Raymond B. answered 05/21/23
Math, microeconomics or criminal justice
2x^2 -5x+3 =0
factor
(2x-3)(x-1)=0
x-1=0, 2x-3=0
x =1 or 3/2
use the quadratic formula
x=5/4 +/-(1/4)sqr(25-4(2)(3))
= (5+/-1)/4 = 6/4, 4/4
= 1.5, or 1
complete the square
2(x^2-5x/2) + 3 =0
2(x^2-5x/2 +25/16) = -3 + (25/16)2
2(x-5/4)^2 = (50-48)/16
(x-5/4)^2 = 1/16
x-5/4 = +/-1/4
x = 5/4 +/-14
x = 6/4. 4/4
x = 3/2, 1
or use a graphing calculator
and see where the graph intersects
the x axis
the equation solutions are the same
as the x intercepts, zeros or roots
or use an online equation calculator
it's often useful to solve the problem with more than one method to check your answers
if you get 2 different answers, at least one of them is wrong
y=2x^2-5x +3 is an upward opening parabola since the x^2 coefficient >0
y'(x) = 4x-5=0, x=5/4 = x coordinate of the vertex = minimum point
y coordinate of the vertex = 2(5/4)^2-5(5/4)+3 = -1/16
in vertex form it's
2(x -5/4)^2 -1/16, vertex=(5/4, -1/16) = minimum point
the x coordinate also is the axis of symmetry, x=5/4 = 1.25
the x intercepts are equidistant to the axis of symmetry, 5/4-1 = 3/2-5/4 = 1/4
using descartes rule, you know there is a maximum 2 positive solutions
as the sign changes twice from 2x^2 to -5x to +3
replace x with -x to see the maximum number of negative solutions
2(-x)^2 -5(-x) +3 has no sign changes, therefore there are zero negative solutions
either there are 2 positive real solutions or there are 2 imaginary solutions
for this specific example, there are 2 real positive and 0 negative and 0 imaginary solutions