Question

find three consecutive positive integers such that the sum of the square of the first and the product of the other two is 154​

Answer 1

Answer:

$\large{\underline{\boxed{\sf 8,9 \: and \: 10}}}$

Step-by-step explanation:

Let the three consecutive numbers be x, (x + 1) and (x + 2) respectively.

According to the question;

x² + (x + 1)(x + 2) = 154

⇒ x² + x² + 2x + x + 2 = 154

⇒ 2x² + 3x + 2 - 154 = 0

⇒ 2x² + 3x - 152 = 0

On splitting the middle term ;

⇒ 2x² + 19x - 16x - 152 = 0

⇒ 2x(x - 8) + 19( x - 8) = 0

⇒ (x - 8)(2x + 19) = 0

⇒ x = 8 or x = -$\sf \dfrac{19}{2}$

So, taking the positive value.

• x = 8

Thus, the required numbers are ;

• x = 8
• x + 1 = 8 + 1 = 9
• x + 2 = 8 + 2 = 10

Hence, the required numbers are 8, 9 and 10.

Answer 2

Step-by-step explanation:

let the three consecutive number be- x, x+1 and x+2

$x { }^{2} + (x + 1)(x + 2) = 154\\ \\ x {}^{2} + x {}^{2} + 2x + x + 2 = 154 \\ 2x {}^{2} + 3x - 152 = 0 \\ 2x {}^{2} + (19 + 16)x - 152 = 0 \\ 2x {}^{2} + 19x - 16x - 152 = 0 \\ x(2x + 19) - 8(2x + 19) = 0 \\ (2x + 19) (x - 8) = 0 \\ x = - 19 \div 2 \: \: and \: \: x = 8$

we cannot take -19/2 as it is negative

so x=8

three consecutive number are 8, 9 and 10.

best of luck for 10th