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

2 Answers
May 2, 2018

-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

-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]}