## Soni-kay 3 years ago When 4 inches are added to each side of square, its area increases by 64 square inches. What is the area of the initial square?

1. amruta1997

36

2. AlexUL

\[A=x ^{2}\] with x being the length of the initial sides \[A+64=(x+4)\]

3. amruta1997

the area of previous square was 36 sq. inches

4. AlexUL

\[(x+4)^{2}\] sorry

5. tcarroll010

(x + 4)^2 = x^2 + 64. This will be a quadratic in "x".

6. Soni-kay

how did you get 36 so i can do the rest by myself?

7. amruta1997

it can be done by trial and error also

8. tcarroll010

(x + 4)^2 = x^2 + 64 will become x^2 + 8x + 16 = x^2 + 64. Just combine terms and simplify.

9. Soni-kay

ok one thing why do you square it

10. amruta1997

area of square is its side squared

11. tcarroll010

You have to get the area. Both of the first square and the second.

12. amruta1997

and x is the side

13. Soni-kay

thanks guys

14. tcarroll010

The second area is (x + 4)^2 and the first area is x^2

15. amruta1997

hmm

16. amruta1997

its our pleasure

17. AlexUL

\[A _{1}=x _{1}^{2}\] \[A _{2}=x _{2}^{2}\] \[A _{2}=A _{1}+64\] \[x _{2}=x _{1}+4\] Does that help?

18. Soni-kay

lol i got it

19. Soni-kay

ok so i did x^2 + 8x + 16 = x^2 + 64. and i got 6 what did i do wrong

20. AlexUL

you got \[x _{1}=6\] so what's the area? Never forget what was the initial question