Question

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

Answer 1

Answer:

Step-by-step explanation:

let the 3 consecutive integers be x,x+1,x+2.

x²+(x+1)(x+2)=46.

x²+x²+2x+x+2=46

2x²+3x-44=0

2x²+11x-8x-44=0

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

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

x=4&-11/2

put this value in the integers ..

if x=4

then other are 5,6.

if x=-11/2

then other are -9/2,-7/2

brainlist it

Answer 2

Answer:

Three consecutive positive integers are 4, 5 and 6.

First integer = 4

Second integer = 5

Third integer = 6

Step-by-step explanation:

Let x, (x + 1) and (x + 2) are three consecutive positive integers.

According to question-

The sum of the square of the first and the product of two others is 46.

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

$x^{2} + x^{2} +2x+x+2 = 46$

$2x^{2} + 3x + 2 =46$

$2x^{2} +3x -44 = 0$  

$2x^{2}+(11-8)x - 44 = 0$

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

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

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

If (2x + 11) = 0

Then, 2x = -11

$x = -\dfrac{11}{2}$

If (x - 4) = 0

Then x = 4

Taking x = 4 because the integers are positive.

First integer = x

First integer = 4

Second integer = x + 1

                          = 4 + 1

Second integer = 5

Third integer = x + 2

                      = 4 + 2

Third integer = 6

Hence, three consecutive positive integers are 4, 5 and 6.