The Discriminant

Try this

Move $k$ slowly from $- 3$ to $8$ : the whole parabola rises, and its intersections with the $x$ -axis go from two to one to none. As the intersections move closer together, which way does the $Δ$ needle go?
When the needle stops exactly at $Δ = 0$ , what is $k$ ? Where is the vertex of the parabola then? What has happened to the two roots?
Move $k$ up one more step. $Δ$ becomes negative, and $Δ$ is exactly what sits under the square root in the quadratic formula. Can you take the square root of a negative number?
Here $Δ = 16 - 4 k$ . Work it out first: for which values of $k$ is $Δ > 0$ ? Then check your answer with the slider.

Predict first

The equation $x^{2} + b x + 9 = 0$ has two equal real roots. How many such values of $b$ are there?

Only

1

2

Infinitely manyNone

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

Why

Why is the single number $b^{2} - 4 a c$ enough? Complete the square and the answer is on the right of the equals sign.

The sign of Δ decides the number of roots

Step 1 / 5

Complete the square first
Complete the square for $a x^{2} + b x + c = 0$ ( $a \neq = 0$ ). The left side is a square, so it can never be negative. On the right the denominator $4 a^{2}$ is positive, so the sign of the right side depends only on the numerator $b^{2} - 4 a c$ . Give it a name: $Δ$ .

$(x + \frac{b}{2 a})^{2} = \frac{b ^{2} - 4 a c}{4 a ^{2}}$
Δ > 0: two distinct real roots
Δ = 0: two equal real roots
Δ < 0: no real roots
It works backwards too

The vertex tells the same story: the $y$ -coordinate of the vertex of $y = a x^{2} + b x + c$ is $- \frac{Δ}{4 a}$ . When the parabola opens upward, $Δ > 0$ means the vertex has sunk below the $x$ -axis, and $Δ < 0$ means it floats above.

The discriminant can prove “there are always roots”

Some equations contain a letter, yet you can be sure they have two distinct real roots whatever value the letter takes. The trick is to complete the square once more, this time on $Δ$ .

Example: $x^{2} - (m + 1) x + m - 2 = 0$ .

Δ = (m + 1)^{2} - 4 (m - 2) = m^{2} - 2 m + 9 = (m - 1)^{2} + 8 \geq 8 > 0.

A square plus $8$ can never be less than $8$ . So whatever $m$ is, the equation always has two distinct real roots.

Move

m

through every value: the needle always stays on the

Δ > 0

side, and its smallest value is

8

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: when $Δ > 0$ , the equation has two distinct real roots

The equation $k x^{2} - 4 x + 1 = 0$ with $k = 2$ : $Δ = 16 - 8 = 8 > 0$ , and the parabola meets the $x$ -axis at two points. All normal.

Practice

Exercise 1Basic

Without solving, work out the discriminant $Δ$ of $2 x^{2} - 3 x + 2 = 0$ and say what kind of roots it has. (Enter $Δ$ in the answer box.)

Exercise 2Basic

The equation $x^{2} - 6 x + m = 0$ has two equal real roots. Find $m$ .

Exercise 3Advanced

The equation $k x^{2} - 2 x - 1 = 0$ in $x$ has two distinct real roots. Find the smallest integer $k$ that works.

Exercise 4Challenge

For the equation $x^{2} - (m + 2) x + 2 m - 1 = 0$ , write $Δ$ as an expression in $m$ , complete the square, and find the minimum value of $Δ$ . Use this to show that whatever real number $m$ is, the equation has two distinct real roots.

Try this

Why

The sign of Δ decides the number of roots

The discriminant can prove “there are always roots”

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Working out Δ but forgetting that the x² coefficient cannot be 0

Writing “has real roots” as Δ > 0

Subtracting when c is negative