Question

How do you solve `n^{2} = 13n - 42`?

Answer 1

`n=6`
`n=7`

Explanation:

By Sum & Product
=`n^2-13n+42=0`

=`n^2-7n-6n+42=0`

=`n(n-7)-6(n-7)=0`

=`(n-6)(n-7)=0`

=`n-6=0`

=`n=6`

=`n-7=0`

=`n=7`

Hope this helps!

Answer 2

rewrite the equation in the `Ax^2 + Bx + C = 0` form and factor the resulting trinomial into two binomials

Explanation:

`n^2 -13n + 42 = 13n -13 n - 42 + 42` gives the correct form

`n^2 - 13n + 42 = 0`

The C term is positive so both binomial factors must be the same

The B terms is negative so both binomial factors must be negative

The sum of the binomial must equal 13 so find factors of 42

42 x 1
21 x 2
14 x 3
7 x 6

7 and 6 add to 13 so ( 7 and 6) are the correct set of factors.

`( n-7) xx ( n-6) = 0`

Solving for each binomial gives the answers.

`n -7 = 0` add 7 to both sides

`n-7 +7 = 0 +7` one answer is

`n = 7`

`n -6 = 0` add 6 to both sides

`n- 6 + 6 = 0 +6` the other answer is

`n = 6`

Answer 3

6 and 7

Explanation:

`y = n^2 - 13n + 42 = 0`
The 2 real roots have same positive sign (ac > 0, and ab < 0).
Find 2 real roots, both positive, knowing their sum (-b = 13), and their product (c = 42). They are 6 and 7.

Note . This method is simple and fast. It avoids doing factoring by grouping and solving the 2 binomials.