The equation to be solved is $9^x + 2(1-a) 3^x + a = 0$. We have to find all integral values of $a$ between $1$ and $30$ for which the above equation has roots of opposite sign.

I substituted $3^x = t$ and got the equation $ t^2 + 2(1-a)t + a= 0$. Applied Discriminant $\ge 0$. But I could not get any solution.

  • $\begingroup$ We need $$\frac a1<0$$ $\endgroup$ – lab bhattacharjee May 9 '17 at 14:01
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    $\begingroup$ @labbhattacharjee Not quite. Note that we don't want the solutions in $t$ to have opposite signs, we want the solutions in $x$ to have opposite signs, which means that the solutions in $t$ must both be positive, and on either side of $1$. $\endgroup$ – Arthur May 9 '17 at 14:04

Your substitution is good. Now you can apply the quadratic formula to solve for $t$ (in terms of $a$, of course). As noted in the comments above, you want your two values for $t$ to be both positive, one on the interval $(0,1)$, and the other on $(1,\infty)$.

I'm getting a discriminant: $4(a^2-3a+1)$, or $4((a-\frac32)^2-\frac54)$, which is negative for $a=1,2$, so we only really need to consider values of $a$ from $3$ to $30$.

The solutions for $t$ are $a-1 \pm \sqrt{a^2-3a+1}$. When $a=3$, one of these solutions is $1$, implying a root of $x=0$, which is not desired. For $a>3$ the radical satisfies the inequality:

$$a-2 < \sqrt{a^2-3a+1} < a-1.$$

That puts the difference obtained when choosing the minus sign between $0$ and $1$, as desired. Thus, all values from $a=4$ to $a=30$ should work.

  • $\begingroup$ I tried the method suggested. I got the value of $a \ge 4 $ . I will be thankful if you please verify if this is correct.Thanks for answering! $\endgroup$ – Ananth Kamath May 9 '17 at 14:16

Obviously $a>0$ since both roots must be positive. Let $f(t)=t^2+2(1-a)t+a$

Note that

$$f(0^+)>0$$ $$f(\infty)>0$$

So all you need is that $f(1)<0$ for $f$ to has a root in $(0,1)$ and in a root in$(1,\infty)$

So setting $f(1)<0$,


that implies

$$a \in \{4,5,...,30\}$$

  • $\begingroup$ Thanks for answering. But on what basis do we assume $a=3$ ? $\endgroup$ – Ananth Kamath May 9 '17 at 14:26
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    $\begingroup$ The sum and the product of the roots are positive. So the only thing that matters is to check that one root is greater than 1 and another is less than 1. So set $t=1$ to find out for which value of $a$ it will happen. So $1+2(1-a)+a=0$ implies that $a=3$ $\endgroup$ – PILAR May 9 '17 at 14:31
  • $\begingroup$ @AnanthKamath I wrote the proof completely. It's fairly short $\endgroup$ – PILAR May 9 '17 at 15:01
  • $\begingroup$ thanks for attatching the proof. $\endgroup$ – Ananth Kamath May 9 '17 at 15:04

Everything is fine until substitution .As roots of given equation are of opposite sign this implies $0$ should lie between the roots .Therefore ,$a.f(k) \lt 0$ where $f(x)=ax^2+bx +c $ and $k$ is the number which lies between roots.

Solution $$f(x)=t^2 +2(1-a)t+a \space \space\text{ where k=0,a=1}$$ $$\implies 1.(1+2(1-a)+a) \lt 0$$ $$\implies 3-a \lt 0 $$ $$\implies a \gt 3 $$


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