Q.Find two numbers whose sum is $24$ and whose product is as large as possible.

1. Solve the inequality to find $[\alpha,\beta]$.

   $x^2-10x+24=(x-4)(x-6)\le0$ holds for $x\in[4,6]$ (the parabola opens upward, so it's $\le0$ between its roots). This is the interval referred to in the problem, so $\alpha=4$ and $\beta=6$.

2. Find $b$ from $\alpha=4$.

   $f(x)=x^2+bx+5$ opens upward, with vertex minimum

Setting this equal to $\alpha=4$:

3. Find $a$ from $\beta=6$. $g(x)=-x^2+ax+5$ opens downward, with vertex maximum

Setting this equal to $\beta=6$: …

The problem gives a quadratic expression $6x - x^2 + 12$ and says it attains its extreme value $\beta$ at $x = \alpha$. The word "extreme" here means the maximum or minimum — for a quadratic, that happens at the vertex. Since the coefficient of $x^2$ is negative ($-1$), the parabola opens downward, so the extreme is a maximum.

The key idea: for any quadratic $ax^2 + bx + c$, the vertex (where the extreme occurs) is at $x = -\frac{b}{2a}$. Here the expression is $-x^2 + 6x + 12$, so $a = -1$, $b = 6$, $c = 12$. The extreme value $\beta$ is simply the value of the expression at that $x$.

Let’s work through it.

1. Find $\alpha$ — the $x$-coordinate of the vertex. Using $x = -\frac{b}{2a}$:

So $\alpha = 3$.

2. Find $\beta$ — the extreme value, i.e., the value of the expression at $x = 3$. Substitute: …

When dealing with polynomial equations, especially cubic or quartic ones, and information about their roots is given (like being in arithmetic progression (AP) or geometric progression (GP)), Vieta's formulas are your primary tool. These formulas relate the coefficients of a polynomial to sums and products of its roots.

For a cubic equation $Ax^3 + Bx^2 + Cx + D = 0$, if the roots are $\alpha, \beta, \gamma$, Vieta's formulas state:

• Sum of roots: $\alpha + \beta + \gamma = -B/A$
• Sum of products of roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = C/A$
• Product of roots: $\alpha\beta\gamma = -D/A$

The key insight for roots in GP is how to represent them to simplify calculations. If we let the roots be $a/r, a, ar$, their product becomes simply $a^3$, which makes finding $a$ very straightforward.

Here's how we solve the problem:

1. Identify the equation and its coefficients:

   The given equation is $3x^3 - 26x^2 + 52x - 24 = 0$.

   Comparing this to the standard form $Ax^3 + Bx^2 + Cx + D = 0$, we have:

   $A = 3$

   $B = -26$

   $C = 52$

   $D = -24$

2. Represent the roots in geometric progression:

   Let the roots of the equation be $\dfrac{a}{r}, a, ar$. This choice is strategic because it simplifies the product of the roots.

3. Apply Vieta's formulas:

   For a cubic equation $Ax^3 + Bx^2 + Cx + D = 0$ with roots $\alpha, \beta, \gamma$:

   $\alpha + \beta + \gamma = -\frac{B}{A}$

   $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{C}{A}$

   $\alpha\beta\gamma = -\frac{D}{A}$

   Using our chosen roots $\dfrac{a}{r}, a, ar$:

   • Sum of roots:

     $\dfrac{a}{r} + a + ar = -\dfrac{(-26)}{3} = \dfrac{26}{3}$

     $a\left(\dfrac{1}{r} + 1 + r\right) = \dfrac{26}{3} \quad \text{(Equation 1)}$

   • Sum of products of roots taken two at a time:

     $\left(\dfrac{a}{r}\right)(a) + (a)(ar) + (ar)\left(\dfrac{a}{r}\right) = \dfrac{52}{3}$

     $\dfrac{a^2}{r} + a^2r + a^2 = \dfrac{52}{3}$

     $a^2\left(\dfrac{1}{r} + r + 1\right) = \dfrac{52}{3} \quad \text{(Equation 2)}$

   • Product of roots:

     $\left(\dfrac{a}{r}\right)(a)(ar) = -\dfrac{(-24)}{3}$

     $a^3 = \dfrac{24}{3}$

     $a^3 = 8$

4. Solve for 'a' using the product of roots:

   From $a^3 = 8$, we find $a = 2$. This is the middle root of the GP.

