Which of the following quadratic equations has roots as two consecutive integers.?

Option 4 : a2 + 39a – 70 = 10a – 280

Calculation:

Option 1:

13m + m2 – 250 = 4m – 30

⇒ m² + 13m - 4m - 250 + 30 = 0

⇒ m² + 9m - 220 = 0

⇒ m² + 20m - 11m - 220 = 0

⇒ m(m + 20) - 11(m + 20) = 0

⇒ (m - 11) (m + 20) = 0

So, m = 11, -20

Option 2:

t2 + 20t + 250 = 50 – 10t

⇒ t² + 20t + 10t + 250 - 50 = 0

⇒ t² + 30t + 200 = 0

⇒ t² + 10t + 20t + 200 = 

⇒ t(t + 10) + 20 (t + 20) = 0

⇒ (t + 10) (t + 20) = 0

So, t = -10, -20

Option 3,

10p + 280 = p2 + 23p – 50

⇒ p² + 23p - 10p - 50 - 280 = 0

⇒ p² + 13p - 330 = 0

No integer solution is possible for this equation.

Option 4:

a2 + 39a – 70 = 10a – 280

⇒ a² + 39a - 10a - 70 + 280 = 0

⇒ a² + 29a + 210 = 0

⇒ a² + 15a + 14a + 210 = 0

⇒ a(a + 15) + 14(a + 15) = 0

⇒ (a + 15) (a + 14) = 0

So, a = -14, -15

∴ a2 + 39a – 70 = 10a – 280 has roots as two consecutive integers.