Question

three consecutive positive integers are such that the sum of the square of the first and the product of other two numbers is 46. find the integers​

Answer 1

Answer:

4, 5, 6

Step-by-step explanation:

let the first number be x

since numbers are consecutive

then ,

next number = x + 1 , x + 2

Given that,

the square of the first number + the product of other two numbers = 46

ATQ

x² + (x + 1)(x + 2) = 46

x² + x² + 2x + x + 2 = 46

2x² + 3x + 2 = 46

2x² + 3x + 2 - 46 = 0

2x² + 3x - 44 = 0

2x² - 8x + 11x - 44 = 0

2x(x - 4) +11(x - 4) = 0

(2x + 11)(x - 4) = 0

2x + 11 = 0

2x = -11

x = -11/2 [given that the number is positive]

now,

x - 4 = 0

x = 4

so,

first number = 4

second number = x + 1 = 4 + 1 = 5

third number = x + 2 = 4 + 2 = 6

so, the required integers are 4 , 5 , 6

Answer 2

Answer:

Hey mate I will you

Step-by-step explanation:

Let the number be ' a'

Therefore the convective number will be

a,a+1and a+2

According to question

It is given that

The that the sum of the square of the first and the product of other two numbers is 46

Therefore,

${a}^{2} + ( a + 1 \times +( a + 2)) = 46 \\ \\ {a}^{2} + {a}^{2} + 2a \: + a + 2 = 46 \\ \\ {2a}^{2} + 3a + 2 = 46 \\ \\ {2a}^{2} +3a - 44 = 0 \\ by \: using \: middle \: term \: spliting \\ 2a(a - 4) + 11(a - 4) = 0 \\ (2a + 11) \: (a - 4) = 0 \\ \\ 2a + 11 = 0 \: or \: a - 4 = 0 \\ a = - \frac{11}{2} and \: a \: = 4 \\ \\ but \: \: a \: is \: not \: equal \: to \: 0 \: because \: it \: given \: positive \: only$

The fore

Therefore,

X=4

X+1=4+1=5

X+2=4+2=6

the integers are 4,5,6