5. Substitute 'a' into Equation 2 to find 'r':

   Substitute $a=2$ into Equation 2:

   $(2)^2\left(\dfrac{1}{r} + r + 1\right) = \dfrac{52}{3}$

   $4\left(\dfrac{1}{r} + r + 1\right) = \dfrac{52}{3}$

   Divide both sides by 4:

   $\dfrac{1}{r} + r + 1 = \dfrac{52}{3 \times 4}$

   $\dfrac{1}{r} + r + 1 = \dfrac{13}{3}$

   Subtract 1 from both sides:

   $\dfrac{1}{r} + r = \dfrac{13}{3} - 1$

   $\dfrac{1}{r} + r = \dfrac{13 - 3}{3}$

   $\dfrac{1}{r} + r = \dfrac{10}{3}$

6. Solve the quadratic equation for 'r':

   Multiply the equation $\dfrac{1}{r} + r = \dfrac{10}{3}$ by $3r$ (assuming $r \neq 0$):

   $3 + 3r^2 = 10r$ …

The key idea is that for a rational function of the form $\frac{ax^2+bx+c}{dx^2+ex+f}$, when we set it equal to a constant $y$, we get a quadratic in $x$. For real $x$, this quadratic must have real roots, so its discriminant must be non-negative. That inequality gives the range of $y$, and the boundary values of $y$ (the minimum and maximum) occur exactly when the discriminant is zero — which also gives the $x$-values where those extremes happen.

Let’s work through it.

1. Let $y = \dfrac{x^2 + x + 1}{x^2 - x + 1}$. Cross-multiply (note $x^2 - x + 1 > 0$ for all real $x$, so no sign issues):

Bring all terms to one side:

2. This is a quadratic in $x$. For $x$ to be real, the discriminant must be $\ge 0$:

3. Expand and simplify:

Multiply by $-1$ (reversing inequality):

4. Factor the quadratic in $y$:

So the range of $y$ is $\frac{1}{3} \le y \le 3$. The minimum value is $\frac{1}{3}$ and the maximum is $3$, as given.

5. The $x$-values where these extremes occur are found by setting the discriminant to zero (the boundary of the inequality). For $y = \frac{1}{3}$ and $y = 3$, the quadratic in $x$ becomes a perfect square, giving a single (repeated) root. …

The key idea is geometric: each modulus equation describes a circle in the complex plane.

$|Z_1 - 3 - 4i| = 5$ means $Z_1$ lies on a circle centered at $C_1 = 3 + 4i$ with radius $r_1 = 5$.

$|Z_2| = 15$ means $Z_2$ lies on a circle centered at the origin $C_2 = 0$ with radius $r_2 = 15$.

We want the sum of the largest and smallest possible distances between a point on the first circle and a point on the second circle. That’s a classic geometry problem: the distance between two circles’ points ranges from the distance between centers minus the sum of radii (if they overlap) to the distance between centers plus the sum of radii. But here the circles are separate, so we need the exact range.

1. Find the distance between the centers.

   Center $C_1 = 3 + 4i$ corresponds to point $(3,4)$. Center $C_2 = (0,0)$.

   Distance $d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

2. Determine the minimum distance $|Z_1 - Z_2|_{\min}$.

   The smallest distance between points on two circles occurs along the line joining their centers.

   Since $d = 5$, $r_1 = 5$, $r_2 = 15$, the circles are positioned such that the smaller circle (radius 5) is entirely inside the larger circle (radius 15) — because $d + r_1 = 5 + 5 = 10 < r_2 = 15$.

   So the minimum distance is the gap from the outer edge of the small circle to the inner edge of the large circle along the line:

   $|Z_1 - Z_2|_{\min} = r_2 - (d + r_1) = 15 - (5 + 5) = 5$. …

The problem asks for the sum of the squares of all possible values of $\alpha$ such that the given quadratic equation has equal roots. The key concept here is the discriminant of a quadratic equation.

For any quadratic equation of the form $Ax^2 + Bx + C = 0$, the nature of its roots is determined by a value called the discriminant, denoted by $D$.

The condition for a quadratic equation to have equal roots is that its discriminant must be zero.

Let's apply this concept step-by-step to the given equation.

1. Expand the given equation into standard quadratic form:

   The given equation is $(x-\alpha)(x-3)+1=0$.

   First, we expand the product $(x-\alpha)(x-3)$:

   $(x-\alpha)(x-3) = x(x-3) - \alpha(x-3)$

   $= x^2 - 3x - \alpha x + 3\alpha$

   $= x^2 - (3+\alpha)x + 3\alpha$

   Now, substitute this back into the original equation:

   $x^2 - (3+\alpha)x + 3\alpha + 1 = 0$

   This is now in the standard quadratic form $Ax^2 + Bx + C = 0$.

2. Identify the coefficients $A$, $B$, and $C$:

   Comparing $x^2 - (3+\alpha)x + (3\alpha+1) = 0$ with $Ax^2 + Bx + C = 0$, we have:

   $A = 1$

   $B = -(3+\alpha)$

   $C = 3\alpha+1$

3. Apply the condition for equal roots ($D=0$):

   For equal roots, the discriminant $D = B^2 - 4AC$ must be zero.

   Substitute the identified coefficients:

   $(-(3+\alpha))^2 - 4(1)(3\alpha+1) = 0$

4. Solve the resulting equation for $\alpha$:

   Simplify the equation from the previous step: …

We are given the quartic

with roots $\alpha, \beta, \gamma, \delta$. We need

Concept and intuition:

If we want the product $(\alpha+4)(\beta+4)(\gamma+4)(\delta+4)$, we can think of it as the value of the polynomial’s shifted version. Specifically, if we define a new variable $y = x+4$, then $x = y-4$. Substituting into the original polynomial gives a polynomial in $y$ whose roots are exactly $\alpha+4, \beta+4, \gamma+4, \delta+4$. The constant term of that new polynomial (up to sign) is the product of those roots. Then we just multiply by 4.

