Solving quadratic equation with a parameter

Question

I'm currently stuck on a problem that asks me to solve the equation $(a-1)x^2-ax+1=0$ and find the smallest integer value of $a$ such that all the solutions are positive. I've tried using the quadratic formula to find the solutions, but I'm not sure how to proceed from there to find the smallest value of $a$.

Here's what I've tried so far:

$$ \begin{split} (a-1)x^2-ax+1 &= 0 \\ x^2 - \frac{ax}{a-1} + \frac{1}{a-1} &= 0 \end{split} $$

Using the quadratic formula, we get: $$ x=\frac{a}{2(a-1)} \pm \sqrt{\frac{a^2}{4(a-1)^2}-\frac{1}{a-1}} $$

Simplifying the second term: $$ \begin{split} x &=\frac{a}{2(a-1)} \pm \sqrt{\frac{a^2-4(a-1)}{4(a-1)^2}} \\ &= \frac{a}{2(a-1)} \pm \sqrt{\frac{(a-2)^2}{4(a-1)^2}} \\ &= \frac{a}{2(a-1)} \pm \frac{a-2}{2(a-1)} \end{split} $$

Since we want all the solutions to be positive, we need $\frac{a}{a-1}$ to be positive. Therefore, $a$ must be greater than 1.

However, I'm not sure how to find the smallest integer value of $a$ that satisfies this condition. Can someone please help me out?

Thanks in advance!

Answer 1

I think the answer would just be $1$ in that case, since when we set $a=1$ we have $-x+1=0 \implies x =1$ which is the only solution, and it's positive. Your analysis correctly shows that $a$ cannot be less than $1$, as that would yield a negative solution, meaning that $a=1$ is the smallest possible integer value of $a$ in which all the solutions are positive.

Answer 2

Your Line of Reasoning

So you have $$ x_+ = \frac{a}{2(a-1)} + \frac{a-2}{2(a-1)} = \frac{a}{a-1} - \frac{2}{2(a-1)} = 1 > 0. $$ The other solution is $$ x_- = \frac{a}{2(a-1)} - \frac{a-2}{2(a-1)} = \frac{2}{2(a-1)} $$ so $x_- > 0 \iff a > 1$.

Now if $a=1$, your equation is not a quadratic equation anymore, so if that answer works for you, note that it yields $x=1$ as the only solution.

However, if it does not work for you, you must use $a=2$, the next smallest.

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Alternate Argument

A much simpler argument can be constructed from noting that the equation $$ (a-1)x^2-ax+1 = 0 $$ has coefficients which add up to $0$, so it always has $x=1$ as the solution. Then, assuming $a\ne 1$, you have $$ \begin{split} (a-1)x^2-ax+1 &= (a-1)x^2 - (a-1)x - (x-1) \\ &= (a-1)x(x-1) - (x-1) \\ &= ((a-1)x - 1)(x-1) \end{split} $$ which is easier to analyze

Answer 3

The equation can be written as $\,ax(x-1) = x^2 - 1 = (x+1)(x-1)\,$. The root $\,x=1\,$ is positive, and if there is a second positive root then after canceling the factor $\,(x-1)\,$ and dividing by $\,x \ne 0\,$ it follows that $\,a = \frac{x+1}{x} = 1 + \frac{1}{x} \gt 1\,$, which proves that $\,a \gt 1\,$ is a necessary condition for $2$ positive roots. It is also a sufficient condition, since $\,a-1 \gt 0\,$ implies $\,a\gt 0\,$, and that means both the sum and the product of the roots are positive by Vieta's relations, so both roots are positive. When $a$ is an integer $\,a \gt 1 \iff a \ge 2\,$, so $\,a=2\,$ is the minimum value for which the equation has two positive roots.

The remaining case is when the equation only has one single root, which happens when $\,a=1\,$ and the quadratic reduces to a linear equation with the root $x=1$. If that case is allowed, then the minimum integer value of $a$ for which all roots (either one or two roots) are positive is $a=1$.