Let a, b, c ∈ R and a ≠ 0. If α is a root of a2 x2+b x+c=0, β is a root a2 x2-b x-c=0 and 0

Question

Let a, b, c ∈ R and a ≠ 0. If α is a root of a2 x2+b x+c=0, β is a root a2 x2-b x-c=0 and 0<α<β, then the equation a2 x2+2 b x+2 c=0 has a root &gamma; that always satisfies (A) α<&gamma;<β (B) &gamma;=α+(β/2) (C) &gamma;=α (D) &gamma;=(α+β/2)

Tardigrade Admin · Accepted Answer

text Let f(x)=a2 x2+2 b x+2 c f(α)=a2 α2+2 b α+2 c=-a2 α2 f (β)= a 2 β2+2 β b+2 c =3 a 2 β2 ∴ f (α) ⋅ f (β)<0 Hence, &gamma; lies between α and β.

Q. 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 $a^{2} x^{2} - b x - c = 0$ and $0 < α < β$ , then the equation $a^{2} x^{2} + 2 b x + 2 c = 0$ has a root $γ$ that always satisfies

$α < γ < β$

$γ = α + \frac{β}{2}$

$γ = α$

$γ = \frac{α + β}{2}$

$Let f (x) = a^{2} x^{2} + 2 b x + 2 c$
$f (α) = a^{2} α^{2} + 2 b α + 2 c = - a^{2} α^{2}$
$f (β) = a^{2} β^{2} + 2 β b + 2 c = 3 a^{2} β^{2}$
$∴ f (α) \cdot f (β) < 0$
Hence, $γ$ lies between $α$ and $β$ .

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

α<γ<β

γ=α+2β​

γ=α

γ=2α+β​