Show that $ Ax^2 + x + 1 = 0 $ has roots greater than $1$ given $-1<A<0$

Question

Not sure how to answer this question, i tried putting the numbers in the quadratic formula and got $x = 2-2\sqrt{A}/2a$? Any help would be much appreciated.

Answer 1

The problem was changed!

The original problem was:

Show that $Ax^2 + x + 1 = 0$ has roots greater than 1 given $-1<A<0$

One of roots it's $$\frac{-1-\sqrt{1-4a}}{2a}.$$ We'll prove that

$$\frac{-1-\sqrt{1-4a}}{2a}>1$$ or $$-1-\sqrt{1-4a}<2a$$ or $$-\sqrt{1-4a}<2a+1,$$ which is obvious for $1+2a>0,$ but for $1+2a\leq0$ we need to prove that $$1-4a>(2a+1)^2$$ or $$a(a+2)<0,$$ which is obvious.

The second problem:

Show that $-Ax^2 - x + 1 = 0$ has roots greater than 1 given $-1<A<0$

It's wrong, of course. Try $A=-\frac{1}{2}.$

Answer 2

The posted question was revised a few times.

First assume the equation (as it was given initially, before the edits) is $$ax^2+x+1=0$$ with $-1 < a < 0$, and that the goal is to prove that the equation has a root greater than $1$.

Since the discriminant is $$D=(1)^2-4(a)(1)=1-4a$$ and $a < 0$, it follows that $D > 0$, hence the equation has two distinct real roots.

Let $u,v$ be the roots.

By Vieta's formulas,

• $u+v=-{\large{\frac{1}{a}}}$.$\\[4pt]$
• $uv={\large{\frac{1}{a}}}$.

From $a < 0$, we get ${\large{\frac{1}{a}}} < 0$.

Thus $uv < 0$, hence one of $u,v$ is positive, and the other is negative.

From $-1 < a < 0$, we get $-{\large{\frac{1}{a}}} > 1$.

Thus, $u+v > 1$, hence, since one of $u,v$ is negative, the other one must be greater than $1$.

That resolves the original version of the problem.

Next assume the equation is $$-ax^2-x+1=0$$ but to make it work out (and to show a few more ideas), assume the restriction on $a$ is $-{\large{\frac{1}{4}}} < a < 0$.

For this version of the problem, the goal is to show that both roots are greater than $1$.

Some of the reasoning is the same as in the previous version . . .

Since the discriminant is $$D=(1)^2-4(-a)(1)=1+4a$$ and $a > -{\large{\frac{1}{4}}}$, it follows that $D > 0$, hence the equation has two distinct real roots.

Let $u,v$ be the roots.

By Vieta's formulas,

• $u+v=-{\large{\frac{1}{a}}}$.$\\[4pt]$
• $uv=-{\large{\frac{1}{a}}}$.

From $a < 0$, we get $-{\large{\frac{1}{a}}}> 0$, so $u+v > 0$, and $uv > 0$.

Since $u+v > 0$, at least one of $u,v$ must be positive,

Since $uv > 0$, and at least one of $u,v$ is positive, they must both be positive.

Without loss of generality, assume $0 < u \le v$.

From $-{\large{\frac{1}{4}}} < a < 0$, we get $-{\large{\frac{1}{a}}} > 4$, so $uv > 4$, hence we must have $v > 2$.

Noting that $u+v=uv$, we get $u={\large{\frac{v}{v-1}}}$, which is greater than $1$, since $v > v-1 > 0$.

Thus, both roots are greater than $1$, as was to be shown.

Answer 3

the assertion is equivalent to the statement that: $1 \lt b$ is a sufficient condition that for the line $y = b(1+x)$ to meet the parabola $y = x^2$ at a point to the right of the line $x =1$

In fact, since for all values of $b$ the line passes through $(-1,0)$, the geometric condition requires the line to have a slope greater that $\frac12$, so that it lies above the line joining $(-1,0)$ to $(1,1)$