How do you solve #21n ^ { 2} + 5n - 6= 0#?

2 Answers

Answer:

#n=3/7, -2/3#

Explanation:

Given:
#color(white)(xxx)21n^2+5n-6=0#

#rArr 21n^2-9n+14n-6=0# [Broke up #5n# as #-9n +14n#]

#rArr 3n(7n-3)+2(7n-3)=0# [Grouping the Like Terms]

#rArr (7n - 3)(3n + 2)=0# [Group Again]

Now, If :

#3n+2=0#, then #3n=-2 rArr n= -2/3#

And If :

#7n-3=0#, then #7n=3 rArr n = 3/7#

Hence Explained.

Mar 28, 2018

Answer:

#n=-2/3" or "n=3/7#

Explanation:

#"factorise the quadratic using the a-c method"#

#"that is factors of "21xx-6=-126" which sum to "+5#

#"the factors of "-126" which sum to + 5 are + 14 and - 9"#

#color(blue)"split the middle term using these factors"#

#rArr21n^2+14n-9n-6=0larrcolor(blue)"factor by grouping"#

#rArrcolor(red)(7n)(3n+2)color(red)(-3)(3n+2)=0#

#"take out the common factor "(3n+2)#

#rArr(3n+2)(color(red)(7n-3))=0#

#"equate each factor to zero and solve for n"#

#3n+2=0rArrn=-2/3#

#7n-3=0rArrn=3/7#