Only real roots!

Consider the quadratic $x^2-47x+k=0$ . For this equation to have real roots, its discriminant must be greater than or equal to zero. Hence

$\displaystyle 47^2-4k \geqslant 0 \Rightarrow k \geqslant 552.25$ .

The real roots occur if $k$ varies from 1 to 552. The product of roots of each quadratic is $k$ . Hence, the product of all real roots is

$\displaystyle 1\cdot 2 \cdot 3\cdot 4........\cdot 552=\fbox{552!}$ .

The product the two real roots of x^2 − 47x + k = 0 is k. So, The product the two real roots x^2 − 47x + 1 = 0 is 1 and x^2 − 47x + 2 = 0 is 2. x^2 − 47x + k = 0. Delta = 47^2 - 4k > 0 <=> k < 552,25 => k <= 552 > the product of all real roots of the polynomial above is 552!

we wil find till wat value of k will the quadratic eqtns will give real roots ,,here (b b-4ac) 47 47-4k>0; solve above v get k<552.25 so the real soltns till 552 now v se the product , its simple all the last terms of quad. eqtns multiply to give the constant term or the products of roots

A quadratic $\displaystyle x^{2} - ax + b$ has real roots if and only if its determinant $\Delta$ is positive, which means $a^2 \geq 4b$ . This shows that $x^2 - 7x + k$ will only have real roots until ${47}^2 \geq 4k \Rightarrow k \leq 552$ .

The product of the roots in $x^2 - 47x + k$ is $k$ . Because $k$ goes from 1 to 552 (last integer such that the $\Delta$ inequality holds), the product of all real roots of the polynomial $\displaystyle \prod_{k=1}^{999} (x^2 -47x + k)$ is $1 \times 2 \times 3 \times \cdots \times 552 = 552!$ It is obvious now that $\boxed{n = 552.}$

Roots of each of the product are also the roots of the polynomial. For real roots, $b^2 -4ac \geq 0$ . The only value that changes for every quadratic polynomial is $c$ , so for integers values of $c$ where $1 \leq c \leq 999$ , $47^2-4(1)(c) \geq 0$ $4c \leq 47^2$ $c \leq 552.25$ $\Rightarrow c \leq 552$ Since every product is a quadractic, and complex roots only occurs in conjugate pairs, when $c \geq 553$ , the quadractic equations will have 2 complex roots. From Vieta's formula, the product of real roots required $= 1*2*3*...*552 = 552!$ $n=552$

All the polynomials on the R.H.S are quadratic, each of whose roots are real when the determinant $b^{2} - 4ac \geq 0$ .

So, $47^{2} - 4(1)(k) \geq 0$ , where $k$ is the constant term in each quadratic polynomial.

We get $k \leq 552.25$

Which mean $x^{2} - 47x + 1$ , $x^{2} - 47x + 2$ , ........, $x^{2} - 47x + 552$ will give real roots, and all the polynomials after this will give imaginary roots.

Therefore, Product of their roots = $1. 2. 3. 4. .... 552$

$= 552!$

Hence, $n = 552$

That's the answer!

Note that for very large values of $k$ , $x$ will be imaginary. We have to find the threshold at which $x$ becomes imaginary. Using the concept of discriminant of a quadratic,

$D = b^{2} - 4ac > 0$

I'm omitting the case of repeated roots ( $D = 0$ ) because 47 is odd. Also, I don't know the LaTeX symbol for greater than or equal to.

$(-47)^{2} - 4(1)(k) > 0$

Solving for $k$ gives $k < 552.25$

So, for all integer values above 552, $x$ is imaginary. What is remaining is finding the product of the real roots, which is trivial, because in a quadratic, the constant term is equal to the product of the roots.

Hence, the product of the real roots is computed by $1 \times 2 \times 3 \times ... \times 552 = 552!$

We need to find the product of real roots .
Let the roots be $(a_1,a_2);(a_3,a_4);(a_5,a_6); \cdots ;(a_ {1997},a_ {1998})$ of the equations $(x^2-47x+1);(x^2-47+2 ); \cdots ;(x^2-47x+999)$ .

We observe, $(47^2-4 \times 999) < 0$ .

Hence, there must exist some $k$ such that $47^2-4 \times (a_{k},a_{k-1} \cdots a_{2},a_{1}) > 0$ .

If we find that $k$ our job's done as we need to find:

$(a_{1} \times a_{2}) \times (a_{3} \times a_{4})$ $\cdots \cdots$ $(a_{k-1} \times a_{k})$ which is simply equal to:

$1 \times 2 \times 3 \times 4 \cdots \cdots \times k$ ( By Vieta's relations and as $a$ is $1$ )

The discriminant must be positive or $\bigtriangleup \geq 0$ for the roots to be real .

We need, $47^{2}=2209 > 4 \times c$ (Since $4 \nmid 2209)$

And $\lfloor \frac{2209}{4} \rfloor=552$

Hence, maximum value of $c$ or $k$ is $552$ .

Hence, the required product is $1\times 2\times \times \cdots \times 552=552!=n!$ .So, our answer is $n=\boxed{552}$ .

Discriminant is b^2 - 4 a c > 0 (only real roots). So 47^2 - 4* k >0 ===> k< 552.5

The product of the roots is obtained by multiplying the constant termf of the individual expressions. so ,

552 * 551 550 .... *3 2*1 === > 552!

for the factor (x²-47x+k) to have real roots,

discriminant,d=47² -4k ≥ 0

==>k ≤ 552 .25

===>factors from (x²-47x+1) to (x²-47x+552) only have real soln.s

product of real roots for (x²-47x+k) ,P_k = k -----Vieta's formula

====>product of all real roots = product of k from 1 to 552 = 1 2 *3.... 552 =552 !

====>n=552 is the soln. to express n! as the product of real roots for the eqn.

square of 47 = 2209 for real roots, determinant should not be negative. thus, 2209-4k>=0 which gives k<=552 product of roots of each quadratic factor of the polynomial is the k term itself. thus, the required product is 552!