Discriminant and Vieta

In one sentence

The three coefficients of a quadratic equation $a x^{2} + b x + c = 0$ ( $a \neq = 0$ ) carry two messages about the roots, and neither needs you to solve for the roots:

sign of Δ = b^{2} - 4 a c \Rightarrow how many real roots x_{1} + x_{2} = - \frac{b}{a}, x_{1} x_{2} = \frac{c}{a}

The discriminant answers “are there any?”; Vieta’s formulas answer “what do they add up to and multiply to?”. Very often that is all a problem asks.

Watch: y = x² − 4x + c as c grows

Step 1 / 4

c = 3
The graph meets the $x$ -axis at two points, and $Δ = 16 - 12 = 4 > 0$ . The roots are $1$ and $3$ : their sum is $4$ and their product is $3$ .

$Δ = 4 > 0$
c = 4
c = 5
What did you see?

Predict first

The equation $x^{2} - 6 x + k = 0$ has two equal real roots. What is $k$ ?

Choose an answer first; then you can see the result.

Stop 1 · The discriminant (Vol. 2)

“Has two equal real roots” translates into one equation: $Δ = 0$ . A question about $x$ becomes an equation in the letter $k$ .

Example: for which k does x² + kx + (k + 3) = 0 have two equal real roots?

Step 1 / 5

Try one value first
When $k = 0$ the equation is $x^{2} + 3 = 0$ . The graph is entirely above the $x$ -axis: no roots.

$k = 0 : Δ = - 12$
Write the discriminant
Set it equal to 0
Check k = 6
Check k = −2

Stop 2 · Vieta’s formulas (Vol. 2)

The two roots of $x^{2} - 5 x + 3 = 0$ are not integers: written out, they involve $13$ . But as long as the expression you want does not change when the two roots swap, you can avoid the roots altogether and use only their sum $5$ and product $3$ .

Below, call the two roots $a$ and $b$ , so $a + b = 5$ and $ab = 3$ .

Example: without solving, find a² + b², (a − b)² and a³ + b³

Step 1 / 4

First check there are roots
$Δ = 25 - 12 = 13 > 0$ , so two real roots really exist. Do not skip this step.

$Δ = 13 > 0$
Sum of squares
Square of the difference
Sum of cubes

Why can every expression that is unchanged by the swap be rewritten like this? That story belongs to another line: the Symmetry line (M5).

Stop 3 · Functions and equations (Vol. 3)

“How many common points do a line and a parabola have?” is the same as “how many roots does the quadratic equation you get by combining them have?”. A question about graphs is handed to the discriminant.

Example: the line y = 2x + m and the parabola y = x² have only one common point. Find m

Step 1 / 5

Combine them
The $x$ -coordinate of a common point satisfies $x^{2} = 2 x + m$ , that is, $x^{2} - 2 x - m = 0$ . The figure shows $y = x^{2} - 2 x - m$ : its roots are the $x$ -coordinates of the common points.

$x^{2} - 2 x - m = 0$
When m = 3
Only one common point
Read off the point of tangency
Further down

Stop 4 · Integer roots (Vols. 2 and 3)

When both roots must be integers, Vieta’s formulas turn “a quadratic equation” into “the sum and product of two integers”. After that it is an integer game.

Example: find all integers m for which both roots of x² − mx + m + 2 = 0 are integers

Step 1 / 5

Write down Vieta’s formulas
Let the two integer roots be $x_{1} \leq x_{2}$ .

$x_{1} + x_{2} = m, x_{1} x_{2} = m + 2$
Eliminate m
Make a product
List the cases
Back to m

Practice

Exercise 1Basic

$x_{1}, x_{2}$ are the two roots of $x^{2} - 7 x + 5 = 0$ . Without solving the equation, find $x_{1}^{2} + x_{2}^{2}$ .

Exercise 2Advanced

The equation $x^{2} + (k - 1) x + k + 2 = 0$ in $x$ has two equal real roots. Find all possible values of $k$ . (Separate them with commas.)

Exercise 3Challenge

The real numbers $x_{1}, x_{2}$ are the two roots of the equation $x^{2} - 2 k x + k + 2 = 0$ in $x$ . Find the minimum value of $x_{1}^{2} + x_{2}^{2}$ .

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

Using Vieta without checking the discriminant

The real numbers

x_{1}, x_{2}

are the two roots of

x^{2} - 2 k x + k + 6 = 0

x_{1}^{2} + x_{2}^{2} = (2 k)^{2} - 2 (k + 6) = 4 k^{2} - 2 k - 12

, which is smallest when

k = \frac{1}{4}

, with minimum

- \frac{49}{4}

What went wrong? Think first, then open

Diagnosis The sum of the squares of two real numbers can never be negative: the answer itself is shouting “something is wrong”. When $k = \frac{1}{4}$ , $Δ = 4 k^{2} - 4 k - 24 < 0$ , so the equation has no real roots at all, and “the sum of the roots” and “the product of the roots” are promises nobody can keep.

First make sure there are roots:

Δ = 4 (k - 3) (k + 2) \geq 0

, that is,

k \leq - 2

k \geq 3

. In this range,

4 k^{2} - 2 k - 12

is smallest at

k = - 2

, with minimum

16 + 4 - 12 = 8

“First make sure the roots exist, then talk about their sum and product”

Exhibit

Forgetting it must be a quadratic equation first

The equation

k x^{2} - 4 x + 2 = 0

has two different real roots, so

Δ = 16 - 8 k > 0

, which gives

k < 2

What went wrong? Think first, then open

Diagnosis $k = 0$ satisfies $k < 2$ , but then the equation is $- 4 x + 2 = 0$ , which has only one root. The discriminant is a tool for quadratic equations: before using it, check that the coefficient of $x^{2}$ is not $0$ .

k < 2

and

k \neq = 0

“First check whether a is 0, then compute Δ”

In one sentence

Stop 1 · The discriminant (Vol. 2)

Stop 2 · Vieta’s formulas (Vol. 2)

Stop 3 · Functions and equations (Vol. 3)

Stop 4 · Integer roots (Vols. 2 and 3)

Practice

Mistake Museum · Exhibits from this stop

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