The discriminant — how many solutions — Mathematics

Before this

This leads to

Sketching quadratic graphs
One linear and one quadratic equation
What is a complex number · coming soon

Three equations that look almost the same

Look at these three equations. They differ only in the last number, and only by 1 each time.

x^{2} - 6 x + 8 = 0 x^{2} - 6 x + 9 = 0 x^{2} - 6 x + 10 = 0

Yet they behave completely differently. The first has two solutions, $x = 2$ and $x = 4$ . The second has only one, $x = 3$ . The third has no real solutions at all. A change of 1 in one number changes the whole story.

The factorising lecture asked a question and left it open: can you tell which case you are in before you start solving? The last lecture ended by pointing at the answer. It is the part of the quadratic formula under the square root, $b^{2} - 4 a c$ . This lecture shows why that one number decides everything, and what else it can tell you.

The number under the root

The last lecture split the quadratic formula into two parts:

x = - \frac{b}{2 a} \pm \frac{b ^{2} - 4 a c}{2 a}

The first part is the centre: the $x$ -value of the vertex, halfway between the roots. The second part is a distance. One root lies that distance to the right of the centre, and the other lies the same distance to the left. Everything about how many roots there are lives in that second part. More exactly, it lives in the number under the square root.

That number, $b^{2} - 4 a c$ , is called the discriminant. The word comes from the Latin discriminare, "to distinguish", because it distinguishes between the cases. The English mathematician James Joseph Sylvester introduced the name in 1851. In this course we write it out in full, or call it $D$ when we need a letter. Some books write $Δ$ instead.

There are only three possibilities for any number: it is positive, zero or negative. Take them one at a time.

$D > 0$ . A positive number has a square root that is not zero. So the distance $\frac{D}{2 a}$ is not zero. Adding it and subtracting it give two different answers. There are two distinct real roots.
$D = 0$ . The square root of zero is zero. So the distance is zero, and adding it or subtracting it changes nothing. Both roots are the same number, $x = - \frac{b}{2 a}$ . There is one repeated root.
$D < 0$ . No real number squares to a negative number. So $D$ does not exist among the real numbers, and the formula cannot produce a real answer. There are no real roots.

The discriminant

For $a x^{2} + b x + c = 0$ , with $a \neq = 0$ , the discriminant is $D = b^{2} - 4 a c$ .

If $D > 0$ , there are two distinct real roots. If $D = 0$ , there is one repeated real root, $x = - \frac{b}{2 a}$ . If $D < 0$ , there are no real roots.

Try it on the three opening equations. Each has $a = 1$ and $b = - 6$ , so $b^{2} = 36$ . Then $D = 36 - 32 = 4$ , $D = 36 - 36 = 0$ and $D = 36 - 40 = - 4$ . Positive, zero, negative: two roots, one root, none. One short calculation each, and no solving at all.

Stop and think

A student says: "When the discriminant is 0, the equation has no solution, because 0 means nothing." Is the student right?

Show the answer

No. When $D = 0$ , the ± adds and subtracts zero, so the two roots land on the same number. There is still a solution: exactly one. For $x^{2} - 6 x + 9 = 0$ , $D = 0$ and the solution is $x = 3$ . Check: $9 - 18 + 9 = 0$ . It is "no real solutions" that needs $D$ to be negative, not zero.

Why it works: the height of the vertex

The formula explains the three cases. But there is a deeper reason, and it comes from the graph. This is the most important part of the lecture, so take your time here.

The last lecture completed the square on $a x^{2} + b x + c$ with letters. Halfway through, it reached this line:

(x + \frac{b}{2 a})^{2} = \frac{b ^{2} - 4 a c}{4 a ^{2}}

Multiply both sides by $a$ , and move everything to one side. This gives the completed-square form of every quadratic:

a x^{2} + b x + c = a (x + \frac{b}{2 a})^{2} - \frac{D}{4 a}

Now read it. The bracket is zero when $x = - \frac{b}{2 a}$ . At that point the graph is at its vertex, and its height is $- \frac{D}{4 a}$ . So the discriminant tells you how high or low the vertex is.

Take $a > 0$ first, so the parabola opens upwards, like a U. Then $4 a$ is positive, and the vertex height has the opposite sign to $D$ .

If $D > 0$ , the vertex is below the $x$ -axis. A U-shape whose lowest point is below the axis must cross it twice: once going down, and once coming back up.
If $D = 0$ , the vertex is on the axis. The graph touches the axis at one point and turns back.
If $D < 0$ , the vertex is above the axis. The lowest point of the U is already above zero, so the graph never reaches the axis.

