Basic quadratic calculus

Question

Given $$ AX^2+2X-1=0 \,\ \text{where} \,\ A>0$$ What value of A would make the absolute value of both roots bigger than $1$? I found the roots using quadratic formula and also found that A has to be both $[0,3]$ and $[-1,0]$ and $A>0$, does it mean such an $A$ doesn't exist? Thanks!

Answer 1

Note: The original question stated "both roots are greater than 1", without any mention of absolute value. This solution is for that question.

We want $$\frac{-2\pm \sqrt{4+4A}}{2A}=\frac{-1\pm\sqrt{1+A}}{A}>1$$

If $A>0$ then $-1+\sqrt{1+A}>-1-\sqrt{1+A}$ so both roots are bigger than 1 if $$-1-\sqrt{1+A}>A$$ which rearranges to $$0>1+A+\sqrt{1+A}$$ This is plainly impossible, so no solutions exist for $A>0$.

If $A<0$ we must in fact have $-1\le A<0$ else we are taking an impossible square root. Now $$\frac{-1-\sqrt{1+A}}{A}>\frac{-1+\sqrt{1+A}}{A}(>0)$$ so both roots are bigger than 1 if $$\frac{-1+\sqrt{1+A}}{A}>1$$ We multiply both sides by the negative $A$, getting $$-1+\sqrt{1+A}<A$$ which rearranges to $$\sqrt{1+A}<1+A$$ However $0\le A+1<1$, so $\sqrt{1+A}$ is never less than $1+A$. Hence no solutions exist for $A<0$ either.

Answer 2

The roots are

$$ X_1=\frac{-1+\sqrt{1+A}}{A},\;\;\; \text{and}\;\;\; X_2=\frac{-1-\sqrt{1+A}}{A}. $$

Notice that $X_2<0$ for any $A>0$. So there exists no $A$ such that both roots are positive.

Answer 3

Note: The original question stated "both roots are greater than 1", without any mention of absolute value. This solution is for that question.

If you're familiar with Vieta's formulas, then the product of the two roots will be $\frac{-1}{A}$. If both roots are greater than $1$, then their product is as well, so $\frac{-1}{A}>1$ and hence $A<-1$. However, the sum of the two roots will be $-\frac{2}{A}$. Since $A<-1$, the sum of the two roots will be less than $-\frac{2}{-1}=2$. But it's not possible for two numbers, each greater than $1$, to have sum less than $2$.

Answer 4

For $A>0$ we have \begin{align} 0&=X^2 + 2 \frac{1}{A}X - \frac{1}{A} \\ &= (X + 1/A)^2 -1/A^2 -1/A \\ &= (X - (-1/A))^2 + (- (A+1)/A^2) \iff \\ X &= \frac{\pm\sqrt{A+1}-1}{A} \end{align} This is a parabola with vertex at $(-1/A, -(A+1)/A^2)$.

Then $$ \frac{\sqrt{A+1}-1}{A} \le \lvert X\rvert = \left\lvert \frac{\pm\sqrt{A+1}-1}{A} \right\rvert \le \frac{\sqrt{A+1}+1}{A} $$ So we need \begin{align} 1 &< \frac{\sqrt{A+1}-1}{A} \iff \\ A + 1 &< \sqrt{A+1} \Rightarrow \\ (A + 1)^2 &< A + 1 \iff \\ A^2 + 2A + 1 &< A+1 \iff \\ A^2 + A &< 0 \end{align} This is not possible for positive $A$, as $A^2$ is positive as well.