Step-by-step solution:

1. Set up the shift. Let $y = x+4$, so $x = y-4$. The original polynomial is

We want $Q(y) = P(y-4)$. The roots of $Q(y)$ are the numbers $y$ such that $x = y-4$ is a root of $P$, i.e., $y = \alpha+4, \beta+4, \gamma+4, \delta+4$.

2. Substitute $x = y-4$ into $P(x)$. Compute each term carefully:

3. Multiply by the coefficients.

4. Sum all terms to get $Q(y)$. Combine like powers of $y$:

So

5. Product of the shifted roots. For a monic polynomial, the constant term is $(-1)^n$ times the product of the roots. But here the leading coefficient is 4, not 1. The polynomial $Q(y)$ can be written as

where $r_i = \alpha+4, \beta+4, \gamma+4, \delta+4$. Expanding, the constant term is

The core of this problem lies in understanding the relationship between the arithmetic mean (AM) and geometric mean (GM) of a set of positive numbers, combined with Vieta's formulas for polynomial roots.

The Arithmetic Mean-Geometric Mean (AM-GM) inequality states that for any set of $n$ non-negative real numbers $x_1, x_2, \ldots, x_n$, their arithmetic mean is always greater than or equal to their geometric mean:

Vieta's formulas provide a direct link between the roots of a polynomial and its coefficients. For a polynomial of degree $n$, $P(x) = kx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0$, with roots $r_1, r_2, \ldots, r_n$:

• Sum of roots: $\sum r_i = -\frac{a_{n-1}}{k}$
• Sum of products of roots taken two at a time: $\sum_{i<j} r_i r_j = \frac{a_{n-2}}{k}$
• ...and so on, with alternating signs.
• Product of roots: $r_1 r_2 \ldots r_n = (-1)^n \frac{a_0}{k}$

Let's apply these concepts to the given problem.

1. Identify the roots and apply AM-GM.

   The given polynomial is $x^5 - ax^4 + bx^3 - cx^2 + dx - 1 = 0$.

   Let its five roots be $r_1, r_2, r_3, r_4, r_5$.

   We are given that all roots are positive. Therefore, we can apply the AM-GM inequality to these five roots.

   The arithmetic mean (AM) of the roots is $\frac{r_1 + r_2 + r_3 + r_4 + r_5}{5}$.

   The geometric mean (GM) of the roots is $\sqrt[5]{r_1 r_2 r_3 r_4 r_5}$.

2. Use the condition AM = GM.

   The problem states that the arithmetic mean and geometric mean of the roots are equal.

   This means $\frac{r_1 + r_2 + r_3 + r_4 + r_5}{5} = \sqrt[5]{r_1 r_2 r_3 r_4 r_5}$.

   As discussed, the equality in the AM-GM inequality holds if and only if all the numbers are equal.

   Therefore, we must have $r_1 = r_2 = r_3 = r_4 = r_5$. Let's call this common root $r$.

3. Determine the value of each root.

   Now that we know all roots are equal to $r$, we can use Vieta's formulas.

   For the polynomial $x^5 - ax^4 + bx^3 - cx^2 + dx - 1 = 0$, the product of the roots is given by the constant term divided by the leading coefficient, with a sign adjustment.

   Watch out

   Remember the alternating signs in Vieta's formulas. For a polynomial $P(x) = A_n x^n + A_{n-1} x^{n-1} + \ldots + A_1 x + A_0$, the product of roots is $(-1)^n \frac{A_0}{A_n}$.

   Here, $n=5$ (odd), the leading coefficient is $1$, and the constant term is $-1$.

   So, the product of roots $r_1 r_2 r_3 r_4 r_5 = (-1)^5 \frac{-1}{1} = (-1)(-1) = 1$.

   Since all roots are equal to $r$, we have $r \cdot r \cdot r \cdot r \cdot r = r^5 = 1$.

   As the roots are positive, $r$ must be $1$.

   Thus, all five roots of the polynomial are $1$.

4. Find the values of $a, b, c, d$ using Vieta's formulas.

   Now we know $r_1 = r_2 = r_3 = r_4 = r_5 = 1$. We can find the coefficients $a, b, c, d$ using Vieta's formulas for the polynomial $x^5 - ax^4 + bx^3 - cx^2 + dx - 1 = 0$.

   • Coefficient $a$ (related to sum of roots): The sum of the roots is $r_1 + r_2 + r_3 + r_4 + r_5 = -(-a)/1 = a$. …

We form 4-digit numbers using each of $0,3,6,9$ exactly once. A leading $0$ does not give a 4-digit number, so those cases must be removed.

Step 1 — Sum of all $4!$ arrangements (ignoring the leading-zero rule).

Across all $4!=24$ permutations, each of the four digits occupies each place value exactly $3!=6$ times. With digit-sum $0+3+6+9=18$:

Step 2 — Subtract the arrangements that begin with $0$. …