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

3 Answers
Faisal
Mar 30, 2018

#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!

David Drayer
Mar 30, 2018

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 #

Nghi N
Mar 30, 2018

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.

Related questions
Impact of this question
2046 views around the world
You can reuse this answer
Creative Commons License