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find the set of values of x for which 2x^2+3x-5 > or = 0

 Apr 16, 2016

Best Answer 

 #1
avatar
+5

Solve for x:
2 x^2+3 x-5 = 0

The left hand side factors into a product with two terms:
(x-1) (2 x+5) = 0

Split into two equations:
x-1 = 0 or 2 x+5 = 0

Add 1 to both sides:
x = 1 or 2 x+5 = 0

Subtract 5 from both sides:
x = 1 or 2 x = -5

Divide both sides by 2:
Answer: |  x = 1                     or                 x = -5/2

 Apr 16, 2016
 #1
avatar
+5
Best Answer

Solve for x:
2 x^2+3 x-5 = 0

The left hand side factors into a product with two terms:
(x-1) (2 x+5) = 0

Split into two equations:
x-1 = 0 or 2 x+5 = 0

Add 1 to both sides:
x = 1 or 2 x+5 = 0

Subtract 5 from both sides:
x = 1 or 2 x = -5

Divide both sides by 2:
Answer: |  x = 1                     or                 x = -5/2

Guest Apr 16, 2016
 #2
avatar+37146 
+5

Now if you graph the function you will find that guest1 got the correct answers for where the function EQUALS zero.   The graph will show you that for 1  < = x  <= -5/2   the value will be GREATER than or EQUAL to  ZERO

 Apr 16, 2016

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