How do you solve #2x^2+5x-3<=0#?

2 Answers
May 2, 2018

Answer:

#-3<=x<=1/2#

Explanation:

First find the roots of the answer
#2x^2+5x-3#=#(2x-1)(x+3)#
graph{(2x-1)(x+3) [-5.546, 5.55, -2.773, 2.774]}
Since the equation is below the x-axis,
the answer will be between #-3 and 1/2#
Therefore #-3<=x<=1/2#

May 2, 2018

Answer:

#-3<=x<=1/2#

Explanation:

#"factorise the quadratic"#

#(x+3)(2x-1)<=0#

#"solve "(x+3)(2x-1)=0#

#rArrx=-3" or "x=1/2larrcolor(blue)"zeros"#

#"the coefficient of "x^2" term ">0rArruuu#

#rArr-3<=x<=1/2#
graph{2x^2+5x-3 [-10, 10, -5, 5]}