First slide

Theory of equations

Question

Suppose A,B,C are defined as $A = a^{2} b + {ab}^{2} - a^{2} c - {ac}^{2}$ , $B = b^{2} c + {bc}^{2} - a^{2} b - {ab}^{2}$ and $C = a^{2} c + {ac}^{2} - b^{2} c - {bc}^{2}$ , where $a > b > c > 0$ and the equation ${Ax}^{2} + Bx + C = 0$ has equal roots, then a, b, c are in

Moderate

A

A.P
B

G.P
C

H.P
D

A.G.P

Solution

$\begin{aligned} A = a (b - c) (a + b + c) \\ B = b (c - a) (a + b + c) \\ C = c (a - b) (a + b + c) \end{aligned}$
Now, ${Ax}^{2} + Bx + C = 0$
$\Rightarrow (a + b + c) \{a (b - c) x^{2} + b (c - a) x + c (a - b)\} = 0$
Given that roots are equal. Hence,
D = 0
$\begin{array}{l} or b^{2} c^{2} - 2 {ab}^{2} c + b^{2} a^{2} - 4 a^{2} bc + 4 {acb}^{2} + 4 a^{2} c^{2} - 4 {abc}^{2} = 0 \\ or b^{2} c^{2} + b^{2} a^{2} + 4 a^{2} c^{2} + 2 {ab}^{2} c - 4 a^{2} bc - 4 {abc}^{2} = 0 \\ or (bc + ab - 2 ac)^{2} = 0 \\ or bc + ab = 2 ac \end{array}$
$\Rightarrow \frac{1}{a} + \frac{1}{c} = \frac{2}{b}$
Hence, a, b, c are in H.P.

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