The three opening equations as graphs, y = x² − 6x + c. Raising c lifts the whole parabola. With c = 8 the vertex is 1 below the axis (two roots, 2 and 4). With c = 9 it sits on the axis (one root, 3). With c = 10 it is 1 above the axis (no real roots). In each case the vertex height is −D ÷ 4.

What if $a < 0$ ? Then the parabola opens downwards, like an upside-down U, and its vertex is its highest point. Now $4 a$ is negative, so the vertex height has the same sign as $D$ . When $D > 0$ , the highest point is above the axis, and the graph comes down through it on both sides: two roots. When $D < 0$ , the highest point is below the axis, so the graph never reaches it. Both changes of sign happen together, and the rule stays exactly the same. That is why the rule needs no "if $a$ is positive" condition.

That is it. That is the whole idea. The discriminant measures how far the vertex is from the axis, on the side that matters. The number of roots is just the question "does the curve reach the axis?"

See it for yourself below.

Explore — lifting a parabola through the axis

The graph is y = x² − 6x + c. Move c. Watch the vertex move up and down, and watch the discriminant change sign at the moment the vertex crosses the axis. Try c = 5, then c = 9, then c = 12.

c = 5

The graph y = x² − 6x + c for five values of c. The vertex is always at x = 3, and its height is c − 9, which is −D ÷ 4.

c	D = 36 − 4c	Vertex height	Roots
5	16	−4	two: 1 and 5
8	4	−1	two: 2 and 4
9	0	0	one: 3
10	−4	1	none
13	−16	4	none

Did you see what happened as $c$ rose towards 9? The two roots slid towards each other, met at $x = 3$ , and then disappeared together. Roots of a quadratic never vanish one at a time. They merge into one and then go together.

More than a count

The discriminant tells you more than how many roots there are.

How far apart the roots are

The two roots are $\frac{D}{2 a}$ either side of the centre, so the gap between them is twice that: $\frac{D}{∣ a ∣}$ . (The bars mean the size of $a$ , without its sign. A distance is never negative.) For $x^{2} - 6 x + 8 = 0$ , the gap is $4 = 2$ , and indeed the roots 2 and 4 are 2 apart. As $D$ shrinks to zero, the gap shrinks to zero. This is why the single root is called a repeated root: it is two roots in the same place. Some books say "two equal roots" for the same thing.

Whether the roots are whole numbers or fractions

The last lecture's quick test belongs here. If $D$ is a perfect square (0, 1, 4, 9, …) and $a$ , $b$ , $c$ are integers, the roots are whole numbers or fractions, so the quadratic factorises. If $D$ is positive but not a perfect square, the roots contain a surd. Then go straight to the formula.

A shortcut: when a and c have opposite signs

Look at $3 x^{2} + 2 x - 7 = 0$ . Here $a = 3$ is positive and $c = - 7$ is negative. So $- 4 a c = - 4 \times 3 \times (- 7) = + 84$ , which is positive. And $b^{2}$ is never negative. So $D = b^{2} + 84$ must be positive, whatever $b$ is. There are two roots, and you knew it before you even looked at $b$ .

The graph says the same thing. When $x = 0$ , the height is $c$ . If the parabola opens upwards ( $a > 0$ ) but passes through a point below the axis ( $c < 0$ ), it must cross the axis on both sides of that point.

Stop and think

Does the shortcut work the other way? If $a$ and $c$ have the same sign, must there be no real roots?

Show the answer

No. If $a$ and $c$ have the same sign, $- 4 a c$ is negative, but $b^{2}$ can still be larger. For $x^{2} - 6 x + 8 = 0$ , $a$ and $c$ are both positive, yet $D = 36 - 32 = 4$ and there are two roots. The shortcut works in one direction only: opposite signs guarantee two roots, but the same signs guarantee nothing. You still have to calculate $D$ .

Worked examples

Worked Example 1 — Counting roots without solving

How many real roots does each equation have? (a) $x^{2} - 5 x + 4 = 0$ (b) $x^{2} + 6 x + 9 = 0$ (c) $2 x^{2} - 3 x + 5 = 0$

(a) $a = 1$ , $b = - 5$ , $c = 4$ . $D = (- 5)^{2} - 4 (1) (4) = 25 - 16 = 9$ . Positive, so two distinct real roots. Since 9 is a perfect square, they are whole numbers: $x = \frac{5 \pm 3}{2}$ , so 4 and 1.

(b) $D = 6^{2} - 4 (1) (9) = 36 - 36 = 0$ , so one repeated root, $x = - \frac{6}{2} = - 3$ . The left side is a perfect square, $(x + 3)^{2}$ . A zero discriminant and a perfect-square quadratic are the same thing.

