Question

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

6x² + 7x – 10 = 0

Answer 1

Given: 6x² + 7x – 10 = 0

Step-wise calculation:

1. Compare with ax² + bx + c = 0: a = 6, b = 7, c = –10.

2. Discriminant Δ = b² – 4ac

= 7² – 4 × 6 × (–10) 

= 49 + 240

= 289

By the discriminant test Δ = b² – 4ac the sign of Δ determines whether roots are real/equal/imaginary.

3. Here Δ = 289 > 0 and Δ = 17² is a perfect square.

Δ > 0 ⇒ The roots are real and unequal (since Δ ≠ 0).

4. When the coefficients are integers or rationals and the discriminant is a perfect square, the square-root term in the quadratic formula is rational, so the two roots given by `(−b ± sqrt(Δ))/(2a)` are rational.

By contrast, a non-square positive Δ gives irrational roots e.g., `sqrt(48)` leads to irrational expressions.

Because Δ = 289 = 17² > 0, the equation has two distinct (unequal) real roots and since Δ is a perfect square and coefficients are integers, those roots are rational.

Hence, rational and unequal is correct.