Polynomials path · A06

Before this lesson: Solving polynomial equations

Your goal: Find root expressions without calculating each root separately.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Names you may know: Vieta’s theorem; Viète’s formulas.

Vieta’s Formulas: Roots and Coefficients: the key idea

For ax²+bx+c=a(x−r)(x−s), expansion gives r+s=−b/a and rs=c/a. These are Vieta’s relations. For a monic cubic with roots r,s,t, the coefficients are −(r+s+t), rs+rt+st and −rst. The signs alternate in higher degrees. Use these symmetric combinations when explicit radicals would obscure the structure. To form a new equation from transformed roots, compute their new sum and product, and check any denominator restrictions. Roots are counted with multiplicity.

A worked example

The roots of x²−6x+7=0 are r,s. Find r²+s² without finding either root.

Vieta gives r+s=6 and rs=7. Since r²+s²=(r+s)²−2rs, the answer is 36−14=22. The useful structure is symmetry in the roots.

Your turn: change one thing

Find a monic quadratic whose roots are the reciprocals of the roots of x²−5x+3.

Try this on paper before opening the explanation.

Compare your reasoning

The original roots are nonzero because their product is 3. The reciprocal sum is (r+s)/(rs)=5/3 and their product is 1/3. The monic equation is x²−(5/3)x+1/3=0, or equivalently 3x²−5x+1=0.

Pause and check

A trap to avoid: Wrong signs, missing the leading coefficient, or ignoring repeated roots.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

If r,s are roots of x²−ux+v=0, prove that r³+s³=u³−3uv.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

Expand (r+s)³=r³+s³+3rs(r+s). Substitute r+s=u and rs=v and rearrange. No assumption that r and s are distinct is needed, so the identity also covers repeated roots.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. The roots of x²−7x+10 have what sum?

Foundation

  1. 7
  2. -7
  3. 10
  4. -10
Hint

The sum is the negative coefficient of x in a monic quadratic.

Answer and reasoning

7. Here −b=−(−7)=7.

2. The roots of 2x²−3x+8 have what product?

Foundation

  1. 8
  2. -4
  3. 3
  4. 4
Hint

Divide the constant coefficient by the leading coefficient.

Answer and reasoning

4. The product is c/a=8/2=4.

3. r+s=5 and rs=4. Find r²+s².

Core

  1. 17
  2. 9
  3. 21
  4. 25
Hint

Expand the square of the sum.

Answer and reasoning

17. 25−2×4=17.

4. A monic quadratic has roots 2 and −3. Which is it?

Core

  1. x²−5x+6
  2. x²+x−6
  3. x²−x−6
  4. x²+x+6
Hint

Form (x−2)(x+3).

Answer and reasoning

x²+x−6. Expansion gives x²+x−6; the root sum is −1 and product −6.

5. When may you form the reciprocal-root equation using Vieta?

Stretch

  1. When every original root is nonzero
  2. Only when roots are integers
  3. Only when roots are distinct
  4. Only when the sum is zero
Hint

Reciprocals require nonzero denominators.

Answer and reasoning

When every original root is nonzero. For a quadratic this is equivalent to the product c/a being nonzero.

6. r+s=3 and rs=−2. Find r³+s³.

Stretch

  1. 45
  2. 9
  3. 21
  4. 33
Hint

Use (r+s)³−3rs(r+s).

Answer and reasoning

45. 27−3(−2)(3)=45.

7. For x²−sx+p=0, the product of its two roots is what?

Foundation

  1. s
  2. −p
  3. −s
  4. p
Hint

Compare with (x−α)(x−β).

Answer and reasoning

p. The constant term is αβ=p.

8. Roots of x²−5x+3 have sum of squares equal to what?

Core

  1. 31
  2. 19
  3. 25
  4. 22
Hint

Use (α+β)²−2αβ.

Answer and reasoning

19. 25−2·3=19.

9. Why can Vieta avoid solving for each root?

Stretch

  1. It assumes roots are integers
  2. Symmetric expressions depend on sums and products of roots
  3. It changes the roots
  4. It removes all equations
Hint

Use the coefficients as known symmetric quantities.

Answer and reasoning

Symmetric expressions depend on sums and products of roots. Many required expressions reduce directly to elementary symmetric sums.

Choose your next step

Continue to Symmetric polynomials. If this felt difficult, return to a prerequisite above. Every lesson stays open.

Open my revision list →

Original teaching material · IMOolympiad.com. Send a specific correction through our contact page. Learning progress is optional and stays in this browser.