📝 Explanation:

x² – x = 72

x = 12

Roots are imaginary (Discriminant < 0)

📝 Explanation: Only option A is in standard quadratic form ax² + bx + c = 0.

📝 Explanation:

D = b² – 4ac

= 9 – 40

= –31 ⇒ no real roots.

If a quadratic equation ax² + bx + c = 0 has two equal real roots, we know that, discriminant b² – 4ac = 0

(i) 2x² + kx + 3 = 0

a = 2, b = k, c = 3

b² – 4ac = 0

(k)² – 4(2)(3) = 0

k² – 24 = 0

k² = 24

k = √24

k = ± √2 × 2 × 2 × 3

k = ± 2√6

Roots imply sum = 2,

product = –1 ⇒ x² – 2x – 1 = 0

Let the speed of the train be s km/hr and the time taken be t hours.

Distance = Speed × Time

360 = s × t

⇒ t = 360 / s

Increased speed of the train can be written as s + 5

New time to cover the same distance = t – 1

(s + 5) × (t – 1) = 360 ….(1)

st – s + 5t – 5 = 360

360 – s + 5(360/s) – 5 = 360 [Since, st = 360 and t = 360 / s]

– s + 1800/s – 5 = 0

– s² + 1800 – 5s = 0

s² + 5s – 1800 = 0

We will solve this quadratic equation by quadratic formula

ax² + bx + c = 0

X = -b ± √ (b2 – 4ac) / 2a

Comparing s² + 5s – 1800 = 0 with ax2 + bx + c = 0, we get a = 1, b = 5, c = – 1800

b² – 4ac = (5)2 – 4(1)(- 1800)

= 25 + 7200

= 7225 > 0

Hence, the real roots exist.

x = [-b ± √ (b2 – 4ac)] / 2a

s = (- 5 ± √ 7225) / 2

s = (- 5 ± 85) / 2

s = (- 5 + 85) / 2 and s = (- 5 – 85) / 2

s = 80 / 2 and s = – 90 / 2

s = 40 and s = – 45

Speed of the train cannot be a negative value.

Therefore, speed of the train is 40 km /hr.