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

Answer 1

Answer: (D)

$\alpha <\gamma <\beta$

Take $f\left(x\right)=a^2{x}^2+2bx+2c$
Now, $f\left(\alpha \right)=2(a^2{\alpha }^2+b\alpha +c)-a^2{\alpha }^2\dots \dots \left(1\right)$
Since $\alpha$ is a root of $a^2{x}^2+bx+c$
So $\left(1\right)$ becomes $-a^2{\alpha }^2$
Then $f\left(\beta \right)=-2(a^2{\beta }^2-b\beta -c)+4a^2{\beta }^2\dots \dots \left(2\right)$ which similarly becomes $f\left(\beta \right)=2a^2{\beta }^2$
Since $f\left(\alpha \right)f\left(\beta \right)=-4{\alpha }^2{\beta }^2{a}^4<0$
Hence the root of $f\left(x\right)$ ie $\gamma$ lies between $\alpha$ and $\beta$
i.e. $\alpha <\gamma <\beta$.