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There are three consecutive integers such that the square of the first increased by the product of the other two gives 154. What are the integers?

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7,8,11

8,9,10

11,12,13

12,14,15

answer is B.

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Detailed Solution

Given, the sum of the square of the number and product of the other two numbers is 154. Also, the three numbers are consecutive integers respectively.
Assume the consecutive integer numbers as x, x + 1 and x + 2.
According to the given condition,

⇒ x 2 + (x + 1) (x + 2) = 154 ⇒ x 2 + x 2 + x + 2 x + 2 = 154 ⇒ 2 x 2 + 3 x + − 152 = 0

Comparing with general form,

a x^{2} + bx + c = 0

a = 2

b = 3

c = - 152

We know that the quadratic formula to find the value of x,
⇒

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

⇒ x = − 3 ± 32 − 4 × 2 × (− 152) 2 × 2 ⇒ x = − 3 ± 9 + (8 × 152) 4 ⇒ x = − 3 ± 9 + 1216 4

⇒ x = − 3 ± 1225 4 ⇒ x = − 3 ± 35 4

That is, x can be

− 3 + 35 4 or − 3 − 35 4

x = − 3 + 35 4 ⇒ x = 324 ⇒ x = 8

x = − 3 − 35 4 ⇒ x = − 38 4 ⇒ x = − 9.5

Since -9.5 is not an integer, x = 8 is the only solution.
Thus, the three consecutive integers are 8, 9 and 10 and the correct option is 2.

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