Quadratic Roots–Coefficients (Vieta)

GCSE Algebra quadratic roots
Practice this formula All formulas
\( y=a(x-r_1)(x-r_2)\;\Rightarrow\; r_1+r_2=-\tfrac{b}{a},\; r_1r_2=\tfrac{c}{a} \)

Statement

If \(ax^2+bx+c=0\) has roots \(r_1, r_2\), then:

\[ r_1+r_2=-\frac{b}{a}, \quad r_1r_2=\frac{c}{a} \]

Why it’s true

  • Factorising gives \(ax^2+bx+c=a(x-r_1)(x-r_2)\).
  • Expanding: \(ax^2 - a(r_1+r_2)x + ar_1r_2\).
  • Comparing coefficients: \(-a(r_1+r_2)=b\), \(ar_1r_2=c\).
  • Thus: \(r_1+r_2=-b/a\), \(r_1r_2=c/a\).

Recipe (how to use it)

  1. Identify \(a,b,c\) from quadratic equation.
  2. Root sum = -b/a.
  3. Root product = c/a.
  4. Use these to find relationships or check solutions.

Spotting it

Used in questions asking for sum/product of roots without explicitly solving.

Common pairings

  • Quadratic formula.
  • Problems involving symmetric expressions of roots.
  • Algebraic manipulation (finding new quadratic with given root relations).

Mini examples

  1. Given: \(x^2-5x+6=0\).
    \(r_1+r_2=-b/a=5\), \(r_1r_2=6\).
  2. Given: \(2x^2+3x-2=0\).
    \(r_1+r_2=-3/2\), \(r_1r_2=-1\).

Pitfalls

  • Forgetting the minus sign in \(r_1+r_2=-b/a\).
  • Assuming works only for integer roots—it works always (real/complex).

Exam strategy

  • Quickly calculate sum/product before solving to check answers.
  • Great shortcut for symmetric expressions like \(r_1^2+r_2^2\).

Summary

Vieta’s formulas link coefficients to roots: \(r_1+r_2=-b/a\), \(r_1r_2=c/a\).