Question

In a set of 3 numbers the first is a positive integer, the second is 3 more than the first and the 3rd is a square of the second

Answer 1

Given:

In a set of 3 numbers the first is a positive integer, the second is 3 more than the first and the 3rd is a square of the second.

To find:

The equation for the given situation if the sum of the numbers is 77.

Solution:

a. Let the first number in the set is x.

The second is 3 more than the first. So, the second number is $(x+3)$.

The 3rd number is a square of the second. So, the third number is $(x+3)^2$.

Therefore, the first, second and third numbers are $x,(x+3),(x+3)^2$ respectively.

b. The sum of the numbers is 77.

First number + Second number + Third number = 77

c. So, the equation in terms of x is:

$x+(x+3)+(x+3)^2=77$

Therefore, the required equation is $x+(x+3)+(x+3)^2=77$.

d. On simplification, we get

$x+x+3+x^2+6x+9=77$                   $[\because (a+b)^2=a^2+2ab+b^2]$

$x^2+(6x+x+x)+(3+9)=77$

$x^2+8x+12=77$

Subtract 77 from both sides.

$x^2+8x+12-77=77-77$

$x^2+8x-65=0$

Therefore, the simplified form of the required equation is $x^2+8x-65=0$.