Minimum Monic

Suppose the quadratic $f(x)$ is $x^2 + ax + b$ .

We know, its roots are $\dfrac{-a\pm\sqrt{a^2 - 4b}}{2}$ It has positive roots so a must be negative ( we will require this at last)(see the bottom of solution for a broader explaination) and $a^2 \geq 4b$ (we we need this a little later)

Now, $f(1) = 441$

$\implies$ $b-a+1= 441$

$\implies$ $b - 440= - a$

$\implies$ $(b-440)^2 = a^2$

$\implies$ $b^2 - 880b+ 440^2 \geq 4b$

$\implies$ $b^2 - 884b + 440^2 \geq 0$

$\implies$ $b^2 - 400b - 484b + 440^2 \geq 0$

$\implies$ $(b-400)(b-484) \geq 0$

Therefore, either $b \geq 400$ or $b \geq 484$ . In other words, either the minimum value of $f(0)$ is $400$ or $484$

Now, if $b=400$ then we have by the given condition

$1+a+400=441$ $\implies$ $a = 40$ . But this contradicts the negativity of $a$ . Thus $b=400$ as the minimum value of $f(0)$ is ruled out. Now we will show that the minimum value of $f(0) = 484$ indeed holds.

With $f(1) = 441$ and $b = 484$ , we have $1 + a + 484 = 441$ $\implies$ $a = -44$ . This satisfies our requirement that a is negative and we can easily verify that plugging $a = -44$ in $f(x)$ indeed gives positive real roots. Thus the minimum value of $f(0)$ is $\boxed{484}$

$\large{**Note**}$ : The roots of $f(x)$ must be real and positive. We know the roots are $\dfrac{-a\pm\sqrt{a^2 - 4b}}{2}$ . Thus the discriminant is always positive . If $D$ denotes the discriminant then we observe one root has numerator $-a -D$ . If $a$ is not negative then $-a$ is negative and thus we will end up having a negative root.

Let

$f(x)=x^2-(\alpha+\beta)x+(\alpha\beta)$

where $\alpha, \beta$ is the root of the polynomial and $(\alpha,\beta)>0$ .

Given that,

$f(1)=441$

$1-(\alpha+\beta)+\alpha\beta=441$

We want to find the minimum value of $f(0)$ , which is $\alpha\beta$ . Noticed that

$\alpha \beta-\alpha-\beta-440=0$

and

$\alpha \beta-\alpha-\beta-440 \geq \alpha\beta-2\sqrt{\alpha\beta}-440$

Shows us that the minimum value occurs when $\alpha=\beta$ , so note

$1-2\alpha+\alpha^2=441$

$(\alpha-1)^2=441$

$\alpha=22$

Hence, the minimum value of $f(0)=\alpha\beta=22^2=\boxed{484}$ .

First, we let $f(x) = x^2 + bx + c.$ Since $f(1) = 441,$ we can write the following: $$ 1^2 + b(1) + c = 441 $$ $$ 1 + b + c = 441 $$ $$ b + c = 440 $$ $$ c = 440 - b. (1) $$

In addition, since $f(x)$ has real roots, its discriminant must be greater than or equal to 0. That is, $b^2 - 4c \geq 0.$ We can substitute equation (1) here, to get $b^2 - 4(440 - b) \geq 0.$

This then can be evaluated as follows: $$ b^2 + 4b − 1,760 \geq 0 $$ $$ (b - 40)(b + 44) \geq 0. $$ Here, using the critical points, it can be shown that $b \leq -44$ or $b \geq 40.$

By looking at a graph, we note that if $b$ is positive, the root(s) of the function will be negative. Therefore, $b \leq -44.$ We now find a general value for $f(0).$ This turns out to be $c.$ This shows that we want to minimize $c.$ Since $c = 440 - b,$ and $b \leq -44,$ the minimum value for $c$ occurs when $b = -44.$ This minimum value is $484.$

Therefore, the minimum value of $f(0)$ is $484.$

Let $f(x)=x^2+ax+b$ . First, note that $f(1)=1+a+b=441$ implies that $a+b=440$ .

Next, note that by the Quadratic Formula, we know that the roots of this quadratic are $\frac{-a \pm \sqrt{a^2-4b}}{2}$ .

If $a \geq 0$ , then $\frac{-a-\sqrt{a^2-4b}}{2}$ will be non-positive. Hence, we must have $a<0$ . Therefore, it is sufficient to minimize the absolute value of $a$ , since as $a$ decreases $b$ must increase due to the first equation.

