Question

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

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that three consecutive positive integers as

First positive integer = x

Second positive integer = x + 1

Third positive integer = x + 2

According to statement, the sum of the square of the first and the product of the other two is 46.

$\rm \: {x}^{2} + (x + 1)(x + 2) = 46$

$\rm \: {x}^{2} + {x}^{2} + 2x + x + 2 = 46$

$\rm \: 2{x}^{2} + 3x + 2 = 46$

$\rm \: 2{x}^{2} + 3x + 2 - 46 = 0$

$\rm \: 2{x}^{2} + 3x - 44 = 0$

Now, on splitting the middle terms, we get

$\rm \: 2{x}^{2} + 11x - 8x - 44 = 0$

$\rm \: x(2x + 11) - 4(2x + 11) = 0$

$\rm \: (2x + 11)(x - 4) = 0$

$\rm\implies \:x = 4 \: \: or \: \: x = - \frac{11}{2} \: \{rejected \: as \: its \: not \: an \: integer \}$

So,

First positive integer = x = 4

Second positive integer = x + 1 = 4 + 1 = 5

Third positive integer = x + 2 = 4 + 2 = 6

$\rule{190pt}{2pt}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac

Answer 2

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

(x)² + [(x+1)(x+2) = 46

x² + [x²+2x+1x+2] = 46

x² + [x²+3x + 2] = 46

2x² + 3x + 2 = 46

2x² + 3x = 46 - 2

2x² + 3x = 44

2x² + 3x - 44 = 0

2x² + 11x - 8x - 44 = 0

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

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

x - 4 = 0

x = 4

2x + 11 = 0

2x = -11

x = - 11/2

The numbers are positive integers. Hence, the value of x = 4

Hence, the numbers are :

4, (4+1) & (4+2)

which are 4, 5 and 6.