If I add the square of a number to 2 times the number, I get 63. The number can be 7 or -9
Solution:
Given that if i add the square of a number to 2 times the number, i get 63
We have to find the number.
Let the number be "n"
Square of number + 2 times the number = 63
Hence we get,
![n^2 + 2n = 63\\\\n^2 + 2n - 63 = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pmp040txbopm4sba2oe62jp7kudlfr32os.png)
Let us factorize the expression
![n^2 + 2n - 63 = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/mrc31a8gx6vttgz4cu9k8w5xmuczqhws3e.png)
"2n" can be written as "-7n + 9n"
![n^2 -7n + 9n - 63 = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/4zl1a9y4106jyyhtm4ubrfcd8fws64s5ga.png)
Now 63 can be written as "9 x 7"
![n^2 -7n + 9n - (9 * 7) = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/d8ihdb0kv9w158e5t6guc1lya14a1u94eo.png)
Take "n" as common term from two terms and "9" as common from last two term
![n(n - 7) + 9(n - 7) = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/a9n6ki9hu6v5mdftl9a5eq5hj1bf3dqyl3.png)
Now take "n - 7" as common term
![(n-7)(n + 9) = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/7c773fr0spyi5e78seqacm9yqncwp0u6a6.png)
Equating to zero we get,
n - 7 = 0 or n + 9 = 0
n = 7 or n = -9
Hence the number can be 7 or -9