(c) $D = (- 3)^{2} - 4 (2) (5) = 9 - 40 = - 31$ . Negative, so no real roots. That is the complete answer. There is nothing more to solve.

Check (c) with the vertex: its height is $- \frac{D}{4 a} = \frac{31}{8} \approx 3.9$ , above the axis, and the parabola opens upwards. So it never reaches zero.

Worked Example 2 — A ball in the air

A ball is thrown upwards. Its height after $t$ seconds is $h = 20 t - 5 t^{2}$ metres. Does it reach 15 m? 20 m? 25 m?

"Does it reach 15 m?" means "does $20 t - 5 t^{2} = 15$ have a solution?" Rearrange to zero: $5 t^{2} - 20 t + 15 = 0$ . Then $D = 400 - 4 (5) (15) = 400 - 300 = 100$ . Positive, so two times: the ball passes 15 m once on the way up and once on the way down. (They are $t = 1$ and $t = 3$ .)

For 20 m: $5 t^{2} - 20 t + 20 = 0$ , so $D = 400 - 400 = 0$ . One time only, $t = 2$ . The ball just reaches 20 m at the top of its flight and turns back.

For 25 m: $5 t^{2} - 20 t + 25 = 0$ , so $D = 400 - 500 = - 100$ . Negative: the ball never reaches 25 m.

Look at what the three cases mean here. Two roots: it passes that height twice. One root: that height is the very top. No roots: that height is out of reach. The discriminant has found the maximum height, 20 m, without any graph.

Worked Example 3 — Working backwards to find k

The equation $k x^{2} + 4 x + 1 = 0$ has a repeated root. Find $k$ , and find the root.

A repeated root means $D = 0$ . Here $a = k$ , $b = 4$ , $c = 1$ , so $D = 16 - 4 k$ .

16 - 4 k = 0 \Rightarrow k = 4

One more check is needed. The equation is only a quadratic if $a \neq = 0$ , that is $k \neq = 0$ . Here $k = 4$ , so that is fine.

The root is $x = - \frac{b}{2 a} = - \frac{4}{8} = - \frac{1}{2}$ . Check: $4 x^{2} + 4 x + 1 = (2 x + 1)^{2}$ , which is zero only when $x = - \frac{1}{2}$ .

The pattern for these questions: turn the condition on the roots into a condition on $D$ , then solve for the unknown.

Worked Example 4 — An inequality with two branches

Find the values of $k$ for which $x^{2} + k x + 9 = 0$ has two distinct real roots.

Two distinct roots means $D > 0$ . Here $D = k^{2} - 4 (1) (9) = k^{2} - 36$ . So we need $k^{2} > 36$ .

This is the step where students most often go wrong. It is tempting to write $k > 6$ and stop. But $k$ can be negative too: $(- 7)^{2} = 49$ , which is also greater than 36. The square of $k$ is large when $k$ is far from zero in either direction. So:

k^{2} > 36 \Leftrightarrow k > 6 or k < - 6

Test one value in each region. $k = 7$ : $D = 49 - 36 = 13 > 0$ , two roots. $k = - 7$ : $D = 13 > 0$ , two roots. $k = 4$ : $D = 16 - 36 = - 20 < 0$ , none. And at the edges, $k = \pm 6$ , $D = 0$ , which gives a repeated root, not two distinct ones. So the edges are not included.

Worked Example 5 — Proving roots exist for every k

Show that $x^{2} + k x - 3 = 0$ has two distinct real roots for every value of $k$ .

$D = k^{2} - 4 (1) (- 3) = k^{2} + 12$ . A square is never negative, so $k^{2} \geq 0$ . This means $k^{2} + 12 \geq 12$ , which is positive. So $D > 0$ for every $k$ , and there are always two distinct real roots.

Now check someone else's working. The mistake below is one of the most common in exams.

Spot the error — how many roots does 3x² = 5x − 4 have?

One line of this working is wrong. Tap the line you think is wrong. Every line has an explanation.

The working, with the mistake marked and corrected:

Line 1. a = 3, b = 5, c = −4. The mistake: the equation was not rearranged to zero first. As 3x² − 5x + 4 = 0, the values are a = 3, b = −5, c = 4.
Line 2. b² − 4ac = 25 − 4 × 3 × (−4). (This follows correctly from line 1.)
Line 3. = 25 + 48 = 73. (The arithmetic is right.)
Line 4. 73 > 0, so there are two distinct real roots. (This follows from the mistake. The correct discriminant is 25 − 48 = −23, so there are no real roots.)

Three classic mistakes

Trap 1 — "Zero means no solution"

$D = 0$ gives exactly one root, $x = - \frac{b}{2 a}$ . Only a negative discriminant means no real roots.

