Q.Let $S$ be the sum, $P$ the product and $R$ the sum of reciprocals of $n$ terms of a G.P. Prove that $P^2 R^n = S^n$.

The Quadratic Discriminant Condition — First Encounter

Imagine you're asked to solve $x^2 - 5x + 6 = 0$. You factor it: $(x-2)(x-3)=0$, so $x=2$ or $x=3$. Two clean, real answers.

Now try $x^2 - 2x + 5 = 0$. Factor? It doesn't work nicely. You try the quadratic formula and get $x = 1 \pm 2i$ — complex numbers, not real at all.

What about $x^2 - 4x + 4 = 0$? 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 $ax^2 + bx + c = 0$ (with $a \neq 0$) 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 $\Delta = b^2 - 4ac$ is the quantity under the square root in the quadratic formula:

The square root is the gatekeeper. If what's inside is positive, you get two different real numbers. If zero, you get one (the $\pm$ gives the same thing). If negative, the square root is imaginary — no real solutions.

The Precise Statement

For the quadratic equation $ax^2 + bx + c = 0$ where $a, b, c$ are real numbers and $a \neq 0$, define the discriminant:

Then:

That's the entire condition. Three cases, one number.

Why It Works — A Quick Proof

The quadratic formula is derived by completing the square:

The left side is a square — always $\geq 0$ for real $x$. So the right side must also be $\geq 0$ for a real solution. The right side's sign is entirely determined by $b^2 - 4ac$ (since $4a^2 > 0$). Hence:

• If $b^2 - 4ac > 0$, the right side is positive → two real square roots → two real $x$.
• If $b^2 - 4ac = 0$, the right side is zero → one real $x$.
• If $b^2 - 4ac < 0$, the right side is negative → no real square root → no real $x$.

Worked Examples

Example 1: $2x^2 - 4x + 1 = 0$

$a=2$, $b=-4$, $c=1$.

$\Delta = (-4)^2 - 4(2)(1) = 16 - 8 = 8 > 0$ → two distinct real roots.

Example 2: $x^2 + 6x + 9 = 0$

$a=1$, $b=6$, $c=9$.

$\Delta = 36 - 4(1)(9) = 36 - 36 = 0$ → one repeated real root (indeed, $(x+3)^2=0$).

Example 3: $3x^2 - 2x + 5 = 0$

$a=3$, $b=-2$, $c=5$.

$\Delta = 4 - 4(3)(5) = 4 - 60 = -56 < 0$ → no real roots.

Why This Matters for Exams …