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 [tex](x+3)[/tex].
The 3rd number is a square of the second. So, the third number is [tex](x+3)^2[/tex].
Therefore, the first, second and third numbers are [tex]x,(x+3),(x+3)^2[/tex] 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:
[tex]x+(x+3)+(x+3)^2=77[/tex]
Therefore, the required equation is [tex]x+(x+3)+(x+3)^2=77[/tex].
d. On simplification, we get
[tex]x+x+3+x^2+6x+9=77[/tex] [tex][\because (a+b)^2=a^2+2ab+b^2][/tex]
[tex]x^2+(6x+x+x)+(3+9)=77[/tex]
[tex]x^2+8x+12=77[/tex]
Subtract 77 from both sides.
[tex]x^2+8x+12-77=77-77[/tex]
[tex]x^2+8x-65=0[/tex]
Therefore, the simplified form of the required equation is [tex]x^2+8x-65=0[/tex].
