<span>Let p, np be the roots of the given QE.So p+np = -b/a, and np^2 = c/aOr (n+1)p = -b/a or p = -b/a(n+1)So n[-b/a(n+1)]2 = c/aor nb2/a(n+1)2 = cor nb2 = ac(n+1)2
Which will give can^2 + (2ac-b^2)n + ac = 0, which is the required condition.</span>
I see the answers,but where is the problem (equation that needs to be solved)
Answer:
A. Right angle -- if you multiply the slopes of 2 intersecting lines and you "-1" then the lines are perpendicular - in this case the line in quadrant 3 has a slope of "1" and the line in quadrant 4 has a slope of "-1", hence their product is "-1" and the angles formed at the intersection of the lines are right angles.
Step-by-step explanation:
