Equations and Inequalities with a Parameter: Problems with Solutions

Solution:
The equation is quadratic in [tex]x[/tex] (its leading coefficient is [tex]1[/tex]). For its discriminant [tex]D[/tex] we get
[tex]\frac{D}{4}=(2m-3)^2-\left(5m^2-16m+20\right)=4m^2-12m+9-5m^2+16m-20=-m^2+4m-11[/tex]
Completing the square:
[tex]\frac{D}{4}=-\left(m^2-4m+4\right)-7=-(m-2)^2-7[/tex]
Since [tex](m-2)^2\ge 0[/tex] for every real [tex]m[/tex], we get [tex]\frac{D}{4}\le -7\lt 0[/tex]. Hence for every value of [tex]m[/tex] the discriminant is negative and the equation has no real roots.

Solution:
Assume that such an [tex]m[/tex] exists. Then [tex]x_1[/tex] and [tex]x_2[/tex] are real roots of the equation, and by Vieta's formulas
[tex]x_1x_2=1[/tex]
On the other hand, [tex]x_1\gt 2\gt 0[/tex] and [tex]x_2\gt 5\gt 0[/tex], so
[tex]x_1x_2\gt 2\times 5=10[/tex]
This contradicts [tex]x_1x_2=1[/tex]. Hence no such value of [tex]m[/tex] exists.

Solution:
Since [tex]m^2+2\gt 0[/tex], the equation is quadratic for every [tex]m[/tex]. The sum of its coefficients is zero:
[tex](m^2+2)-(m^2+6)+4=0[/tex]
so [tex]x_1=1[/tex] is a root for every [tex]m[/tex]. By Vieta's formulas the other root is
[tex]x_2=\frac{x_1x_2}{x_1}=\frac{4}{m^2+2}[/tex]
Because [tex]m^2+2\ge 2[/tex], we get
[tex]0\lt x_2\le \frac{4}{2}=2[/tex]
with equality only for [tex]m=0[/tex]: then the equation is [tex]2x^2-6x+4=0[/tex] with roots [tex]1[/tex] and [tex]2[/tex]. So no root exceeds [tex]2[/tex], and the value [tex]2[/tex] is attained. The largest possible root is [tex]2[/tex].

Solution:
Case 1: the leading coefficient is zero, [tex]a^2-4=0[/tex]. For [tex]a=2[/tex] the inequality becomes [tex]-3\lt 0[/tex], which is true for every [tex]x[/tex], so [tex]a=2[/tex] works. For [tex]a=-2[/tex] it becomes [tex]8x-3\lt 0[/tex], i.e. [tex]x\lt \frac{3}{8}[/tex], which does not hold for every [tex]x[/tex]; so [tex]a=-2[/tex] is rejected.
Case 2: [tex]a^2-4\ne 0[/tex]. A quadratic trinomial is negative for every [tex]x[/tex] exactly when its leading coefficient and its discriminant are both negative:
[tex]a^2-4\lt 0\iff a\in\left(-2, 2\right)[/tex]
[tex]\frac{D}{4}=(a-2)^2+3(a^2-4)=(a-2)(a-2+3a+6)=4(a-2)(a+1)\lt 0\iff a\in\left(-1, 2\right)[/tex]
Together: [tex]a\in\left(-1, 2\right)[/tex]. (For [tex]a=-1[/tex] the trinomial is [tex]-3(x-1)^2[/tex], which equals [tex]0[/tex] at [tex]x=1[/tex], so [tex]a=-1[/tex] is rightly excluded.) Combining both cases:
[tex]a\in\left(-1, 2\right][/tex]

Solution:
If [tex]a=2[/tex], the inequality becomes [tex]8x+6\gt 0[/tex], which fails, for example, at [tex]x=-1[/tex].
If [tex]a\ne 2[/tex], a quadratic trinomial is positive for every real [tex]x[/tex] exactly when its leading coefficient is positive and its discriminant is negative:
[tex]a-2\gt 0,\quad \frac{D}{4}=16-(a-2)(a+4)\lt 0[/tex]
The second condition is
[tex]a^2+2a-24\gt 0\iff (a+6)(a-4)\gt 0\iff a\lt -6\text{ or }a\gt 4[/tex]
Together with [tex]a\gt 2[/tex] this gives [tex]a\gt 4[/tex]. The value [tex]a=4[/tex] is not allowed: then the trinomial is [tex]2x^2+8x+8=2(x+2)^2[/tex], which equals [tex]0[/tex] at [tex]x=-2[/tex]. Hence the smallest integer is [tex]a=5[/tex]. Check:
[tex]a=5:\quad 3x^2+8x+9,\quad 3\gt 0,\quad \frac{D}{4}=16-27\lt 0[/tex]

Solution:
The denominator is positive for every [tex]x[/tex]:
[tex]x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\gt 0[/tex]
so we may multiply by it without changing the direction of the inequality:
[tex]x^2+ax\lt 3x^2+3x+3\iff 2x^2+(3-a)x+3\gt 0[/tex]
A quadratic trinomial with positive leading coefficient is positive for every real [tex]x[/tex] exactly when its discriminant is negative:
[tex](3-a)^2-24\lt 0\iff |a-3|\lt 2\sqrt{6}\iff 3-2\sqrt{6}\lt a\lt 3+2\sqrt{6}[/tex]
Since [tex]16\lt 24\lt 25[/tex], we have [tex]4\lt 2\sqrt{6}\lt 5[/tex], so
[tex]-2\lt 3-2\sqrt{6}\lt -1,\quad 7\lt 3+2\sqrt{6}\lt 8[/tex]
The integers in this interval are [tex]-1, 0, 1, \ldots, 7[/tex], that is [tex]9[/tex] values.

Solution:
Let [tex]f(x)=\left(a^2+1\right)x^2+(3a-4)x+a^2+5[/tex]. The leading coefficient [tex]a^2+1[/tex] is positive for every [tex]a[/tex], so the parabola opens upward. For such a quadratic, [tex]f(3)\lt 0[/tex] holds exactly when [tex]f[/tex] has two real roots and [tex]x_1\lt 3\lt x_2[/tex] (so the discriminant need not be studied).
If moreover [tex]f(7)\gt 0[/tex], then [tex]7[/tex] lies outside [tex]\left[x_1, x_2\right][/tex], and since [tex]7\gt 3\gt x_1[/tex], this means [tex]x_2\lt 7[/tex]; conversely, [tex]x_1\lt 3\lt x_2\lt 7[/tex] implies [tex]f(7)\gt 0[/tex]. So the condition is equivalent to the system
[tex]\begin{cases} f(3)\lt 0\\ f(7)\gt 0\end{cases}[/tex]
We compute
[tex]f(3)=9\left(a^2+1\right)+3(3a-4)+a^2+5=10a^2+9a+2=(2a+1)(5a+2)[/tex]
[tex]f(3)\lt 0\iff -\frac{1}{2}\lt a\lt -\frac{2}{5}[/tex]
[tex]f(7)=49\left(a^2+1\right)+7(3a-4)+a^2+5=50a^2+21a+26[/tex]
As a quadratic in [tex]a[/tex], the last expression has positive leading coefficient and discriminant [tex]21^2-4\times 50\times 26=441-5200\lt 0[/tex], so [tex]f(7)\gt 0[/tex] for every [tex]a[/tex]. Therefore
[tex]a\in\left(-\frac{1}{2}, -\frac{2}{5}\right)[/tex]

Solution:
The domain is [tex]x\ge 0[/tex], [tex]x\ge a[/tex]. Write the equation as
[tex]\sqrt{x-a}=1-\sqrt{x}[/tex]
The left side is non-negative, so every solution satisfies [tex]1-\sqrt{x}\ge 0[/tex], i.e. [tex]\sqrt{x}\le 1[/tex]. Under this condition squaring is an equivalent step:
[tex]x-a=1-2\sqrt{x}+x\iff \sqrt{x}=\frac{1+a}{2}[/tex]
This requires [tex]\frac{1+a}{2}\ge 0[/tex], i.e. [tex]a\ge -1[/tex], and the condition [tex]\sqrt{x}\le 1[/tex] gives [tex]\frac{1+a}{2}\le 1[/tex], i.e. [tex]a\le 1[/tex]. For such [tex]a[/tex]
[tex]x=\frac{(a+1)^2}{4}\ge 0,\quad x-a=\frac{(1-a)^2}{4}\ge 0[/tex]
so [tex]x[/tex] lies in the domain. Hence for [tex]a\in\left[-1, 1\right][/tex] the equation has the unique root [tex]x=\frac{(a+1)^2}{4}[/tex], and for every other [tex]a[/tex] it has no roots. Check of the endpoints: for [tex]a=-1[/tex], [tex]x=0[/tex] and [tex]\sqrt{1}+\sqrt{0}=1[/tex]; for [tex]a=1[/tex], [tex]x=1[/tex] and [tex]\sqrt{0}+\sqrt{1}=1[/tex].

