How do you solve #6x^2+5x=1#?

2 Answers
Apr 29, 2017

Re-arranging we get:

#6x^2 + 5x - 1 = 0#

We can write this as:

#6x^2 + 6x - x - 1 = 0#

i.e. #6x(x + 1) - (x + 1) = 0#

=> #(6x - 1)(x + 1) = 0#

so, #x = 1/6 or x = -1#

:)>

Apr 29, 2017

Answer:

#x=-1" or " x=1/6#

Explanation:

#"subtract 1 from both sides and equate to zero"#

#6x^2+5x-1=1-1#

#rArr6x^2+5x-1=0#

#"using the a-c method of factorising"#

#"require the factors of the product " -6" which sum to +5"#

#"the factors are " +6, -1#

#rArr6x^2+6x-x-1=0larr" split 5x"#

#rArr6x(x+1)-1(x+1)=0larr" factor pairs"#

#(x+1)(6x-1)=0larr" factor out (x+1)"#

#(x+1)=0rArrx=-1#

#(6x-1)=0rArrx=1/6#

#rArrx=-1" or " x=1/6" are the solutions"#