Also, since both roots must be real, we have that the discriminant must be non-negative-- i.e. we must have $\sqrt{a^2-4b} \geq 0$ . It follows that $a^2 \geq 4b$ . To maximize $|a|$ , we set the two sides equal to each other. In other words, we have $a^2=4b$ .

Solving for $b$ and plugging the value into the equation $a+b=440$ , we see that \(a+\frac{a^2}{4}=440

\Rightarrow a^2+4a=1760

\Rightarrow (a+2)^2=1764=42^2\).

From this we see that $a=-44$ , and thus $b=440-(-44)=484$ .

Let $f(x) = x^2 + ax +b$ , where $a$ and $b$ are real numbers (we know that the coefficients of the quadratic are real because its roots are real). Note that $f(0) = b$ , so we are trying to minimize the value of $b$ . Since $f(1) = 441$ , we have $1 + a + b = 441 \Rightarrow a = 440 - b$ . Thus, our quadratic becomes $f(x) = x^2 + (440-b)x + b$ . By the quadratic formula, the roots of this quadratic are of the form $x = \frac{b-440 \pm \sqrt{(440-b)^2 - 4b}}{2}$ .

Because the roots must be positive, we must have $b-440 \pm \sqrt{(440-b)^2 - 4b} > 0$ . Also, because the discriminant is nonnegative, we must have $b-440 > 0 \Rightarrow b > 440$ .

Furthermore, the roots must be real, so the discriminant must be nonnegative. Thus we have,

$(440-b)^2 - 4b \geq 0$

$b^2 - 884b + 440^2 \geq 0$

We apply the quadratic formula to find the roots of the LHS,

$b = \frac{884 \pm \sqrt{884^2 - 4 \cdot 440^2}}{2}$

$b = \frac{884 \pm \sqrt{884^2 - (880)^2}}{2}$

$b = \frac{884 \pm \sqrt{1764 \cdot 4}}{2}$

$b = \frac{884\pm 84}{2}$

$b = 442 \pm 42$

$b = 400, 484$

Therefore, we have

$(b-400)(b-484) \geq 0$

The values of $b$ that satisfy this inequality are all $b$ such that $b \leq 400$ or $b \geq 484$ . However, for the roots of our original quadratic to be positive, we have found that $b>440$ . Therefore, the minimal value of $f(0) = b$ is $\boxed{484}$ .

We are given a monic quadratic polynomial that satisfies $f(1)=441$ . We can thus write $1^2 +b*1 + c = 441 <-> b + c = 440$ . Now, we are also told that the polynomial has real roots, so the discriminant must at least be zero, i.e $b^2-4*a*c \geq 0$ . The minimum value of $f(0)$ , is the same as the minimum value of $c$ since $0^2+b*0 + c = c$ , so we'll solve $b^2-4*a*c=0$ . So we can write $b^2-4*c=0 <-> \frac{b^2}{4}=c$ . From before we know that $b+c=440$ hence $440=b+\frac{b^2}{4} <->b^2+4*b-4*440=0$ . This equation has two solutions, $b=-44$ or $b=40$ . Since we are told that we are looking for postive real roots, it's $b=-44$ we are looking for (b=40 yields a negative root). So we end up with $1^2 -44*1+c=441 -> c=484$ so the minimum value of $f(0)$ is $484$

Yet another solution!
Write $f(x)=x^{2}+ax+b$ , $a,b$ are real numbers
As roots are real, the discriminant of the above equation which is $a^{2}-4b$ must be greater than $0$ . So $a^{2} \ge 4b$
As roots are positive, their sum and their product are also positive, so by Vieta's formulas, $a<0$ and $b>0$
Also, $f(1)=a+b+1=441$ which yields $a+b=440$ so $b=440-a$
Combining $b=440-a$ and $a^{2} \ge 4b$ , we get $a^2+4a-1760 \ge 0$
that is, $(a-40)(a+44)\ge0$ which gives $a\le-44$ [ $a \ge 40$ is not admissible as $a<0$ ]
Hence, $b \ge 440+44=484$
But $f(0)=b$ so $f(0) \ge 484$
Now, we have to establish that $f(0)=484$ can be achieved.
As $f(0)=b=484$ , we have $a=441-1-484=-44$ . So $f(x)=x^2+ax+b=x^2 -44x+484$ which has a repeated root of $22$ which is positive.
This monic polynomial above satisfies all the conditions of $f(x)$ given in the question and hence, $f(x)=484$ can be achieved.
So our answer is $484$ . Also notice that this is $22^2$ and $f(1)=21^2$ , is this a coincidence?

