How do you solve #-14+ 13x - 3x ^ { 2} = 0#?
2 Answers
Explanation:
multiply L.H.S and R.H.S. by
substitute
substitute
multiply L.H.S and R.H.S. by
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#