If $p, q, r$ are positive and are in A.P., the roots of quadratic equation $px^2+qx+r=0$ are all real for
- $\left |\dfrac {r}{p}-7 \right |\geq 4\sqrt 3$
- $\left |\dfrac {p}{r}-7\right |\geq 4\sqrt 3$
- All $p$ and $r$
- No $p$ and $r$
For px^2 + qx + r = 0 to have real roots, the discriminant D = q^2 - 4pr >= 0. Since p, q, r are in A.P., q = (p+r)/2. Substituting: ((p+r)/2)^2 - 4pr >= 0 -> (p+r)^2 - 16pr >= 0 -> p^2 + 2pr + r^2 - 16pr >= 0 -> p^2 - 14pr + r^2 >= 0. Dividing by r^2: (p/r)^2 - 14(p/r) + 1 >= 0. Solving this inequality leads to |p/r - 7| >= 4*sqrt(3).
Because p, q, and r are in an arithmetic progression, we can write 2q = p + r. For the quadratic equation px^2 + qx + r = 0 to have real roots, its discriminant D = q^2 - 4pr must be non-negative. Substituting q = (p+r)/2 into the discriminant formula gives ((p+r)/2)^2 - 4pr >= 0, which simplifies to (p^2 - 14pr + r^2)/4 >= 0. Dividing by r^2 and letting k = p/r, we get k^2 - 14k + 1 >= 0, which corresponds to the inequality (k - 7)^2 >= 48. Solving this yields |p/r - 7| >= 4*sqrt(3).