The Quadratic Discriminant Condition — First Encounter
Imagine you're asked to solve x2−5x+6=0. You factor it: (x−2)(x−3)=0, so x=2 or x=3. Two clean, real answers.
Now try x2−2x+5=0. Factor? It doesn't work nicely. You try the quadratic formula and get x=1±2i — complex numbers, not real at all.
What about x2−4x+4=0? That's (x−2)2=0, so only x=2 (a repeated root).
Three different behaviours from three quadratics. The discriminant is the single number that tells you, before you solve, which case you're in.
The Intuition
A quadratic equation ax2+bx+c=0 (with a=0) represents a parabola. The solutions are where this parabola crosses the x-axis.
If it crosses at two distinct points → two real roots.
If it just touches the axis at one point → one repeated real root.
If it never touches the axis → no real roots (two complex roots).
The discriminant Δ=b2−4ac is the quantity under the square root in the quadratic formula:
x=2a−b±b2−4ac
The square root is the gatekeeper. If what's inside is positive, you get two different real numbers. If zero, you get one (the ± gives the same thing). If negative, the square root is imaginary — no real solutions.
Note
The name "discriminant" comes from Latin discriminare — to distinguish. It discriminates between the three possible root types.
The Precise Statement
For the quadratic equation ax2+bx+c=0 where a,b,c are real numbers and a=0, define the discriminant:
Δ=b2−4ac
Then:
Condition on Δ
Nature of roots
Real?
Δ>0
Two distinct real roots
Yes
Δ=0
One real root (repeated)
Yes
Δ<0
Two complex conjugate roots
No
Δ=b2−4ac
That's the entire condition. Three cases, one number.
Why It Works — A Quick Proof
The quadratic formula is derived by completing the square:
ax2+bx+c=0⟹(x+2ab)2=4a2b2−4ac
The left side is a square — always ≥0 for real x. So the right side must also be ≥0 for a real solution. The right side's sign is entirely determined by b2−4ac (since 4a2>0). Hence:
If b2−4ac>0, the right side is positive → two real square roots → two real x.
If b2−4ac=0, the right side is zero → one real x.
If b2−4ac<0, the right side is negative → no real square root → no real x.
Watch out
A common mistake: forgetting that a must be non-zero. If a=0, it's not a quadratic — it's linear, and the discriminant formula doesn't apply.
Worked Examples
Example 1:2x2−4x+1=0
a=2, b=−4, c=1.
Δ=(−4)2−4(2)(1)=16−8=8>0 → two distinct real roots.
Example 2:x2+6x+9=0
a=1, b=6, c=9.
Δ=36−4(1)(9)=36−36=0 → one repeated real root (indeed, (x+3)2=0).
The key idea is that the given quadratic expression in p is always non-positive, which forces its discriminant to be non-positive (since the coefficient of p2 is positive).
Step 1: Treat the expression as a quadratic in p:
(a2+b2+c2)p2−2(ab+bc+cd)p+(b2+c2+d2)≤0.
The leading coefficient a2+b2+c2>0 (since a,b,c are real and not all zero; if they were all zero the inequality would force b=c=d=0, contradicting "different real numbers").
Step 2: For a quadratic with positive leading coefficient to be ≤0 for some real p, its discriminant must be ≥0 (to have real roots). But here the inequality holds for all real p? Actually, the condition is that there exists some p satisfying it — the most restrictive case is when the quadratic is a perfect square (discriminant =0), giving a single p where the expression equals zero.
The inequality is a quadratic in p that is always non-positive, so its discriminant must be non-positive. This forces a condition that makes a,b,c,d consecutive terms of a geometric progression.
The problem gives you an inequality involving p, but p itself is just a real number — it's not fixed. The trick is to see the left-hand side as a quadratic expression in p:
(a2+b2+c2)p2−2(ab+bc+cd)p+(b2+c2+d2)≤0
Since a,b,c,d,p are real, this quadratic in p is never positive. For a quadratic Ax2+Bx+C to be ≤0 for some real x, its discriminant must be ≥0 (so it has real roots). But here the inequality holds for a particularp — we don't know which one. However, the coefficients themselves are sums of squares, so A=a2+b2+c2>0 (since a,b,c are different real numbers, at least one is non-zero). A quadratic with positive leading coefficient can be ≤0 only if its discriminant is non-negative and the value at the vertex is ≤0. But the key insight is different: we can complete the square or treat it as a perfect square condition.
Let's work through it step by step.
Recognise the structure.
The expression looks like it might be a perfect square of something like (ap−b)2+(bp−c)2+(cp−d)2. Let's check:
That's exactly the left-hand side! So the inequality becomes:
(ap−b)2+(bp−c)2+(cp−d)2≤0
Sum of squares is non-negative.
Each term (ap−b)2, (bp−c)2, (cp−d)2 is ≥0. Their sum is ≤0. The only way this can happen is if each term is exactly zero: