# Given $x^2+x-a<3$, solve for $a$ so that there is at least one solution $<0$

Given $x^2+x-a<3$, solve for $a$ so that there is at least one solution $<0$

I have the idea how to solve this problem but my solution seems weird and unfinished.

Firstly, I move $3$ to the left hand side.

The graphic of the solution is open upwards and has an axis of symmetry at $-1/2$. This means that in order for the inequality to have any solution, its discriminant must be $>0$. So $4a+8>0$, which means $a>-2$. But that seems to be it. Due to the axis and the graph, there must be at least one negative solution. Am I missing something?

The inequality should be satisfied for at least a negative value of $x$.

We know that, when the discriminant is positive, the inequality is satisfied for $$x_1<x<x_2$$ where $x_1$ and $x_2$ are the roots. Hence the problem is to find when the smaller root is negative: this happens when $-a-3<0$, that is, $a>-3$.

Note that considering the discriminant is irrelevant: when $-a-3<0$ the discriminant is certainly positive.

Hint:

$1+4(a+3) \gt 1$

$a \gt -3$

This ensures atleast one of the roots is negative as the roots are

$x_1 = \dfrac{-1+\sqrt{1+4(a+3)}}{2}$

$x_2 = \dfrac{-1-\sqrt{1+4(a+3)}}{2}$

The inequality can be written as $$x^2+x=x(x+1)<3+a$$ On differentiating $y=x^2+x$ and equating to $0$, we find that $y$ has a minimum of $-\frac14$. This means that to satisfy the criterion that at least one solution of the inequality is less than $0$, we must have $$3+a>-\frac14$$ so that $a>-\frac{13}4$.

You took a very efficient and simple track in analyzing the problem geometrically.
So as you say, the parabola $x(x+1)$ is vertical, open upwards, and with symmetry axis at $x=-1/2$.
Then just add that the ordinate of the parabola on that axis, i.e. the vertex, has $y=-1/4$.
Thus until $y=a+3 > -1/4$ you will have that part of the parabola will lay below that, for an interval of $x$ centered around $x=-1/2$ and thus always comprising negative values.

The inequality is equivalent to $$x^2+x-(a-3)<0$$. This is satisfied if and only if

• The polynomial has two real roots, i.e. $$\Delta=1+4(a+3)>0\iff a>-\dfrac{13}4$$.
• One of these roots is negative. However, note that when these roots have the same sign (in the extended sense), i.e. when $$a+3 <0$$, this sign is negative, since the sum of the roots is $$-1$$. When they don't, well… one of them is negative.

As a conclusion, if the polynomial has two roots, one is negative, so the inequation has solutions. The condition is thus $$a>-\dfrac{13}4.$$

• the determinanat should greater than or equal to 0 for real roots. But the question is asking if one of the roots could be less than 0. So it is already half a negative. thus "a" just needs be little greater than -3 for atleast one of the roots to be just less than 0. @egreg correctly analysed it. Nov 18, 2017 at 12:35
• @SatishRamanathan: What I proved (unless I made a stupid mistake) is that if $a>-3$, one root is negative, and if $-13/4<a<-3$ both are negative – in the latter case, all solutions to the inequation are negative. Nov 18, 2017 at 12:53