Revision: Complex Numbers and Quadratic Equations JEE Main Complex Numbers and Quadratic Equations

For the quadratic equation ax² + bx + c = 0, a ≠ 0; the expression b² − 4ac is called the discriminant and is, in general, denoted by the letter 'D'.

Thus, discriminant D = b² − 4ac.

The quadratic equation whose roots are α and β is

x² − (α+β)x + αβ = 0

\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Given the roots of the equation (q – r)x² + (r – p)x + (p – q) = 0 are equal.

∴ Discriminant (D) = 0

⇒ b² – 4ac = 0

⇒ (r – p)² – 4 × (q – r) × (p – q) = 0

⇒ r² + p² – 2pr – 4[qp – q² – rp + qr] = 0

⇒ r² + p² – 2pr – 4qp + 4q² + 4rp – 4qr = 0

⇒ r² + p² + 2pr – 4qp – 4qr + 4q² = 0

⇒ (p + r)² – 4q(p + r) + 4q² = 0

Let (p + r) = y

⇒ y² – 4qy + 4q² = 0

⇒ (y – 2q)² = 0

⇒ y – 2q = 0

⇒ y = 2q

⇒ p + r = 2q

Hence proved.

D = b² – 4ac 

1. Write the given equation in the standard form

   ax² + bx + c = 0

2. Identify the values of a, b, and c.

3. Find the value of the discriminant

   D = b² − 4ac

4. Substitute the values of a, b, and D in the formula

   \[x=\frac{-b-\sqrt{b^2-4ac}}{2a}\]​

5. Simplify to obtain the roots.