Question

Guys, Prove \(x^4 + ax^3 + 2x^2 + bx + 1 = 0\) has real solutions if \(a^2 + b^2 \le 8\).

Answer 1

Sorry, \(a^2 + b^2 \ge 8\)

Answer 2

Do you mean that ALL solutions are real if that condition is satisfied?

Answer 3

yeah!

Answer 4

I can't yet get a proof, but this might help.

\(x^4+ax^3+2x^2+bx+1=(x^2+1)^2+ax^3+bx=0\)
\(\implies (x^2+1)^2=-(ax^3+bx).\)

For real solutions we have to have \(ax^3+bx \le 0\).

Answer 5

I did the same

Answer 6

but I couldn't get ahead of this

Answer 7

given:\[x^4+ax^3+2x^2+bx+1=0\tag{a}\]if this has all real solutions, then it can be written as:\[(x-r_1)(x-r_2)(x-r_3)(x-r_4)=0\]where \(r_1,r_2,r_3,r_4\) are the real solutions. If we expand this we get:\[\begin{align}
x^4-(r_1+r_2+r_3+r_4)x^3+(r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4)x^2\\
-(r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4)x+r_1r_2r_3r_4=0\tag{b}
\end{align}\]comparing equations (a) and (b) we can deduce the following:\[\begin{align}
r_1r_2r_3r_4&=1\tag{c}\\
r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4&=2\tag{d}\\
r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4&=-b\tag{e}\\
r_1+r_2+r_3+r_4&=a\tag{f}
\end{align}\]squaring (f) gives us:\[\begin{align}
a^2&=r_1^2+r_2^2+r_3^2+r_4^2+2(r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4)\\
&=r_1^2+r_2^2+r_3^2+r_4^2+2(2)\qquad\text{[using (d)]}\\
&=r_1^2+r_2^2+r_3^2+r_4^2+4\tag{g}
\end{align}\]squaring (e) gives us:\[\begin{align}
b^2&=(r_1r_2r_3)^2+(r_1r_2r_4)^2+(r_1r_3r_4)^2+(r_2r_3r_4)^2+\\
&\qquad2(r_1^2r_2^2r_3r_4+r_1^2r_2r_3^2r_4+r_1^2r_2r_3r_4^2+r_1r_2^2r_3^2r_4+r_1r_2r_3^2r_4^2)\\
&=(r_1r_2r_3)^2+(r_1r_2r_4)^2+(r_1r_3r_4)^2+(r_2r_3r_4)^2+\\
&\qquad2r_1r_2r_3r_4(r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4)\\
&=(r_1r_2r_3)^2+(r_1r_2r_4)^2+(r_1r_3r_4)^2+(r_2r_3r_4)^2+\\
&\qquad2r_1r_2r_3r_4(2)\qquad\text{[using (d)]}\\
&=(r_1r_2r_3)^2+(r_1r_2r_4)^2+(r_1r_3r_4)^2+(r_2r_3r_4)^2+\\
&\qquad2*1(2)\qquad\text{[using (c)]}\\
&=(r_1r_2r_3)^2+(r_1r_2r_4)^2+(r_1r_3r_4)^2+(r_2r_3r_4)^2+4\tag{h}\\
\end{align}\]adding (g) and (h) gives us:\[a^2+b^2=r_1^2+r_2^2+r_3^2+r_4^2+(r_1r_2r_3)^2+(r_1r_2r_4)^2+(r_1r_3r_4)^2+(r_2r_3r_4)^2+8\]since the non-constant terms are all squares, their minimum value must be 0, therefore:\[a^2+b^2\ge8\]

Answer 8

equation got cut-off at the right hand side!
adding (g) and (h) gives us:\[a^2+b^2=r_1^2+r_2^2+r_3^2+r_4^2+\]\[\hspace{2cm}(r_1r_2r_3)^2+(r_1r_2r_4)^2+(r_1r_3r_4)^2+(r_2r_3r_4)^2+8\]

Answer 9

asnaseer wow your latex!

Answer 10

:)

Answer 11

Very nice asnaseer! :-)