Quadratic EquationHard
Question
If the equation $(6a + 3b + 4c)x^{2} + (11a + 8b + 7c)x + (3c + 5a + 5b) = 0$ has equal real roots, where $a,b,c$ are positive real numbers. Then $a,b,c$ are in
Options
A.A.P.
B.G.P.
C.H.P.
D.nothing can be said
Solution
By observation put $x = - 1$
$$(6a + 3b + 4c) - (11a + 8b + 7c) + (3c + 5a + 5b) = 0$$
$\therefore\ $ Equation has roots $- 1, - 1$
$$\begin{matrix} \therefore & ( - 1)( - 1) = \frac{3c + 5a + 5b}{6a + 3b + 4c} = 1 \\ \Rightarrow & a + c = 2b \end{matrix}$$
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