The value the interval has to throw out

k = 0.300

2.92Δ = 4 − 12k²

2real roots

two distinctverdict

Watch what happens as k passes through zero. The discriminant stays positive and the curve stops being a curve.

The figure needs JavaScript. It draws y = 3kx² + 2x + k as k changes, with the discriminant 4 − 12k² alongside. At k = 0.3 the discriminant is 2.92 and there are two roots; at k = 0.577 it is 0 and there is one repeated root; at k = 1 it is −8 and there are none. At k = 0 the discriminant reads 4 and the curve has become the straight line y = 2x, which has one root. Everything below stands on its own.

Three routes to a solution

Method	Use it when	Example
Factorising	The numbers are friendly. Fastest when it works.	x² − 5x + 6 = (x − 2)(x − 3), so x = 2 or 3.
Completing the square	You also want the vertex, or you are asked for an exact answer.	x² − 4x + 4 = (x − 2)², so x = 2 twice.
The formula	Always works. Reach for it when factorising fails.	x = (−b ± √Δ)/(2a).

The words roots, zeros and solutions all mean the same thing here, and all three appear in questions. A root of the equation is a zero of the function and a solution of both.

What the discriminant tells you

Inside the formula sits Δ = b² − 4ac, under the square root. That is the whole reason it matters:

Δ > 0: the square root term is a real number and the ± gives two distinct real roots. For x² − 5x + 6, Δ = 25 − 24 = 1.
Δ = 0: the square root term is zero, the ± adds nothing, and there are two equal real roots, which is one value counted twice. For x² − 4x + 4, Δ = 0 and the root is 2.
Δ < 0: there is no real square root and there are no real roots. For x² + x + 1, Δ = 1 − 4 = −3.

Δ = 0 means equal roots, not none. That is the commonest misreading, and the geometry says it plainly: the curve touches the x-axis rather than missing it.

The guide's own question, and the trap in it

For 3kx² + 2x + k = 0, read off a = 3k, b = 2, c = k:

Δ = 2² − 4(3k)(k) = 4 − 12k²

Then:

Two distinct real roots needs 4 − 12k² > 0, so k² < 1/3, so −0.577 < k < 0.577, and k ≠ 0.
Two equal real roots at k = ±0.577, exactly ±√3/3.
No real roots outside that interval, |k| > 0.577.

Why k = 0 has to go. Put it in and the x² term disappears: the equation becomes 2x = 0, a straight line through the origin with exactly one root. The discriminant still reads 4, happily positive, because the formula b² − 4ac does not know it is being asked about a quadratic. A discriminant cannot tell you whether you have a quadratic; it can only tell you about one you already have.

So the habit is: check a before you check Δ. If a contains the unknown, say so first. "For the equation to be quadratic we need k ≠ 0" is one line, it is a mark, and it is the line that separates a complete answer from a nearly complete one.

Where completing the square comes from, and why it is worth keeping. Apply it to ax² + bx + c = 0 in general and you get the quadratic formula; that is the derivation. It is also what turns an equation into a vertex, which is why 2.6 and 2.7 are really one idea. And on Paper 1 it is how you give an exact answer: x² − 4x + 1 = 0 becomes (x − 2)² = 3, so x = 2 ± √3, which no decimal will earn you.

On the GDC: the discriminant and the roots

Both machines solve a quadratic from its coefficients, which makes them a fast check on an algebraic answer. Neither will tell you that your a might be zero, so the thinking that this sub-topic tests is the part the machine cannot do.

When you may use it. Analysis Paper 1 is non-calculator, and discriminant questions live there: they want k = ±√3/3, not 0.577. Paper 2 allows the machine, and that is where a numerical root is an acceptable answer.

