt{3}\) then another root will be \(3+ 2\sqrt{3}\) .
I hope this answers your query.
Regards,
Ashutosh
Thanks for your quick response, Ashutosh. I had considered this pattern, but still do not see why it is
necessary that our roots are these.
Let me rephrase:
I understand that \(3 + 2\sqrt{3}\) & \(3- 2\sqrt{3}\) COULD be our roots.
But what says that they HAVE to?
i.e.
For example
Couldn't I have a quadratic with the roots (1) \(3+ 2\sqrt{3}\) and (2) 11 ?
This is a real quadratic with x-intercepts at \(3+ 2\sqrt{3}\) and 11...
in fact, just look at this graph for visual reference
https://www.wolframalpha.com/input/?i=y%3D(x-(3%2B2*sqrt(3)))*(x-11)Just because I know the first root (1), it doesn't seem to me that I can make assumptions about what the second root (2) is, as that could be anything.
Or, I can know one of the x intercepts, but that doesn't tell me anything about what the other is, unless I know something else about the form of the quadratic.
... I can "peg" one of the x-intercepts (roots) and then move the other one to wherever I want and still have a valid parabola/quadratic.
I think your answer is in some way is affirming the consequent; or pulling in info from the output to make a judgment on the input, but the output is not given information, we can only start with the input of the question.
Please let me know me if I am missing something.