How do I use a sign chart to solve #2x^2+5x-3>0#?

1 Answer
Jun 30, 2015

Answer:

#f(x)# is positive when #x < -3# and #x > 0.5#

Explanation:

We have #f(x) = 2x^2 + 5x -3#, and we want to find the values of #x# for which #f(x) > 0#.

We start by finding the critical numbers.

Set #f(x) = 2x^2 + 5x -3 =0# and solve for #x#.

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

#2x-1 = 0# or #x+3 = 0#

#2x = 1#

#x = ½ = 0.5# or # x = -3#

The critical numbers are #-3# and #0.5#.

Now we check for positive and negative regions.

We have three regions to consider: (a) #x < -3#; (b) #-3 < x < 0.5#; and (c) #x > 0.5#.

Case (a): Let #x = -4#.

Then #f(4) = 2×(-4)^2 + 5(-4) -3 =32 – 20 -3 = 9#

#f(x)# is positive when #x < -3#.

Case (b): Let #x = 0#.

Then #f(0) = 2×0^2 + 5×0 -3 = 0 + 0 -3 = -3#

#f(x)# is negative when #-3 < x < 0.5#

Case (c): Let #x = 1#.

Then #f(1) = 2×1^2 + 5×1 -3 = 2 +5 – 3 = 4#

#f(x)# is positive when #x > 0.5#.

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Answer: #f(x)# is positive when #x < -3# and #x > 0.5#.