# equation $\displaystyle a_{n+1}.x^2-2x\sqrt{a^2_{1}+a^2_{2}+…+a^2_{n}+a^2_{n+1}}+\left(a_{1}+a_{2}+…+a_{n}\right) = 0$ has real roots

The Largest positive integer $n$ such that for all real no. $a_{1},a_{2},.......,a_{n},a_{n+1}.$ The equation

$\displaystyle a_{n+1}.x^2-2x\sqrt{a^2_{1}+a^2_{2}+.......+a^2_{n}+a^2_{n+1}}+\left(a_{1}+a_{2}+.......+a_{n}\right) = 0$ has real roots.

My Try:: If Given equation has real roots. Then its Discriminant $\geq 0$

So $\displaystyle 4\left(a^2_{1}+a^2_{2}+.......+a^2_{n}+a^2_{n+1}\right)-4.a_{n+1}.\left(a_{1}+a_{2}+.......+a_{n}\right)\geq 0$

So $\left(a^2_{1}+a^2_{2}+.......+a^2_{n}+a^2_{n+1}\right)-a_{n+1}.\left(a_{1}+a_{2}+.......+a_{n}\right)\geq 0$

after that How can i solve it,plz explain me Thanks

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Try playing around with the equation. What happens if all $a_k$ are equal? What happens if $a_1 = \ldots = a_n = 0$ and $a_{n + 1} \ne 0$? I'm sure you will find other values that are easy to work with, and that could hint at how a solution has to look like. –  vonbrand Mar 2 '13 at 3:42
What you need to use is the converse: if $\Delta\geq 0$, then the quadratic equation has real roots. –  1015 Mar 2 '13 at 16:27

Let $x=a_{n+1}$ then the last inequality is $x^2-xT+S \ge 0$ where $T=a_1+a_2+\cdots+a_n$ and $S=a_1^2+a_2^2+\cdots+a_n^2$. This requires its own discriminant $\Delta=T^2-4s\le 0$. As you try increasing the number of $a_i$'s you see at 4 elements the inequality can be made to work but not after that. Here is how it turns out after a bit of algebra

for $n=1$ we have

$-\Delta=3a_1^2\ge 0$.

for $n=2$ we have

$-\Delta=2(a_1^2+a_2^2)+(a_1-a_2)^2 \ge 0$.

for $n=3$ we have

$-\Delta=1(a_1^2+a_2^2+a_3^2)+(a_1-a_2)^2 +(a_2-a_3)^2+(a_3-a_1)^2\ge 0$.

for $n=4$ we have

$-\Delta=0(a_1^2+a_2^2+a_3^2+a_4^2) \\ +(a_1-a_2)^2+(a_1-a_3)^2+(a_1-a_4)^2 +(a_2-a_3)^2+(a_2-a_4)^2+(a_3-a_4)^2\ge 0$.

for $n=5$ the inequality does not work for all possible $a_i$'s. Use the suggestion of @vonbrand, above, of setting all of them equal to say $t$ to see $-\Delta=4(5t^2)-(5t)^2=-5t^2 <0$ for $t\ne 0$.

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Thanks Maesumi would you like to explain me in detail –  juantheron Mar 2 '13 at 8:35
@juantheron I added more detail. –  Maesumi Mar 3 '13 at 22:51