Trap 2 — Losing a branch

$k^{2} > 36$ means $k > 6$ or $k < - 6$ : two regions, outside the numbers. And $k^{2} < 36$ means $- 6 < k < 6$ : one region, between them. Test a value in each region to be sure.

Trap 3 — Reading a, b, c too early

Rearrange to $a x^{2} + b x + c = 0$ before you read off the values. Moving a term across the equals sign changes its sign, and the sign changes $D$ .

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Without solving, state how many real roots each equation has. (a) $x^{2} + 3 x - 10 = 0$ (b) $9 x^{2} - 30 x + 25 = 0$ (c) $- 2 x^{2} + x - 3 = 0$

Hint 1
In (a), look at the signs of $a$ and $c$ first.

Hint 2
In (b), $b^{2} = 900$ . In (c), $a = - 2$ , $b = 1$ , $c = - 3$ .

Full answer
(a) $a$ and $c$ have opposite signs, so two roots. Check: $D = 9 + 40 = 49$ . (b) $D = 900 - 4 \times 9 \times 25 = 900 - 900 = 0$ : one repeated root, $x = \frac{30}{18} = \frac{5}{3}$ . (c) $D = 1 - 4 (- 2) (- 3) = 1 - 24 = - 23$ : no real roots. The parabola opens downwards and its highest point is at height $- \frac{D}{4 a} = \frac{23}{- 8} \approx - 2.9$ , below the axis.
Find the value of $k$ for which $x^{2} - 8 x + k = 0$ has a repeated root, and find that root.

Hint 1
A repeated root means $D = 0$ .

Hint 2
$D = 64 - 4 k$ .

Full answer
$64 - 4 k = 0$ , so $k = 16$ . The root is $x = - \frac{b}{2 a} = \frac{8}{2} = 4$ . Check: $x^{2} - 8 x + 16 = (x - 4)^{2}$ .
Find the values of $k$ for which $x^{2} + k x + 4 = 0$ has no real roots.

Hint 1
No real roots means $D < 0$ . Write $D$ in terms of $k$ .

Hint 2
$k^{2} - 16 < 0$ , so $k^{2} < 16$ . Is the answer "between" or "outside"?

Full answer
$D = k^{2} - 16 < 0$ , so $k^{2} < 16$ . The square of $k$ is small when $k$ is close to zero, so $- 4 < k < 4$ . Test: $k = 3$ gives $D = 9 - 16 = - 7$ , no roots. $k = 5$ gives $D = 9$ , two roots. $k = 4$ gives $D = 0$ , a repeated root, so 4 itself is not included.
The straight line $y = m x + 1$ touches the curve $y = x^{2} + 2$ at exactly one point. Find the possible values of $m$ , and the point where each line touches.

Hint 1
Where the line meets the curve, the two $y$ -values are equal: $x^{2} + 2 = m x + 1$ .

Hint 2
Rearrange to $x^{2} - m x + 1 = 0$ . "Exactly one point" means this equation has a repeated root.

Full answer
$D = m^{2} - 4 = 0$ , so $m = 2$ or $m = - 2$ . For $m = 2$ : the repeated root is $x = \frac{m}{2} = 1$ , and $y = 1 + 2 = 3$ . So $y = 2 x + 1$ touches the curve at $(1, 3)$ . Check on the line: $2 \times 1 + 1 = 3$ . For $m = - 2$ : $x = - 1$ , and the line $y = - 2 x + 1$ touches at $(- 1, 3)$ . A line that touches a curve at one point without crossing it is called a tangent. The discriminant finds tangents without any drawing.

The bigger idea

The discriminant is the algebra of the graph. One number, $b^{2} - 4 a c$ , says whether the parabola crosses the axis, touches it or misses it. It says how far apart the crossings are, and whether they are whole numbers. A good calculation does not just give an answer; it tells you what kind of answer to expect.

And "no real roots" is not quite the end of the story. Mathematicians later asked what would happen if $- 1$ were allowed to exist. The answer is the complex numbers, studied at Pre-University level, where every quadratic has two roots (counting a repeated root twice).

Key concepts in this lesson

The discriminant of ax² + bx + c = 0 is D = b² − 4ac, the number under the square root in the quadratic formula.
D > 0 gives two distinct real roots, D = 0 gives one repeated root x = −b/2a, and D < 0 gives no real roots.
The vertex of y = ax² + bx + c is at height −D/4a, so the sign of D says whether the graph reaches the x-axis.
The roots are √D/|a| apart, and if D is a perfect square the quadratic factorises with whole numbers or fractions.
To find an unknown k from a condition on the roots, turn the condition into one on D and solve, keeping both branches of any squared inequality.