Question

Let $a,b,c$ be real numbers $a\ne$ 0. If $\alpha$ is a root of $a^2{x}^2+bx+c=0,\beta$ is a root of $a^2{x}^2-bx-c=0$ and $0<\alpha <\beta$, then the equation $a^2{x}^2+2bx+2c=0$ has a root $\gamma$ that always satisfies:

• A. $\gamma =\alpha +\frac{\beta }{2}$
• B. $\alpha <\gamma <\beta$
• C. $\gamma =\frac{\alpha +\beta }{2}$
• D. $\gamma =\alpha$

Answer 1

The correct option is B

$\alpha <\gamma <\beta$

Since $\alpha \&\beta$ are the roots of given equations. So we have $a^2{\alpha }^2+b\alpha +c=0\&a^2{\beta }^2-b\beta -c=0$

Let $f\left(x\right)=a^2{x}^2+2bx+2c=0$

Then $f\left(\alpha \right)=a^2{\alpha }^2+2b\alpha +2c=a^2{\alpha }^2+2(b\alpha +c)=a^2{\alpha }^2-2a^2{\alpha }^2=-a^2{\alpha }^2=-ve$
and
$f\left(\beta \right)=a^2{\beta }^2+2(b\beta +c)=a^2{\beta }^2+2a^2{\beta }^2+2a^2{\beta }^2=3a^2{\beta }^2=+ve$

since $f\left(\alpha \right),f\left(\beta \right)$ are of opposite signs, therefore by theory of equations there lies a root $\gamma$ of the equation $f\left(x\right)=0$ between α and β i.e., $\alpha <\gamma <\beta$