Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$

Question

Problem: Find the all possible values of $a$, such that $$4x^2-2ax+a^2-5a+4>0$$ holds $\forall x\in (0,2)$.

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My work:

First, I rewrote the given inequality as follows:

$$ \begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned} $$

Then, we have

$$ \begin{aligned} 0<x<2\\ \implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned} $$

Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$.

This leads,

$$ \begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\ \implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned} $$

For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$.

Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$.

We have:

$$ \begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\ \implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned} $$

Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$.

Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$.

This case implies, $\left(2x-\frac a2\right)^2≥0$. This means, $f(x)≥\frac {3a^2}{4}-5a+4$

Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$.

Finally, we have to combine all the solution sets we get.

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I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?

Answer 1

Your attempt is mostly correct, apart from a few fairly minor issues regarding an open set being used, and there's some work you didn't need to do. First, as you pointed out, the inequality can be rewritten as

$$f(x) = \left(2x-\frac{a}{2}\right)^2 + \frac{3a^2}{4} - 5a + 4 \gt 0 \tag{1}\label{eq1A}$$

For any fixed value of $a$, then $\frac{3a^2}{4} - 5a + 4 = c$ is a constant. Also, with $g(x) = 2x-\frac{a}{2}$, then \eqref{eq1A} can be rewritten as

$$f(x) = g^2(x) + c \gt 0 \tag{2}\label{eq2A}$$

To have \eqref{eq2A} being true $\forall \, x \in (0,2)$, with

$$m = \min_{x\in[0,2]}g^2(x) \tag{3}\label{eq3A}$$

requires that $m + c \ge 0$ where $g(0) = 0 \lor g(2) = 0 \lor m \neq 0$ (i.e., points $1$ and $2$ below), and $m + c \gt 0$ otherwise (i.e., where $g(x) = 0$ for an $x \in (0,2)$, as described in point $3$ below). The first case is because the minimum value $m$ is not actually reached on the open set $(0,2)$, i.e., $g^2(x) \gt m \; \forall \; x \in (0,2)$, so $g^2(x) + c \gt 0$ means that $m + c \ge 0$.

To determine this minimum value $m$ in \eqref{eq3A} involves $3$ basic cases:

1. $g(x) \lt 0 \; \forall \; x \in (0,2)$, which is your case $1$, with $m$ coming from the upper limit value of $g(x)$, i.e., $m = (4 - \frac{a}{2})^2$.
2. $g(x) \gt 0 \; \forall \; x \in (0,2)$, which is your case $2$, with $m$ coming from the lower limit value of $g(x)$, i.e., $m = (-\frac{a}{2})^2$.
3. $g(x)$ having both negative and positive values (so $0$ as well), which is basically your case $3$ (except you should've used $\gt$ instead of $\ge$ to avoid overlapping with your previous cases, as well to ensure that $g(x) = 0$ for some $x \in (0,2)$), with $m = 0$.

Regarding your analysis of these cases, note that $g(x)$ is a strictly increasing function, with its upper limit value of $4 - \frac{a}{2}$ being $4$ more than its lower limit value of $-\frac{a}{2}$ (and, correspondingly, the lower limit is $4$ less than the upper limit). Thus, for example, with your case $1$, having $4 - \frac{a}{2} \le 0 \; \to \; 4 \le \frac{a}{2} \; \to \; a \ge 8$, i.e., a stronger condition than your $a \ge 0$, so this didn't need to be mentioned. Further, as I explained in my point $1$, you only needed to deal with handling $4 - \frac{a}{2}$.

Similarly, with your case $2$, having $a \le 0 \; \to \; -\frac{a}{2} \ge 0$ means that $4-\frac{a}{2} \ge 4$, which is a stronger condition than your $4-\frac{a}{2} \ge 0$.

Finally, as you wrote, the solution is obtained by combining (more precisely, taking the union of) the sets of values of $a$ resulting from each of your $3$ cases.

Answer 2

Suggestion

In case$-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$ simply means $a \ge 8$.

After re-writing the function in vertex form and plugging in some valid values of a (like 8, 10 etc) will give a graph on quadrant I for x in (0, 2). This means the function is greater than 0 for all x in that domain under that restriction.

Answer 3

I suggest a solution different from OP; albeit simpler.

First, transform the condition $\forall x\in (0,2)$. Equivalent to this condition, introduce $x = \frac{2}{y}$ with $y >1$, so only one boundary has to be inspected and the term remains quadratic in $y$ and $a$:

The inequality $4x^2-2ax+a^2-5a+4>0$ becomes $16-4ay+(a^2-5a+4)y^2>0$.

First note that only $a>0$ has to be inspected, since for $y > 1$ and $a \le 0$ we have $16-4ay+(a^2-5a+4)y^2 > 16 - 9/4 + (-5/2 + a)^2 > 0$ so the condition is always obeyed for $a \le 0$.

Rewrite the question as $$ 16 - 4 a y + (a^2 - 5 a + 4) y^2 = (a y - 2 - \frac{5y}{2})^2 - \frac14 (9 y^2 + 40 y - 48) > 0 \tag{1} $$ Since $9 y^2 + 40 y - 48 > 0$ always holds for $y >1$, we can discuss $g = (a y - 2 - \frac{5y}{2})^2$. For $a \to 0$, $g \to \frac14 (5y +4)^2 = \frac14 (16y^2 + 64) + \frac14 (9 y^2 + 40 y - 48)$ so the question's condition is always obeyed. Likewise, for very large $a$ it is clear that (1) will also be obeyed.

Conversely, for $a=4$ we have $16 - 4 a y + (a^2 - 5 a + 4) y^2 = 16(1-y)$ which will be negative for all $y >1$.

Hence the allowed regions for $a$ will be $a_1 > a $ and $a > a_2$ with $a_2 \ge 4 \ge a_1 > 0$.

What remains is to derive $a_1$ and $a_2$ from the roots of (1) and their dependence on $y$. This is standard curve discussion and we give the results.

We have for the smaller root:

$$\frac{1}{y}\Big(2 + \frac{5y}{2} - \frac12 \sqrt{9 y^2 + 40 y - 48}\Big) \ge \frac{8}{5 + \sqrt{13}} = a_1 \simeq 0.93 $$ which holds since the minimum occurs at $ y = 5 + \sqrt{13} > 1$

We have for the larger root:
$$\frac{1}{y}\Big(2 + \frac{5y}{2} + \frac12 \sqrt{9 y^2 + 40 y - 48}\Big) \le \frac23 (5 + \sqrt{13}) = a_2 \simeq 5.74 $$ which holds since the maximum occurs at $ y = 5 - \sqrt{13} > 1$. $\qquad \Box$

Answer 4

Yes, (though strangely written) your method is correct. It can be reformulated noticing more explicitely that $f(x):=4x^2-2ax+a^2-5a+4$ is decreasing for $x\le a/4$ and increasing for $x\ge a/4$, since $f'(x)=8x-2a.$ We thus recover your 3 cases (and the same numerical results as Andreas, but more directly):

• Case 1: $a/4\ge2$. Then, $f>0$ on $(0,2)$ iff $f(2)\ge0$, i.e. $a^2-9a+20\ge0$, i.e. $a\le2$ or $a\ge4$. This or-condition is always satisfied since $a\ge8$ by hypothesis.
• Case 2: $0<a/4<2$. Then, $f>0$ on $(0,2)$ iff $f(a/4)>0$, i.e. $\frac34a^2-5a+4>0$, i.e. $a<a_1:=\frac{10-2\sqrt{13}}3\approx0.93$ or $a>a_2:=\frac{10+2\sqrt{13}}3\approx5.74.$ This or-condition is satisfied iff $0<a<a_1$ or $a_2<a<8.$
• Case 3: $a/4\le0$. Then, $f>0$ on $(0,2)$ iff $f(0)\ge0$, i.e. $a^2-5a+4\ge0,$ i.e. $a\le1$ or $a\ge4.$ This or-condition is always satisfied since $a\le0$ by hypothesis.

Putting your three cases together gives the whole set of solutions: $a<a_1$ or $a>a_2.$

Answer 5

Might be calculative but this works

Find the roots using quadratic formula then if the condition satisfies then there are 3 cases

$$\text{Case1}:-$$first the smaller of the 2 roots (the one with the negative sign) is greater than 2 this gives the quadratic $a^2-13a-68\ge0$ and $3a^2-20a+16\le0$

$$\text{Case2}:-$$ Again the greater root is less than or equal to 0 $$a^2-5a+4\le0$$ and $3a^2-20a+16\le0$

$$\text{Case3}:-$$ Discriminant is negative ,ie $3a^2-20a+16\ge0$

Solving case by case we get all solutions

Answer 6

Let $g(x)=Ax^2+Bx+C$,if $A>0$, then $g(x)\ge 0$

Case 1: $\forall x \in \Re$, if $B^2\le 4AC$.

Case 2: For $x \in [p,q]$, if the position of the min (bottom) of $g(x)$, $x_0=-\frac{B}{2A}$ should be s.t $x_0 \le p$ or $x_0\ge q$.

Here $$f(x)=4x^2-2ax+a^2-5a+4>0$$ in $x\in (0,2)$ can be met in both the cases.

Case 1: $4a^2 \le 16(a^2-5a+4) \implies 3a^2-20a+16 \ge 0 \implies (x-x_1)(x-x_2)\ge 0$ where $x_1=\frac{2}{3}(5-\sqrt{13}) \approx 0.92), x_2=\frac{2}{3}(5+\sqrt{13}) \approx 5.73$. So $$a\le \frac{2}{3}(5-\sqrt{13})~ \text{or}~ a\ge \frac{2}{3}(5+\sqrt{13}).....(1)$$

Case 2: $x_0=a/4$, $x_0<0 \implies a\le 0$ and $x_0\ge 2 \implies a\ge 8.$ So in this case $$a\le 0 ~\text{or}~ a\ge 8........(2)$$.

Interestingly, the final domain of $a$ needs to be the union of (1) and (2).

Lastly, $f(x)> 0, ~\text{ONLY}~\forall x \in(0,2)$, when $a\in (-\infty,0)\cup (8,\infty).$

Otherwise, $f(x)>0, \forall x \in \Re $, when $a \in (-\infty,\frac{2}{3}(5-\sqrt{13}) \cup (\frac{2}{3}(5+\sqrt{13}), \infty)$

Answer 7

I found the solution sets in your cases:

Case-1) $((-\infty,4)\cup(5,\infty))\cap[8,\infty)=[8,\infty)$

Case-2) $((-\infty,1)\cup(4,\infty))\cap[-\infty,0)=(-\infty,0]$

Case-3) $((-\infty,\frac{10-2\sqrt{13}}{3})\cup(\frac{10+2\sqrt{13}}{3},\infty))\cap[0,8]=[0,\frac{10-2\sqrt{13}}{3})\cup(\frac{10+2\sqrt{13}}{3},8]$

The union of sets we found in all cases is $S=(-\infty,\frac{10-2\sqrt{13}}{3})\cup(\frac{10+2\sqrt{13}}{3},\infty)$. S is the solution of the problem.

My solution:

$f'(x)=0$ gives the minumum of the quadratic function $f(x)=4x^2-2ax+a^2-5a+4$ at $x=\frac{a}{4}$ and the solution of the inequality $f(\frac{a}{4})=\frac{3}{4}a^2-5a+4>0$ is the set $S$ above.

If $a$ were in $[\frac{10-2\sqrt{13}}{3},\frac{10+2\sqrt{13}}{3}]$ then $x=\frac{a}{4}$ would be in $(0,2)$ and $f(\frac{a}{4})$ would be non-positive. So, we can conclude that $S$ is the solution set of the problem.

Answer 8

I've tried to answer in a way which skips over some of the tedious calculation so as to make it easier to understand what this question means. $$f(a)=4x^2-2ax+a^2-5a+4>0$$ Consider, for the moment, $x$ to be a constant. Then we have a quadratic expression in $a$ with roots at (details omitted): $$a_1=- \sqrt{-3x^2+5x+\tfrac94}+x+\tfrac52$$ $$a_2= \sqrt{-3x^2+5x+\tfrac94}+x+\tfrac52$$ In the range of interest of $x$, it turns out that the expression under the square root is always positive, so we always have two real roots. We can plot the roots as follows:

We want $f(a)>0$ for all $x \in (0,2)$. It is easy to see that this requires that $a$ is less than the minimum value of $a_1$ or $a$ is greater than the maximum value of $a_2$. We therefore differentiate $a_1$ and $a_2$ with respect to $x$ (details omitted) to find the turning points which are, for $a_1$: $$x=\tfrac16(5-\sqrt{13}), \;\; a=-\tfrac23(\sqrt{13}-5)$$ and for $a_2$: $$x=\tfrac16(5+\sqrt{13}), \;\; a=\tfrac23(\sqrt{13}+5)$$

And our full solution is therefore: $$a<-\tfrac23(\sqrt{13}-5) \text{ and } a>\tfrac23(\sqrt{13}+5) $$

Answer 9

I've revised your work as follows.

The first step is fine

$$ 0<x<2 \implies -\frac a2<2x-\frac a2<4-\frac a2$$

As a minor issue we should use $\iff$ instead of $\implies$.

• Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$

The condition corresponds to $a\ge8$.

The inequality you reach $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0 \iff (a-4)(a-5)>0$ holds.

• Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$

The condition corresponds to $a\le0$.

The inequality you reach $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0 \iff (a-1)(a-4)>0$ holds.

• Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$

The condition corresponds to $0\le a\le8$. So we could also consider $0<a<8$ but it is not necessary.

The inequality you reach $\frac {3a^2}{4}-5a+4>0 \iff a<\frac{10}3-\frac{2\sqrt{13}}3\approx0.93 \land a>\frac{10}3+\frac{2\sqrt{13}}3\approx5.74 $ holds.

The method looks fine to me since you have covered all the cases!

Answer 10

I want to bring the most straightforward way to calculate this problem systematically. Also, you can check the attached graph for better representation. I write the quadratic equation $f_x(a)=4x^2-2ax+a^2-5a+4$ w.r.t. $a$ as an unknown parameter. We have:

$f_x(a)=a^2-(2x+5)a+(4x^2+4)>0$.

So, $a$ must be greater than the bigger root of $f_x(a)=0$ or less than the smaller root, which are:

$f_x(a)=0 \rightarrow a=x+2.5\pm\sqrt{-3x^2+5x+2.25}=g^\pm(x)$ for $x\in(0,2)$.

So, we have $a>g^+(x)$ or $a<g^-(x)$ for $\forall x \in (0,2)$, i.e.:

$a>\max_{x\in(0,2)}g^+(x)$ or $a<\min_{x\in(0,2)}g^-(x)$. We have:

$dg^+(x)/dx=0\rightarrow 1+\frac{-6x+5}{2\sqrt{-3x^2+5x+2.25}}=0\rightarrow -12x^2+20x+9=36x^2-60x+25$ $\rightarrow 3x^{2}-5x+1=0\rightarrow x=\frac{5+\sqrt{13}}{6}\rightarrow g^+(\frac{5+\sqrt{13}}{6})=\frac{2}{3}(5+\sqrt{13})$