Solution:
Domain: [tex]x^2-3x-4=(x+1)(x-4)\ne 0[/tex], so [tex]x\ne -1[/tex] and [tex]x\ne 4[/tex]. The numerator factors as
[tex]x^2-(3a-1)x+2a^2-2=\left(x-(a+1)\right)\left(x-(2a-2)\right)[/tex]
because [tex](a+1)+(2a-2)=3a-1[/tex] and [tex](a+1)(2a-2)=2a^2-2[/tex]. So the only candidates are [tex]x_1=a+1[/tex] and [tex]x_2=2a-2[/tex]. They coincide only for [tex]a=3[/tex], but then [tex]x_1=x_2=4[/tex] is outside the domain and the equation has no roots. For [tex]a\ne 3[/tex] the two candidates are different, so the equation has exactly one root precisely when one of them equals [tex]-1[/tex] or [tex]4[/tex] and the other does not:
[tex]a+1=-1\Rightarrow a=-2[/tex]; then [tex]x_2=-6[/tex] is admissible;
[tex]2a-2=-1\Rightarrow a=\frac{1}{2}[/tex]; then [tex]x_1=\frac{3}{2}[/tex] is admissible;
[tex]a+1=4[/tex] or [tex]2a-2=4[/tex] gives [tex]a=3[/tex], which is already rejected.
Hence the equation has exactly one root for [tex]a=-2[/tex] (the root [tex]x=-6[/tex]) and for [tex]a=\frac{1}{2}[/tex] (the root [tex]x=\frac{3}{2}[/tex]):
[tex]a\in\left\{-2, \frac{1}{2}\right\}[/tex]

Solution:
The domain is [tex]x\ge 3a[/tex]. For [tex]x=3a[/tex] the square root equals [tex]0[/tex], so the left-hand side is [tex]0[/tex] and [tex]x=3a[/tex] is a solution for every [tex]a[/tex].
For [tex]x\gt 3a[/tex] the square root is positive, so the inequality is equivalent to
[tex]x+a-1\le 0\iff x\le 1-a[/tex]
Hence the solution set is
[tex]\left\{3a\right\}\cup\left(3a , 1-a\right][/tex]
There is exactly one solution if and only if the interval [tex]\left(3a , 1-a\right][/tex] is empty:
[tex]1-a\le 3a\iff a\ge\frac{1}{4}[/tex]
It is not enough to require [tex]3a=1-a[/tex]: for example, for [tex]a=1[/tex] the only solution is [tex]x=3[/tex], since for [tex]x\gt 3[/tex] both factors are positive. Hence [tex]a\in\left[\frac{1}{4} , +\infty\right)[/tex].

Solution:
Let [tex]x_0[/tex] be a common root. Subtracting the second equation from the first gives
[tex](a-1)x_0+8-a=0[/tex]
If [tex]a=1[/tex], this reads [tex]7=0[/tex], which is impossible. So [tex]a\ne 1[/tex] and [tex]x_0=\frac{a-8}{a-1}[/tex]. Conversely, a number that satisfies the second equation and this difference satisfies the first equation as well, so it is enough to substitute [tex]x_0[/tex] into [tex]x_0^2+x_0+a=0[/tex] and multiply by [tex](a-1)^2[/tex]:
[tex](a-8)^2+(a-8)(a-1)+a(a-1)^2=0\iff a^3-24a+72=0[/tex]
Since [tex]a=-6[/tex] is a root,
[tex]a^3-24a+72=(a+6)(a^2-6a+12),\qquad a^2-6a+12=(a-3)^2+3\gt 0[/tex]
Hence [tex]a=-6[/tex] is the only possibility, with [tex]x_0=\frac{-14}{-7}=2[/tex]. Check: for [tex]a=-6[/tex] the equation [tex]x^2-6x+8=0[/tex] has roots [tex]2[/tex] and [tex]4[/tex], and [tex]x^2+x-6=0[/tex] has roots [tex]2[/tex] and [tex]-3[/tex]; the common root is [tex]2[/tex]. The answer is [tex]a=-6[/tex].