Question

Discuss the nature of the roots of the following equation without actually solving it:

9x² – 6x + 1 = 0

Answer 1

Given: 9x² – 6x + 1 = 0

Step-wise calculation:

1. Compare with ax² + bx + c = 0:

a = 9, b = –6, c = 1

2. Discriminant D = b² – 4ac 

= (–6)² – 4 × 9 × 1 

= 36 – 36

= 0

So, D = 0 (Discriminant rule ⇒ Equal roots).

3. When D = 0 the quadratic formula gives `x = (-b)/(2a)`, i.e. both roots equal to `(-b)/(2a)`.

4. Compute the common root:

`x = -(-6)/(2 xx 9)`

= `6/18`

= `1/3`, which is a rational number.

More generally, if a, b are integers and D = 0 then `(-b)/(2a)` is rational.

5. Alternatively note the trinomial is a perfect square:

(3x – 1)² = 9x² – 6x + 1, so the equation is (3x – 1)² = 0 and hence `x = 1/3` (double root).

The equation has two equal real roots (a repeated root) and that root is rational: `x = 1/3`.