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 $- 1, - 1$
$$\begin{matrix} \therefore & ( - 1)( - 1) = \frac{3c + 5a + 5b}{6a + 3b + 4c} = 1 \\ \Rightarrow & a + c = 2b \end{matrix}$$
Create a free account to view solution
View Solution FreeMore Quadratic Equation Questions
If α, β are the roots of ax2 + bx + c = 0; α + h, β + h are the roots of px2 + qx + r = 0, and D1, D...If f (x) = ax2 + bx + c, g (x) = −ax2 + bx + c, where ac ≠ 0, then f (x) g (x) = 0 has -...If α, β are roots of the equation ax2 + bx + c = 0 and α − β = α, β then -...The equation whose roots are is-...If the difference between the roots of the equation x2 + ax + 1 = 0 is less than √5, then the set of possible valu...