Similarly, we have:

$dg^-(x)/dx=0\rightarrow 1-\frac{-6x+5}{2\sqrt{-3x^2+5x+2.25}}=0 \rightarrow x=\frac{5-\sqrt{13}}{6} \rightarrow g^-(\frac{8}{5+\sqrt{13}})=\frac{8}{5+\sqrt{13}}$

So, $a$ range having $f_x(a)=4x^2-2ax+a^2-5a+4>0$ are:

$a>\frac{2}{3}(5+\sqrt{13}) \approx5.7370$ or $a<\frac{8}{5+\sqrt{13}} \approx0.9296$

Answer 11

I think your current description is not satisfactory.

Take Case 3 for example. You said,

This means, $f(x) \ge \frac {3a^2}{4}-5a+4$. Thus, for $f(x) > 0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a \ge 0\wedge 4-\frac a2\ge 0$.

This description does not emphasize the sufficient and necessary condition to include all possible values of $a$. It looks (at least from the description) like a sufficient condition (more precisely, $\frac {3a^2}{4}-5a+4>0$ is a sufficient condition of $f(x) > 0$).

One way to overcome this is given below (Cases 1-2 are omitted here, and given later):

Let $$f(x) := 4x^2 - 2ax + a^2 - 5a + 4 = \left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4. \tag{1}$$

We have $$0 < x < 2 \implies -\frac a2<2x-\frac a2<4-\frac a2. \tag{2}$$

We split into three cases.

Case 3: $-\frac{a}2 < 0$ and $4 - \frac{a}2 > 0$, i.e. $0 < a < 8$

Note that $\frac{a}{4} \in (0, 2)$ and $f(\frac{a}{4}) = \frac {3a^2}{4}-5a+4$. We have, \begin{align*} &f(x) > 0, \quad \forall 0 < x < 2\\ \iff \quad & \frac{3a^2}{4}-5a+4 > 0\\ \iff \quad & a > \frac{10}{3} + \frac23\sqrt{13}, \quad \mathrm{or} \quad a < \frac{10}{3} - \frac23\sqrt{13}. \end{align*} Thus, the set of qualified $a$ is $0 < a < \frac{10}{3} - \frac23\sqrt{13}$ or $\frac{10}{3} + \frac23\sqrt{13} < a < 8$.

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Someone may ask why I took Case 3 for example, rather than Case 1 or 2? The reason is that the conditions in Case 1 and Case 2 (respectively, $a \ge 8$ and $a \le 0$), actually both imply $f(x) > 0$. By contrast, the condition of Case 3 (i.e. $0 < a < 8$) does not imply $f(x) > 0$.

The details are given below.

Case 1: $4 - \frac{a}{2} \le 0$, i.e. $a \ge 8$

From (1) and (2), we have \begin{align*} &0 < x < 2 \\ \implies\quad & \left(2x-\frac a2\right)^2 > \left(4 - \frac{a}2\right)^2\\ \implies \quad & f(x) > \left(4 - \frac{a}2\right)^2 + \frac {3a^2}{4}-5a+4 = (a - 4)(a - 5) > 0. \end{align*} Thus, the set of qualified $a$ is $a \ge 8$.

Case 2: $-\frac{a}2 \ge 0$, i.e. $a \le 0$

From (1) and (2), we have \begin{align*} &0 < x < 2 \\ \implies\quad & \left(2x-\frac a2\right)^2 > \left( - \frac{a}2\right)^2\\ \implies \quad & f(x) > \left(- \frac{a}2\right)^2 + \frac {3a^2}{4}-5a+4 = (a - 1)(a - 4) > 0. \end{align*} Thus, the set of qualified $a$ is $a \le 0$.

Summarily, all possible values of qualified $a$ is $a < \frac{10}{3} - \frac23\sqrt{13}$ or $a > \frac{10}{3} + \frac23\sqrt{13}$.