Quadratic Roots–Coefficients (Vieta)

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: \(r_1+r_2=-b/a\), \(r_1r_2=c/a\).