Quick answer: A symmetric function of the roots α, β is unchanged when α and β are swapped, so it can be written using α+β=−ab and αβ=ac. For example, α2+β2=(α+β)2−2αβ and α3+β3=(α+β)3−3αβ(α+β). For location of roots of f(x)=ax2+bx+c with a>0: a number k lies between the roots exactly when f(k)<0.
This builds on quadratic equations: roots and discriminant.
Standard symmetric functions
Let S=α+β and P=αβ.
| Expression | In terms of S and P |
|---|
| α2+β2 | S2−2P |
| (α−β)2 | S2−4P |
| α3+β3 | S3−3PS |
| α4+β4 | (S2−2P)2−2P2 |
| α1+β1 | PS |
| βα+αβ | PS2−2P |
| α2β+αβ2 | PS |
If α, β are the roots of f(x)=0:
| New roots | New equation |
|---|
| α+k, β+k | f(x−k)=0 |
| kα, kβ | f(kx)=0 |
| α1, β1 | f(x1)=0, that is, cx2+bx+a=0 |
| −α, −β | f(−x)=0 |
Cubic equations
For ax3+bx2+cx+d=0 with roots α, β, γ:
| Relation | Value |
|---|
| α+β+γ | −ab |
| αβ+βγ+γα | ac |
| αβγ | −ad |
Location of roots
Let f(x)=ax2+bx+c with a>0, discriminant D and vertex at −2ab.
| Requirement | Conditions |
|---|
| k lies between the roots | f(k)<0 |
| Both roots greater than k | D≥0, f(k)>0, −2ab>k |
| Both roots less than k | D≥0, f(k)>0, −2ab<k |
| Both roots between k1 and k2 | D≥0, f(k1)>0, f(k2)>0, k1<−2ab<k2 |
| Exactly one root in (k1,k2) | f(k1)f(k2)<0 |
For a<0, apply the same rules to −f(x), or use af(k)<0 for the first row.
Worked examples
Example 1: α² + β²
If α, β are the roots of x2−5x+3=0, find α2+β2 and α3+β3.
Step 1: S=5, P=3.
Step 2: α2+β2=25−6=19.
Step 3: α3+β3=125−45=80.
Example 2: new equation
Form the equation whose roots are the squares of the roots of x2−5x+3=0.
New sum =19, new product =P2=9, so x2−19x+9=0.
Example 3: k between the roots
For what m does 2 lie between the roots of x2−mx+1=0?
f(2)=4−2m+1<0, so m>25.
Example 4: both roots greater than 1
Find k so that both roots of x2−2kx+k2−1=0 exceed 1.
Step 1: D=4k2−4(k2−1)=4>0 always.
Step 2: f(1)=1−2k+k2−1=k2−2k>0, so k<0 or k>2.
Step 3: Vertex k>1.
Answer: k>2. (Indeed the roots are k±1, and k−1>1.)
Example 5: cubic relation
If α, β, γ are the roots of x3−6x2+11x−6=0, find α2+β2+γ2.
(α+β+γ)2−2(αβ+βγ+γα)=36−22=14.
Summary table
| Idea | Key result |
|---|
| α2+β2 | S2−2P |
| α3+β3 | S3−3PS |
| Reciprocal roots | Swap a and c |
| k between roots (a>0) | f(k)<0 |
| One root in an interval | f(k1)f(k2)<0 |
Exam tips
- Rewrite every expression in S and P before substituting.
- For location problems, think of the parabola's graph: sign of f(k), position of the vertex, and D≥0.
- A quick check: find the roots directly if they are simple, as in Example 4.
Next, see arithmetic progression formulas, or browse the NIMCET Algebra chapter.
Common mistakes
- Writing α2+β2=S2.
- Forgetting the vertex condition in "both roots greater than k".
- Using αβγ=ad for a cubic; it is −ad.
- Ignoring the sign of a in location-of-roots conditions.
Frequently Asked Questions
What is α² + β² in terms of the coefficients?
α² + β² = (α + β)² − 2αβ = (b² − 2ac)/a².
What is a symmetric function of roots?
An expression in the roots that does not change when the roots are interchanged, such as α + β or α²β + αβ².
Replace x by 1/x. For ax² + bx + c = 0, the new equation is cx² + bx + a = 0.
When does a number k lie between the roots of a quadratic?
For ax² + bx + c with a > 0, when f(k) < 0. In general, when a·f(k) < 0.
What is the product of the roots of a cubic equation?
For ax³ + bx² + cx + d = 0, the product of the roots is −d/a.