Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — symmetric functions of the roots
The roots of 2x3−5x2+3x−7=0 are α, β and γ. Find α2+β2+γ2.
- 1α+β+γ=−ab=25Note the minus sign in front of b/a.
- 2αβ+βγ+γα=ac=23This one is positive.
- 3Use the identity α2+β2+γ2=(α+β+γ)2−2(αβ+βγ+γα)Squaring the sum produces the cross terms, which are then removed.
- 4=(25)2−2(23)
- 5=425−3=413No need to find the roots themselves.
Example 2 — forming a new cubic
The roots of 2x3−5x2+3x−7=0 are α, β, γ. Find the equation whose roots are 2α, 2β, 2γ.
- 1The original sums are 25, 23 and 27.Sum, sum of pairs, and product.
- 2New sum: 2(25)=5One root per term, so one factor of 2.
- 3New sum of pairs: 22(23)=6Two roots per term, so 22.
- 4New product: 23(27)=28Three roots, so 23.
- 5A monic cubic is x3−(sum)x2+(pairs)x−(product)=0Signs alternate.
- 6x3−5x2+6x−28=0The substitution x=y/2 into the original gives the same answer ✓
Example 3 — checking the relationships on a known cubic
Verify the root-coefficient relationships for x3−6x2+11x−6=0.
- 1f(1)=1−6+11−6=0, so (x−1) is a factor.Factorising gives (x−1)(x−2)(x−3).
- 2The roots are 1,2,3.
- 3Sum: 1+2+3=6, and −ab=−1−6=6 ✓
- 4Pairs: (1)(2)+(2)(3)+(3)(1)=11, and ac=11 ✓
- 5Product: (1)(2)(3)=6, and −ad=−1−6=6 ✓All three relationships confirmed.