Question

Let
x^4 + px^2 + qx + r = 0 be a given quartic with roots α1,...,α4.

Show that α1 + α2 + α3 + α4 = 0 and that α1 · α2 · α3 · α4 = r.

Observe that this implies that if we know any three of the roots, we can deduce what the remaining root is.

Define the following three quantities defined in terms of the roots αi:
s1 = (α1 −α2 + α3 −α4)/2,

s2 = (α1 + α2 −α3 −α4)/2,

s3 = (α1 −α2 −α3 + α4)/2

Answer 1

If roots of an equation are α and β,,

the equation is (x−α)(x−β)=0

i.e. x²−(α+β)+αβ=0

Rewrite this Equation px²+qx+r=0 we get, x²+qpx+rp=0.

Hence, we have sum of the roots α+β=−qp and product of roots αβ=rp

Equation with roots as 2α+1α and 2β+1β is therefore

x²−(2α+1α+2β+1β)x+(2α+1α)(2β+1β)=0

Let us now workout 2α+1α+2β+1β and (2α+1α)(2β+1β)

2α+1α+2β+1β

= 2(α+β)+α+βαβ

= 2×(−qp)+−qprp=−2qp−qr=−2qr−pqrp

and (2α+1α)(2β+1β)

= 4αβ+1αβ+2(αβ+βα)

= 4rp+1rp+2α²+β²αβ

= 4rp+pr+2(α+β)²−2αβαβ

= 4rp+pr+2(−qp)²−2rprp

= 4rp+pr+2q²p²×pr−4

= 4rp+pr+2q²rp−4

= 4r²+p²+2q²−4rprp

and hence equation is

x²−(−2qr−pqrp)x+4r²+p²+2q²−4rprp=0

or rpx²+(2qr+pq)x+(4r²+p²+2q²−4rp)=0