## anonymous 4 years ago Guys, Prove $$x^4 + ax^3 + 2x^2 + bx + 1 = 0$$ has real solutions if $$a^2 + b^2 \le 8$$.

1. anonymous

Sorry, $$a^2 + b^2 \ge 8$$

2. Mr.Math

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

3. anonymous

yeah!

4. Mr.Math

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 $$ax^3+bx \le 0$$.

5. anonymous

I did the same

6. anonymous

but I couldn't get ahead of this

7. asnaseer

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:$a^2+b^2\ge8$

8. asnaseer

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$

9. anonymous

10. asnaseer

:)

11. Mr.Math

Very nice asnaseer! :-)