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SSSSS [86.1K]
3 years ago
11

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

Mathematics
1 answer:
Molodets [167]3 years ago
6 0

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.

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