Quadratic Roots–Coefficients (Vieta)

\( y=a(x-r_1)(x-r_2)\;\Rightarrow\; r_1+r_2=-\tfrac{b}{a},\; r_1r_2=\tfrac{c}{a} \)
Algebra GCSE
Question 3 of 20
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\( For 2x^2-5x+2=0, find r1r2. \)

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
\( r1r2=c/a. \)

Explanation

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

Worked examples

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  1. \( For x^2-5x+6=0, find r1+r2 and r1r2. \)
    1. \( a=1,b=-5,c=6 \)
    2. \( r1+r2=-b/a=5 \)
    3. \( r1r2=c/a=6 \)
    Answer: \( r1+r2=5, r1r2=6 \)
  2. \( For x^2+3x+2=0, find r1+r2 and r1r2. \)
    1. \( a=1,b=3,c=2 \)
    2. \( r1+r2=-3 \)
    3. \( r1r2=2 \)
    Answer: \( r1+r2=-3, r1r2=2 \)
  3. \( For 2x^2+3x-2=0, find r1+r2 and r1r2. \)
    1. \( a=2,b=3,c=-2 \)
    2. \( r1+r2=-3/2 \)
    3. \( r1r2=-1 \)
    Answer: \( r1+r2=-3/2, r1r2=-1 \)
  4. \( For x^2-4x-5=0, find r1+r2 and r1r2. \)
    1. \( a=1,b=-4,c=-5 \)
    2. \( r1+r2=4 \)
    3. \( r1r2=-5 \)
    Answer: \( r1+r2=4, r1r2=-5 \)
  5. \( For 3x^2-2x+1=0, find r1+r2 and r1r2. \)
    1. \( a=3,b=-2,c=1 \)
    2. \( r1+r2=2/3 \)
    3. \( r1r2=1/3 \)
    Answer: \( r1+r2=2/3, r1r2=1/3 \)
  6. \( For 2x^2-7x+3=0, find r1+r2 and r1r2. \)
    1. \( a=2,b=-7,c=3 \)
    2. \( r1+r2=7/2 \)
    3. \( r1r2=3/2 \)
    Answer: \( r1+r2=7/2, r1r2=3/2 \)
  7. \( For x^2+2x+5=0, find r1+r2 and r1r2. \)
    1. \( a=1,b=2,c=5 \)
    2. \( r1+r2=-2 \)
    3. \( r1r2=5 \)
    Answer: \( r1+r2=-2, r1r2=5 \)
  8. \( For 5x^2+6x+1=0, find r1+r2 and r1r2. \)
    1. \( a=5,b=6,c=1 \)
    2. \( r1+r2=-6/5 \)
    3. \( r1r2=1/5 \)
    Answer: \( r1+r2=-6/5, r1r2=1/5 \)
  9. \( For x^2-8x+12=0, find r1+r2 and r1r2. \)
    1. \( a=1,b=-8,c=12 \)
    2. \( r1+r2=8 \)
    3. \( r1r2=12 \)
    Answer: \( r1+r2=8, r1r2=12 \)
  10. \( For 4x^2-3x-5=0, find r1+r2 and r1r2. \)
    1. \( a=4,b=-3,c=-5 \)
    2. \( r1+r2=3/4 \)
    3. \( r1r2=-5/4 \)
    Answer: \( r1+r2=3/4, r1r2=-5/4 \)