If the sum of the roots of the equation $a x ^ { 2 } + bx + c = 0$ is equal to the sum of the square of their reciprocal, then _________________.
- $c ^ { 2 } b , a ^ { 2 } c , b ^ { 2 } a$ are in A.P.
- $c ^ { 2 } b , a ^ { 2 } c , b ^ { 2 } a$ are in G.P.
- $\frac { b } { c } , \frac { a } { b } , \frac { c } { a }$ are in H.P.
- $\frac { b } { c } , \frac { a } { b } , \frac { c } { a }$ are in G.P.
Let roots be p and q. p+q = -b/a and pq = c/a. Given p+q = 1/p^2 + 1/q^2 = (p^2+q^2)/(pq)^2 = ((p+q)^2 - 2pq)/(pq)^2. Substituting values: -b/a = (b^2/a^2 - 2c/a) / (c^2/a^2) = (b^2 - 2ac) / c^2. This simplifies to -bc^2 = a(b^2 - 2ac), or 2a^2c = ab^2 + bc^2. This implies bc^2, a^2c, b^2a are in A.P. as 2(a^2c) = bc^2 + ab^2.
Using the sum and product of roots formulas for ax squared plus bx plus c equals 0, we know the sum of roots is -b over a and the product of roots is c over a. The condition states the sum of roots equals the sum of the squares of their reciprocals, giving the equation -b over a equals (1 over alpha squared) plus (1 over beta squared). Substituting the values and simplifying leads to ab squared plus ac squared equals bc multiplied by (a plus c). This rearranges to show that c squared times b, a squared times c, and b squared times a are in arithmetic progression.