Zeroes and coefficients of a quadratic

For a quadratic $p(x) = ax^2 + bx + c$ ($a \ne 0$), suppose the zeroes are $\alpha$ and $\beta$. There is a stunning shortcut: the sum and product of $\alpha, \beta$ can be read directly off the coefficients, without ever solving the quadratic.

Idea

If $\alpha$ and $\beta$ are zeroes of $p(x) = ax^2 + bx + c$, then $p(x)$ factorises as $p(x) = a(x - \alpha)(x - \beta).$ Expanding the right side, $a(x - \alpha)(x - \beta) = a x^2 - a(\alpha + \beta) x + a \alpha \beta.$ Comparing with $a x^2 + bx + c$: $-a(\alpha + \beta) = b \implies \alpha + \beta = -\frac{b}{a},$ $a \alpha \beta = c \implies \alpha \beta = \frac{c}{a}.$

So:

$\boxed{\alpha + \beta = -\frac{b}{a}, \qquad \alpha \beta = \frac{c}{a}.}$

These two identities are sometimes called Vieta's formulas for the quadratic, and they are the bread and butter of this chapter.

Theorem and consequences

Theorem. For the quadratic $ax^2 + bx + c$ with zeroes $\alpha, \beta$: $\alpha + \beta = -\frac{b}{a}, \qquad \alpha \beta = \frac{c}{a}.$

A few immediate uses.

(i) Constructing a quadratic from given zeroes. Given $\alpha$ and $\beta$, the quadratic with leading coefficient $1$ is $x^2 - (\alpha + \beta) x + \alpha \beta.$ Any non-zero scalar multiple is also a valid answer.

(ii) Finding one zero when the other is known. If you know one zero $\alpha$, then $\beta = -b/a - \alpha$ or $\beta = (c/a)/\alpha$.

(iii) Reading the signs. The signs of $\alpha + \beta$ and $\alpha \beta$ tell you whether the zeroes are positive, negative, or of opposite sign. For example, $\alpha \beta > 0$ and $\alpha + \beta < 0$ means both zeroes are negative.

(iv) Symmetric functions. Expressions like $\alpha^2 + \beta^2$, $\alpha^3 + \beta^3$, $1/\alpha + 1/\beta$, $\alpha/\beta + \beta/\alpha$ are symmetric in $\alpha, \beta$. They can always be written in terms of $\alpha + \beta$ and $\alpha \beta$. For instance, $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta,$ $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3 \alpha \beta (\alpha + \beta),$ $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}.$

So Vieta's formulas let you compute many natural expressions without solving the quadratic.

Worked examples

Example 1. Find the zeroes of $p(x) = x^2 - 5x + 6$ and verify the sum/product relations.

Factor: $x^2 - 5x + 6 = (x - 2)(x - 3)$. Zeroes: $\alpha = 2, \beta = 3$.

Check: $\alpha + \beta = 5 = -(-5)/1$. ✓ $\alpha \beta = 6 = 6/1$. ✓

Example 2. If one zero of $x^2 - 4x + k = 0$ is $3$, find the other and $k$.

By Vieta: $\alpha + \beta = 4$ and $\alpha \beta = k$. With $\alpha = 3$, we get $\beta = 1$. Then $k = 3 \cdot 1 = 3$.

Example 3. Form a quadratic whose zeroes are $-3$ and $4$.

$\alpha + \beta = 1, \alpha \beta = -12$. Quadratic: $x^2 - x - 12$. (Or any non-zero multiple.)

Example 4. If $\alpha$ and $\beta$ are zeroes of $2x^2 - 7x + 3$, find $\alpha^2 + \beta^2$ and $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$.

$\alpha + \beta = 7/2$ and $\alpha \beta = 3/2$.

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta = (7/2)^2 - 2(3/2) = 49/4 - 3 = 37/4$.

$\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha \beta} = \dfrac{7/2}{3/2} = 7/3$.

Example 5. Find a quadratic whose zeroes are the reciprocals of the zeroes of $5 x^2 - 7 x + 1$.

Let $\alpha, \beta$ be zeroes of the given polynomial: $\alpha + \beta = 7/5$, $\alpha \beta = 1/5$.

Reciprocals $1/\alpha, 1/\beta$ have sum $\dfrac{\alpha + \beta}{\alpha \beta} = \dfrac{7/5}{1/5} = 7$, and product $\dfrac{1}{\alpha \beta} = 5$.

Required quadratic: $x^2 - 7x + 5$.

Try it yourself

1. Find the zeroes of $x^2 - 7x + 12$ and verify Vieta.
2. If one zero of $x^2 + (k - 3) x + 2k = 0$ is $-2$, find $k$.
3. Form a quadratic with zeroes $5$ and $-2$.
4. For $p(x) = x^2 - 8x + 15$, find $\alpha^2 + \beta^2$.
5. The zeroes of $3 x^2 - 8 x + 4$ are $\alpha, \beta$. Find $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$.
6. Form a quadratic whose zeroes are the squares of the zeroes of $x^2 - 5x + 4$.
7. If $\alpha + \beta = -3$ and $\alpha \beta = -10$, write the quadratic.
8. The zeroes of $x^2 - p (x + 1) - c$ are $\alpha, \beta$. Show that $(\alpha + 1)(\beta + 1) = 1 - c$.
9. For $p(x) = x^2 + 7 x + 10$, find $\alpha - \beta$ assuming $\alpha > \beta$.
10. Find the value of $k$ such that the quadratic $x^2 - (k+6) x + 2(2k - 1)$ has the sum of its zeroes equal to half their product.

Pitfalls / Insight

• Watch the sign in $\alpha + \beta = -b/a$. It is minus $b$, not $b$.
• Always divide by $a$. If the leading coefficient is not $1$, $b$ and $c$ are not directly the sum and product.
• Symmetric expressions in $\alpha, \beta$ can always be written via $(\alpha + \beta)$ and $\alpha \beta$ , don't try to find each zero separately if you don't have to.

Insight. Vieta's formulas turn algebra into arithmetic on coefficients. Combined with the geometry of the previous topic, you now have two complementary lenses: the graph and the coefficients.