Consider the equation x4+2ax3+x2+2ax+1=0, where a∈R.If equation has at least two distinct positive real roots, then all possible values of a areIf equation has at least two distinct negative real roots, then all possible values of a areIf exactly two roots are positive and two roots are negative, then the number of integral values of a is

Given equation isx4+2ax3+x2+2ax+1=0          (1)or x2+1x2+2ax+1x+1=0or x+1x2+2ax+1x−1=0⇒ t2+2at−1=0                     (2)where t = x + (1/x). Now,x+1x≥2 or x+1x≤−2∴ t≥2 or t≤−2Now, Eq. (1) will have at least two positive roots, when at least one root of Eq. (2) will be greater than 2. From Eq.(2),D=4a2−4(−1)=41+a2>0,∀a∈R              (3)Let the roots of Eq. (2) be α,β. If α,β≤2, then⇒ f(2)≥0 and −B2A<2⇒ 4+4a−1≥0 and −2a2<2⇒ a≥−34 and a>−2⇒ a≥−34Therefore, at least one root will be greater than 2.Then,a<−34                      (4)Combining (3) and (4), we geta<−34Hence, at least one root will be positive if a∈(−∞,−3/4).Given equation isx4+2ax3+x2+2ax+1=0          (1)or x2+1x2+2ax+1x+1=0or x+1x2+2ax+1x−1=0⇒ t2+2at−1=0                     (2)where t = x + (1/x). Now,x+1x≥2 or x+1x≤−2∴ t≥2 or t≤−2Now, Eq. (1) will have at least two roots negative, when at least one root of Eq. (2) will be less than -2. If α,β≥−2, thenf(−2)≥0 and −B2A>−2∴ 4−4a−1≥0 and −2a2>−2∴ a≤34 and a<2∴ a≤34               (5)Combining (3) and (5), at least one root will be less than -2 for Eq. (2) ita>34∴ a∈34,∞Given equation isx4+2ax3+x2+2ax+1=0          (1)or x2+1x2+2ax+1x+1=0or x+1x2+2ax+1x−1=0⇒ t2+2at−1=0                     (2)where t = x + (1/x). Now,x+1x≥2 or x+1x≤−2∴ t≥2 or t≤−2If exactly two roots are positive, then other two roots are negative. Then - 2 and 2 must lie between the roots. So,f(−2)<0 and f(2)<0⇒ a>3/4 and a<−3/4Hence, no such values of a exist.