Question

If a,b,c are 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 lies between $\alpha and\beta .$If true write 0 else 1.

Answer 1

As $\alpha$ is root of $a^2{x}^2+bx+c=0$

$\Rightarrow$ $a^2{\alpha }^2+b\alpha +c=0$ ----- ( 1 )

As $\beta$ is root of $a^2{x}^2-bc-c=0$

$\Rightarrow$ $a^2{\beta }^2-b\beta -c=0$ ---- ( 2 )

Now, we have,

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

$\Rightarrow$ $f\left(\alpha \right)=a^2{\alpha }^2+2b\alpha +2c$

$\Rightarrow$ $f\left(\alpha \right)=a^2{\alpha }^2+2(ba+c)$

Using equation ( 1 ) we get,

$\Rightarrow$ $f\left(\alpha \right)=a^2{\alpha }^2-2a^2{\alpha }^2$

$\Rightarrow$ $f\left(\alpha \right)-a^2{\alpha }^2$

$\Rightarrow$ $f\left(\beta \right)=a^2{\beta }^2+2b\beta +2c$

$\Rightarrow$ $f\left(\beta \right)=a^2{\beta }^2+2(b\beta +c)$

Using equation ( 2 ) we get,

$f\left(\beta \right)=a^2{\beta }^2+2\left(a^2{\beta }^2\right)$

$\Rightarrow$ $f\left(\beta \right)=3a^2{\beta }^2$

$\Rightarrow$ $f\left(\alpha \right)f\left(\beta \right)=-3a^2{\beta }^2{a}^2{\alpha }^2$, which is negative quantity.

So, one root of $f\left(x\right)$ must lie between $\alpha$ and $\beta .$

$\Rightarrow$ $\alpha <\gamma <\beta$

$\therefore$ The answer is $0$