x^2 +2qx+ r =0 , What are the points of intersection of this

Question

x^2 +2qx+ r =0 , What are the points of intersection of this equation with X-axis.

A: q^2>r
B: r^2>q

Thanks.

twixt · Answer

x^2 +2qx+ r =0

Delta = 4(q^2-r)

Solutions = (-2q+-sqrt(Delta))/2= -q+-sqrt(q^2-r)

1 : q^2 > r is OK because solutions do exist
2 : r^2 > q unuseful

Answer is A

intr3pid · Answer

Is this equation lying in an XY-coordinate system? Apparently, not.

Is f(x) = y = x^2 +2qx+ r ?

srijay007 · Answer

Agree with intr3pid
it shud be y = x^2 +2qx+ r and the fn intersects x axis when y = 0

hardworker_indian · Answer

Lets say the question was:
 y = x^2 +2qx+ r, What are the points of intersection of this equation with X-axis. 
A: q^2>r 
B: r^2>q  
What would be your answer? 

I think it should be E, since we will never get to know the exact "values". We can only say whether the line cuts the x axis with A.

intr3pid · Answer

Correct. Unless you change the first condition to q^2<r, in which case there's no real solution, the line does not exist, and there's no intersection.

HOWEVER, and a major one at that, the question put forth is "what are the point s  of intersection?" This means that, theoretically, we can answer this question from the first statement by providing a range of roots (be it in terms of q and r). I just don't know if that's what the question is asking.

x^2 +2qx+ r =0 , What are the points of intersection of this

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