TI-Nspire CX II

On a Calculator page, define it once: Define d(a,b,c)=b x² -4*a*c. Use the x² key rather than typing ^2: the caret opens a superscript box and everything you type next stays inside it. Typed as b^2-4*a*c you define b to the power of 2−4ac instead, and d(1,-5,6) then returns 4.2×10⁻¹⁶ with no error at all, which looks like zero and is not
d(1,-5,6) gives 1, d(1,-4,4) gives 0, d(1,1,1) gives -3
For the k question the machine is no help: the permitted Nspire is non-CAS, so it cannot carry an undefined k through the formula. Work out 4 − 12k² yourself, then use d on numbers to spot-check it, for instance d(3*0.5,2,0.5) against 4 − 12(0.25) = 1
For the roots themselves, nSolve(x x² -5x+6=0,x) gives one root; a Graphs page and Analyze Graph → Zero gives both

Casio fx-CG50

MENU → Equation → F2 Polynomial → degree 2
Enter the three coefficients and F1 SOLVE: for 1, −5, 6 it returns x1=2, x2=3
With no real roots it reports complex values if Complex Mode is a+bi, and says there is no solution if it is Real. Set it deliberately
The discriminant itself is just arithmetic in Run-Matrix: (-5)^2-4×1×6

The mark people lose. Giving the interval and forgetting that a must not be zero. On 3kx² + 2x + k = 0 the discriminant is cheerfully positive at k = 0, so no amount of care with the inequality will catch it; you have to look at the a. The habit: write down a, b and c as three separate things before you write Δ, and if any of them contains the unknown, deal with it there and then.

Inequalities, where the answer is two pieces

The content list says equations and inequalities, and the inequality is the half people lose marks on, because the answer is often not an interval. Solve x² − 6x + 5 > 0.

Find the critical values, which are the roots: (x − 1)(x − 5) = 0, so x = 1 and x = 5.
Decide which side. a = 1 is positive, so the parabola opens upwards: it is below the axis between the roots and above it outside them.
We want above, so the answer is x < 1 or x > 5.

Check it with one number from each region, which takes seconds: at x = 0 the expression is 5, positive, in; at x = 3 it is −4, negative, out; at x = 6 it is 5 again, in. Three substitutions and the answer is settled.

"x < 1 or x > 5" is not "1 < x < 5" written carelessly. They are opposite sets, and only one of them can be right. The sketch is what tells you which: a positive a gives a parabola that is negative between its roots, so a > 0 question asking for > 0 always gives two pieces, and a > 0 question asking for < 0 always gives the single interval between them.

And when a is negative, every direction reverses. Solve −x² + 4x − 3 ≥ 0. Multiply through by −1 and flip the sign: x² − 4x + 3 ≤ 0, which is (x − 1)(x − 3) ≤ 0, so 1 ≤ x ≤ 3. Check: at x = 2 the original is 1, which is indeed at or above zero, and at x = 0 it is −3, which is not.

The safer habit is to not multiply at all: leave it as it is, note that a < 0 means the parabola opens downwards, and read off that a downward parabola is at or above zero between its roots. Same answer, one fewer place to drop a sign.

The mark people lose. Writing 5 < x < 1. It is what comes out of "x < 1 or x > 5" if you try to compress two pieces into one interval, and it describes the empty set. If the two pieces do not join, say "or" and leave them apart.

Your turn

1. Find the discriminant of x² − 4x + 4.

2. How many distinct real roots does x² + x + 1 have?

3. For 3kx² + 2x + k = 0, find the positive value of k giving two equal roots, to 3 decimal places.

4. How many roots does the equation have when k = 0?

5. The discriminant at k = 0 is 4, which is positive. Why is k = 0 still excluded?

Question 5. The discriminant at k equals 0 is 4, which is positive. Why is k equals 0 still excluded? Because 4 is not big enough. Because at k = 0 there is no x² term, so it is not a quadratic and has one root rather than two. Because the discriminant formula is wrong at k = 0. It is not excluded; the interval is correct as it stands.

Where the marks go

1 marka, b and c identified, with any condition on a stated.

1 markThe discriminant, in terms of the unknown.

1 markThe inequality or equation solved.

1 markThe answer as a set of values, exact on Paper 1.

The first mark is the one this page exists for. On any question where a contains the unknown, the condition a ≠ 0 is worth a mark on its own and takes seven words.

Three routes to a solution

What the discriminant tells you

The guide's own question, and the trap in it

On the GDC: the discriminant and the roots

TI-Nspire CX II

Casio fx-CG50

Inequalities, where the answer is two pieces

Your turn

Want a verdict on your own draft?