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Mathematics · Ch 3 — Theory of Equations

Vieta's formula for Quadratic Equations

3.3.1

Vieta's formula for Quadratic Equations

Let α,β\alpha,\beta be the roots of ax2+bx+c=0ax^2+bx+c=0. Since a,α,βa,\alpha,\beta determine the polynomial up to the factor aa,

ax2+bx+c=a(x−α)(x−β)=a(x2−(α+β)x+αβ).ax^2+bx+c = a(x-\alpha)(x-\beta) = a\big(x^2-(\alpha+\beta)x+\alpha\beta\big).

Comparing coefficients of like powers on both sides gives Vieta's formula for a quadratic:

α+β=−ba,αβ=ca.\alpha+\beta = -\frac ba, \qquad \alpha\beta=\frac ca.

Turning this around: a quadratic equation whose roots are α\alpha and β\beta is

x2−(sum of the roots) x+(product of the roots)=0.(1)x^2 - (\text{sum of the roots})\,x + (\text{product of the roots}) = 0. \qquad(1)

Note

The indefinite article "a" quadratic equation (not "the") matters: if P(x)=0P(x)=0 has roots α,β\alpha,\beta, then cP(x)=0cP(x)=0 (any nonzero constant cc) is also a quadratic equation with exactly the same two roots. Formula (1) gives one particular (monic) representative.

Worked illustration (constructing directly from given roots). A quadratic equation with roots 33 and 44 is, by (1), x2−7x+12=0x^2-7x+12=0.

Worked illustration (transforming a known equation's roots — Example 3.1 style). If α,β\alpha,\beta are the roots of 17x2+43x−73=017x^2+43x-73=0, we get α+β=−4317\alpha+\beta=-\tfrac{43}{17} and αβ=−7317\alpha\beta=-\tfrac{73}{17} without solving for α,β\alpha,\beta individually. To build a quadratic with roots α+2, β+2\alpha+2,\ \beta+2, we only need the sum and product of the new roots:

(α+2)+(β+2)=(α+β)+4=−4317+4=2517,(α+2)(β+2)=αβ+2(α+β)+4=−7317−8617+4=−9117.(\alpha+2)+(\beta+2) = (\alpha+\beta)+4 = -\tfrac{43}{17}+4=\tfrac{25}{17}, \qquad (\alpha+2)(\beta+2)=\alpha\beta+2(\alpha+\beta)+4=-\tfrac{73}{17}-\tfrac{86}{17}+4=-\tfrac{91}{17}.

By (1), x2−2517x−9117=0x^2-\tfrac{25}{17}x-\tfrac{91}{17}=0, i.e. 17x2−25x−91=017x^2-25x-91=0, has roots α+2,β+2\alpha+2,\beta+2. …

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