Find the number or integral values for which the inequality $3-|x-a|>x^2$ is satisfied by at least one negative $x$

Question

The given equation is $$x^2-x+a-3<0$$ And $$x^2 + x -(a+3) <0$$

The right answer is 6, which can be obtained by replacing the inequity with an equality, making the discriminant positive and adjusting for a negative root to obtain the interval of $a$

$$x^2-x+a-3=0$$ And $$x^2+x-(a+3)=0$$

So $a>-\frac{13}{4}$ and $a< \frac{13}{4}$

Also $$x=\frac{1\pm \sqrt{13-4a}}{2}$$ So $$\sqrt{13-4a}>1$$ $$a<3$$

That gives $a\in (-\frac{13}{4}, 3)$ ie. 6 integral values

But this method doesn’t sit right with me, since values satisfying the inequality were supposed to be found, not the equality. Maybe it’s just a technicality, but I am not convinced. Can I get an explanation for this?

I know a similar question already exists in the site, but I found it unsatisfactory and it doesn’t answer my specific problem

Answer 1

If $a=3$ (orange line) then we have equality and $x=0$ which is a right boundary case. Hence we need $a<3$.

If $\displaystyle a=-\frac{13}{4}$ (violet line) then we have the left boundary where equality holds for $x<0$. So, we have $\displaystyle a>-\frac{13}{4}$ for inequality.

Answer 2

Draw a parabola $f(x)=3-x^2$ and $g(x)=|x-a|$.

Now, we see that $a_0<a<3$, where $y=x-a_0$ is a tangent to $f$.

Thus, $$(3-x^2)'=1,$$ which gives $x=-\frac{1}{2},$ $f\left(-\frac{1}{2}\right)=\frac{11}{4}$ and $$\frac{11}{4}=-\frac{1}{2}-a_0,$$ which gives $$a_0=-\frac{13}{4}$$ and we got the answer: $$\left(-\frac{13}{4},3\right).$$

Answer 3

You might also approach this geometrically. By manipulating the equation $ \ 3 - |x-a| \ = \ x^2 \ \ , $ you obtained the two quadratic polynomials $ \ x^2-x+a-3 \ $ and $ \ x^2+x-a-3 \ \ . $ "Completing-the-square" for each of these produces $$ x^2 \ - \ x \ + \ a \ - \ 3 \ \ = \ \ \left( x \ - \ \frac12 \right)^2 \ + \ \left(a \ - \ \frac{13}{4} \right) \ \ $$ and $$ x^2 \ + \ x \ - \ a \ - \ 3 \ \ = \ \ \left( x \ + \ \frac12 \right)^2 \ - \ \left(a \ + \ \frac{13}{4} \right) \ \ . $$

We seek the integer value(s) for $ \ a \ $ for which these parabolas lie "below" the $ \ x-$axis $ [ \ 0 \ > \ x^2 + |x-a| - 3 \ ] $ when $ \ x < 0 \ \ . $

Since the latter parabola has its vertex $ \ (h \ , \ k) \ $ at $ \ h \ = \ -\frac12 \ \ , $ there are points on the curve with $ \ x < 0 \ $ as long as the vertex is below the $ \ x-$axis, which requires that $ \ -k \ = \ a + \frac{13}{4} \ > \ 0 \ \Rightarrow \ a \ > \ -\frac{13}{4} \ \ . $

For the former parabola, it will be the case that the vertex $ \ (h' \ , \ k') \ $ is also below the $ \ x-$axis for $ \ k' \ = \ a - \frac{13}{4} \ < \ 0 \ \Rightarrow \ a \ < \ \frac{13}{4} \ \ . $ However, because the $ \ x-$coordinate is $ \ h' \ = \ +\frac12 \ \ , $ the part of this parabola that is below the $ \ x-$axis is only found "to the left" of the $ \ y-$axis when the portion of the parabola "to the left" of its symmetry axis passes through the third quadrant $ \ ( x < 0 \ , \ y < 0 ) \ \ . $ Here, we need
$$ x \ \ = \ \ \frac12 \ - \ \sqrt{\frac{13}{4} \ - \ a} \ \ < \ \ 0 \ \ \Rightarrow \ \ \frac14 \ \ < \ \ \frac{13}{4} \ - \ a \ \ \Rightarrow \ \ a \ \ < \ \ 3 \ \ . $$ For $ \ a \ = \ 3 \ \ , $ the "left arm" of the parabola passes through the origin, so for $ \ a \ < \ 3 \ \ , $ the parabola intersects the $ \ x-$axis at a point $ \ X \ < \ 0 \ \ , $ which guarantees that the parabola has points in the third quadrant.

We have thus established that both of the parabolas have points in the third quadrant only for $ \ \mathbf{ -\frac{13}{4} \ < \ a \ < \ 3} \ \ , $ so the permissible integer values for $ \ a \ $ are $ \ -3 \ , \ -2 \ , \ -1 \ , \ 0 \ , \ 1 \ , \ 2 \ \ . $

[In the graphs below, decreasing the value of $ \ a \ $ lowers the vertex for the family of parabolas on the right $ \ ( h' \ = \ +\frac12 ) \ $ , but raises the vertex for the family on the left $ \ ( h \ = \ -\frac12 ) \ \ . $ Thus both families only have points in the third quadrant for $ \ -\frac{13}{4} \ < \ a \ < \ 3 \ \ . ] $