Let $x_1$ and $x_2$ be the roots of $f(x) = x^2 + bx + c$ . We know that

$f(1) = 1 + b + c = 441$

and

$f(0) = c$

From Vieta's formulae we have that:

$x_1 + x_2 = -b$

and

$x_1x_2 = c$

From these and $f(1) = 1 + b + c = 441$ we get:

$x_1x_2 = 440 + x_1 + x_2$

which can be cast as

$x_2 = \frac{x_1 + 440}{x_1 - 1}$

Multiply this by $x_1$ and get

$x_1x_2 = \frac{x_1^2 + 440x_1}{x_1 - 1}$

We have written $f(0)$ as function of $x_1$ . Its minimum can be found via standard methods of calculus. The minimum is for $x_1 = 22$ which implies

$\boxed{f(0) = 484}$

Here's my solution.

First let us consider a general case.

A quadratic $f(x) = ax^2 + bx + c$ has real roots iff $b^2 -4ac \geq 0$ . The roots are positive iff both the following conditions are satisfied:

$(1)$ $\frac{-b}{2a} > 0$ , i.e., the vertex of the parabola is to the right of the y axis; and

$(2)$ $af(0) > 0$ , i.e., the origin does not lie between the roots.

Now, for our problem,

Let the quadratic be $f(x) = x^2 + bx + c$

From the given conditions we have using the above statements,

$b^2 - 4c \geq 0$ and $\frac{-b}{2} > 0$ and $c > 0$ and $1+ b +c =441$

Now, $1+b+c=441 \iff b+c=440 \iff c=440-b$

Subsitituting this in $b^2 - 4c \geq 0$ we have

$b^2-4(440-b) \geq 0$

$\iff b^2+4b-1760 \geq 0$

$\iff (b+44)(b-40) \geq 0$

$\iff b \leq -44$ since $\frac{-b}{2}>0 \iff b<0$

Therefore,

$-b \geq 44 \iff 440-b \geq 484 \iff c \geq 484$

Therefore the minimum value of $c$ is $\boxed{484}$

Since the equation has two positive real roots, we may represent the monic quadratic equation in the form of: $f(x)=x^2-ax+b$ , where a and b are positive real numbers.

Substitute $x=1$ into the equation, we have $f(1)=1^2-a(1)+b=441$

and $b-a=440$ and $b=a+440$

Hence we can observe that in order to get the minimum value of b, we have to let a be as small as possible.

We also can conclude that $a^2-4b$ has to be as small as possible, i.e =0.(since the equation has real roots and the two roots are not necessarily distinct)

We now have two simultaneous equations:

$a^2-4b=0$ (1)

$b=a+440$ (2)

Substitute $b=a+440$ into equation (1), we have

$a^2-4(440+a)=0$

$a^2-4a-1760=0$

$(a-44)(a+40)=0$

Since a is positive, $a=44$

Substitute a into equation (1), we have $b=44+440=484.$

Let f ( x ) = ( x - a )( x - b ). We know that f (0) = ab . This attains its maximum if a = b . So, f ( x ) = ( x - a )^2.

Using the given from the equation, we have:

f (1) = (1- a )^2 = 441

a = 22 or a = -20

Since a > 0, we take a = 22.

So f (0) = ab = a ^2 = 22^2 =484

Let the two roots be h and k: f(x)= (x-h)(x-k) f(1) = (1-h)(1-k) = 441 and f(0) = (0-h)(0-k) = hk. For hk to be as small as possible, let the difference between h and k be as small as possible. Now, (1-h)(1-k) = 441= (-21)^2 which yields h = k =22. Therefore, hk = 22^2 = 484.

Let the equation of the quadratic be f(x) = x^2+bx+c. Let the positive roots of the quadratic be m and n. The product of the roots is mn=c=f(0). To minimize f(0)=mn, it must be the case that m=n (as if n, say, was greater than m, then their product would be greater than m^2). Therefore, the parabola has a double real zero at x=m and has the equation f(x)=(x-m)^2. As the point (1,441) lies on the curve, then 441=(1-m)^2 which yields the solution m=22, m>0. Therefore, the minimum value of f(0) is 22^2=484.

Same as Christopher Boo's solution. :D