Consider the inequality $9-x^2>|x-a|$ where $a$ is real, then find set of values for $a$ such that at least one negative solution exists

Question

For $x>a$

$$9-x^2 >x-a$$ $$x^2 + x -(9+a)<0$$

So $$1+4(9+a)<0$$ $$a<\frac{-37}{4}$$

Similarly for $x<a$ $$x^2-x+a-9<0$$ So $a>\frac{37}{4}$

Now in both cases, we need at least one $x<0$

In case one, $a<0$ and $x>a$ so $x$ can be both positive and negative, so I don’t know how to deal with that

Similar problem in case 2.

Another inequality I found was $a>-9$, obtained by solving the quadratic for first case, giving $x\in (\frac{1-\sqrt {37+4a}}{2}, \frac{1+\sqrt {37+4a}}{2})$, and then setting the lower limit to $<0$

I can also find another inequality from the second case, but there are way too many inequalities to choose from and the condition that $x>a$ and $x<a$ is throwing me off.

Also is a downward parabola even possible for these cases? The equation demands it to be downward if $f(x)<0$ for all $x$, but the coefficient of $x^2$ is positive here.

Answer 1

ok so as you have said:

Case 1: $x>a$

$$9-x^2 >x-a$$ $$x^2 + x -(9+a)<0$$ Now, $x^2 + x -(9+a)$ is an upward facing parabola and we need to find the region of where it is below the x axis; because after all, that's where it is less than zero. For an upward facing parabola to have a region below the x-axis it must have two real roots, and hence the discriminant must be greater than zero. $\implies b^2-4ac> 0 \implies 1+4(9+a)>0 \implies a>\dfrac{-37}{4}$. So when $a>\dfrac{-37}{4}$ that upward facing parabola has two real roots and hence there exists a region below the x-axis. But this is not the only thing we need. We need this region below the x-axis to have at least one negative value of x. Notice the x coordinate of the vertex of this parabola = $ \dfrac{-b}{2a}= \dfrac{-1}{2}$, which is less than zero. That means at least one of the two roots is negative, and at least one negative value exists in the solution set of: $x^2 + x -(9+a)<0$ iff $a>\dfrac{-37}{4}$.

But we also said $x$ is strictly greater than $a$ initially before removing the absolute value. This means for at least one negative solution, $a$ must be strictly less than zero, because if $a$ exceeds zero, then $x$ must also and then there will be no negative solutions for x... So for $x>a, a\in (\dfrac{-37}{4},0).$

Case 2: $x<a$

$$9-x^2 >-(x-a)$$ $$x^2 - x +(a-9)<0$$

Again discriminant $D>0 \implies a<\dfrac{37}{4}$, but here vertex of parabola is positive = 0.5. This means at least one root is positive. To make sure that the other root is negative we need the parabola to be below the $x$ axis when $x=0.$ So $0^2 - 0 +(a-9)<0 \implies a<9.$ We also need to make sure this negative root is less than $a$ because our initial constraint was $x<a.$ So when does the line x =a, meet the parabola at the negative root? $a^2 - a +(a-9)=0 \implies a = -3 \implies a\in [-3,9)$

So $a \in (\dfrac{-37}{4},0) \cup [-3,9] \implies a \in (\dfrac{-37}{4},9)$

Answer 2

The graphical approach is illuminating - a rough hand drawn graph is sufficient.

As $a$ varies from $A$ to $B$, $y=|x-a|$ intersects the negative branch of $y=9-x^2$. At $A$, $y=|x-a_1|$ is tangent to the parabola at $D$. At $B$, $y=|x-a_2|$ passes through the vertex of parabola, $C=(0,9)$.

Since $\triangle COB$ is right isosceles, $B=(9,0)$.

Slope at $D$ is $-2x=1 \Rightarrow D=(-1/2,35/4)$. Since $\triangle AED$ is right isosceles, $A=(-(35/4+1/2),0)=(-37/4,0)$.

Desired range is $a \in [-37/4,9)$.

Answer 3

No, you started good, then went off the rails, by trying to take the shortcut that the discriminant must be negative.

$\underline{\text{Case 1}:~~ (x-a) \geq 0}$

$$9 - x^2 > x - a. \tag1$$

This implies that

$\displaystyle x^2 + x - (a + 9) < 0 \implies$

$$\left(x + \frac{1}{2}\right)^2 - \left(a + 9 + \frac{1}{4}\right) < 0. \tag2 $$

Examining inequality (2) at the critical value $\left(x = -\frac{1}{2}\right)$:

• $\left(a + 9 + \frac{1}{4}\right)$ must be $> 0$, which implies that the lower bound for $a$ in Case 1 is $a > \frac{-37}{4}.$

• As $a$ approaches $0$ from below, an $x$ will always be available such that $a \leq x < 0$. In this circumstance, it is clear that the LHS of inequality (1) above will be large enough to satisfy the inequality.
  
  Therefore, the upper bound for $a$ in Case 1 is $(a < 0)$.

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$\underline{\text{Case 2}:~~ (x-a) < 0}$

$$9 - x^2 > a - x. \tag3$$

This implies that

$\displaystyle x^2 - x + (a - 9) < 0 \implies$

$$\left[\left(x - \frac{1}{2}\right)^2 - \left(\frac{1}{4}\right)\right] + (a - 9) < 0. \tag4 $$

With $x$ constrained by $x < 0$,

$\displaystyle \left[\left(x - \frac{1}{2}\right)^2 - \left(\frac{1}{4}\right)\right]$ will approach $0$ from above, as $x$ approaches $0$ from below.

Therefore, in Case 2, an upper bound for $a$ is $a < 9.$

So the final question is : what is the lower bound for $a$ in Case 2.

Consider the inequality :

$$x^2 - x + (x - 9) < 0\tag5$$

which is what happens when $x=a$.

Clearly $x$ can never be $\leq -3$, or inequality (5) will be violated. To prove that $(a > -3)$ is a lower bound for $a$ in Case 2, assume that

$\exists ~\delta > 0$ such that $a = (-3 + \delta).$

Then inequality (4) above may be manipulated to

$$\left[\left(x - \frac{1}{2}\right)^2 - \left(\frac{1}{4}\right)\right] + (a - 9) < 0 \implies $$

$$\left(x - \frac{1}{2}\right)^2 < \left(\frac{1}{4}\right) + (9 - [-3 + \delta]) = (12) + \left(\frac{1}{4}\right) - \delta.\tag6$$

Regardless of how small the fixed size of $\delta$ is, as long as $\delta > 0$, you will be able to find a value of $x > -3$, so that inequality (6) above is satisfied.

Therefore, in Case 2, $a$ is constrained by $-3 < a < 9.$

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In summary:

Case 1 bounds on $a$ : $~ -\frac{37}{4} < a < 0$.

Case 2 bounds on $a$ : $~ -3 < a < 9$.

Therefore, the overall constraints on $a$ are $-\frac{37}{4} < a < 9.$