Let a- b- c - R and a - 0- If - is a root of a2 x2 - bx - c - 0- - is a root of a2 x2 - bx - c - 0 and 0 - - - - then the equation- a2x2 - 2bx - 2c - 0 has a root - that always satisfies

$a b c \neq 0$ & $a, b, c \in R$ . If $x_{1}$ is a root of $a^{2} x^{2} + b x + c = 0, x_{2}$ is a root of $a^{2} x^{2} - b x - c = 0$ and $x_{1} > x_{2} > 0$ , then the equation $a^{2} x^{2} + 2 b x + 2 c = 0$ has a root $x_{3}$ such that

Q. Solve this :

I l l u s t r a t i o n 2 : I f α i s a r o o t o f t h e e q u a t i o n a x^{2} + b x + c = 0 a n d β i s a r o o t o f t h e e q u a t i o n - a x^{2} + b x + c = 0, t h e n p r o v e t h a t t h e r e w i l l b e a r o o t o f t h e e q u a t i o n \frac{a}{2} x^{2} + b x + c = 0 l y i n g b e t w e e n α a n d β .

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Let a,b,c∈R and a≠0. If α is a root of a2x2+bx+c=0,β is a root of a2x2−bx−c=0 and 0<α<β, then the equation, a2x2+2bx+2c=0 has a root γ that always satisfies

The correct option is B α<γ<β Let f(x)=a2x2+2bx+2c From the question, a2α2+bα+c=0 and a2β2−bβ−c=0 Now, f(α)=a2α2+2bα+2c=bα+c=−a2α2f(β)=a2β2+2bβ+2c=3(bβ+c)=3a2β2 but 0<α<β ⇒α and β are Positive. ∴f(α)<0, f(β)>0 Hence, α<γ<β.

Let $a, b, c \in R$ and $a \neq 0.$ If $α$ is a root of $a^{2} x^{2} + b x + c = 0, β$ is a root of $a^{2} x^{2} - b x - c = 0$ and $0 < α < β,$ then the equation, $a^{2} x^{2} + 2 b x + 2 c = 0$ has a root $γ$ that always satisfies