How do you solve #-14+ 13x - 3x ^ { 2} = 0#?

2 Answers
Apr 22, 2017

#color(green)(x=7/3# or #color(green)(x=2#

Explanation:

#-14+13x-3x^2=0#

multiply L.H.S and R.H.S. by#-1#

#:.14-13x+3x^2=0#

#:.3x^2-13x+14=0#

#:.(3x-7)(x-2)=0#

#:.3x-7=0 or x-2=0#

#:.3x=7# or #x=2#

#:.color(green)(x=7/3# or #color(green)(x=2#

substitute #color(green)(x=2#

#:.-14+13(color(green)2)-3(color(green)2)^2=0#

#:.-14+26-12=0#

#:.-26+26=0#

substitute #color(green)(x=7/3#

#:.-14+13(color(green)(7/3))-3(color(green)(7/3))^2=0#

#-14+91/3-147/9=0#

#:.(-126+273-147)/9=0#

multiply L.H.S and R.H.S. by#9#

#:.-126+273-147=0#

#:.-273+273=0#

Apr 22, 2017

#x=2" or "x=7/3#

Explanation:

#"multiply through by - 1"#

#rArr3x^2-13x+14=0#

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

#"require the product of factors of 42 which sum to - 13"#

#"these are " -6" and" -7#

#rArr3x^2-6x-7x+14=0#

#"factorise by 'grouping'"#

#color(red)(3x)(x-2)color(red)(-7)(x-2)=0#

#"factor out (x - 2 )"#

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

#x-2=0tox=2#

#3x-7=0tox=7/3#