4+8a^2+9a^4

Question

Middle term break

Answer 1

4+8a²+9a⁴ has only complex zeroes.

To find them we can use the quadratic formula for a²:

a²=(-8±√(64-144))/18=(-8±√-80)/18=(-4±√-20)/9=(-4±2i√5)/9.

We still have to find a. Start with a²=(-4+2i√5)/9.

a=±√((-4+2i√5)/9)=±⅓√(-4+2i√5). If a=p+iq where p, q are real,

a²=p²+2ipq-q²; equate real and imaginary parts:

p²-q²=-4/9; 2pq=2√5/9, pq=√5/9, q=√5/(9p), q²=5/(81p²);

p²-5/(81p²)=-4/9, 81p⁴-5=(-4/9)(81p²)=-36p², 81p⁴+36p²-5=0=(9p²-1)(9p²+5). Since p is real we can reject the factor 9p²+5.

So 9p²=1, p²=1/9, p=±⅓, q²=p²+4/9=5/9, q=±√5/3; but pq=√5/9 so p and q must have the same sign, that is, (p,q)=(⅓,√5/3) or (-⅓,-√5/3).

Therefore, a=p+iq=±⅓(1+i√5).

We continue with a²=(-4-2i√5)/9. Let a=r+is where r, s are real.

r²-s²=-4/9; 2rs=-2√5/9, rs=-√5/9, s=-√5/(9r), so r=p=±⅓ (see above for p and q).

s²=r²+4/9=5/9, s=-√5/3, because rs<0 so when r>0, s<0 and when r<0, s>0, that is, (r,s)=(⅓,-√5/3) or (-⅓,√5/3).

Therefore, a=r+is=⅓-i√5/3 or -⅓+i√5/3, which can be written ±⅓(1-i√5).

So the 4 zeroes are ⅓(1+i√5), ⅓(1-i√5), -⅓(1+i√5), -⅓(1-i√5).