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

Mar 30, 2018

$n = 6$
$n = 7$

#### Explanation:

By Sum & Product
=${n}^{2} - 13 n + 42 = 0$

=${n}^{2} - 7 n - 6 n + 42 = 0$

=$n \left(n - 7\right) - 6 \left(n - 7\right) = 0$

=$\left(n - 6\right) \left(n - 7\right) = 0$

=$n - 6 = 0$

=$n = 6$

=$n - 7 = 0$

=$n = 7$

Hope this helps!

Mar 30, 2018

rewrite the equation in the $A {x}^{2} + B x + C = 0$ form and factor the resulting trinomial into two binomials

#### Explanation:

${n}^{2} - 13 n + 42 = 13 n - 13 n - 42 + 42$ gives the correct form

${n}^{2} - 13 n + 42 = 0$

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.

$\left(n - 7\right) \times \left(n - 6\right) = 0$

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

$n = 6$

Mar 30, 2018

6 and 7

#### Explanation:

$y = {n}^{2} - 13 n + 42 = 0$
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.