Question #78f69

2 Answers
Nov 2, 2017

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

Explanation:

#6x^2+5x-6=0" is in standard form"#

#"that is "ax^2+bx+c color(white)(x);a!=0#

#"to factorise consider the factors of ac which sum to b"#

#"the factors of - 36 which sum to + 5 are + 9 and - 4"#

#6x^2+9x-4x-6=0larrcolor(blue)"split the middle term"#

#"factorise by 'grouping'"#

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

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

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

#2x+3=0rArrx=-3/2#

#3x-2=0rArrx=2/3#

Nov 2, 2017

#(2/3, 0) and (3/2, 0)#

Explanation:

#6x^2+5x-6=0#
#(3x-2)(2x+3)=0#

Therefore, each factor set equal to #0# gives:
#x = 2/3 or -3/2#

Therefore, the roots are #x=2/3 and x=-1.5.#

In the quadratic equation curve of #y=6x^2+5x-6# the two roots result with #y =0#.

Hence, their coordinates are #(2/3,0) and (-1.5,0).#