Question 95037: If 2x^2+5x-3=0 and x>0, then what is the value of x?
I'm having trouble with the test booklet explanation.
It indicates to factor (2x-1)(x-3)=0, but for some reason I am having trouble understanding what happens to the 5x. Maybe it is the factoring portion I am drawing a blank on. Please assist me with step by step details. Thank you.
Answer by williacy(2)   (Show Source):
You can put this solution on YOUR website! 2x^2+5x-3=0
First you must factor 2x^2+5x-3:
1. Multiply the first and last coefficients: 2 x -3 = -6
2. Find two multiples of -6 that can be added to get 5.
3. The two multiples are 6 and -1 {6x-1=-6 and 6+(-1)=5}
4. Use the two multiples to help you write two binomials: (x+6)(x+-1)
5. Since there is a 2 in front of the x^2 term, divide the 6 and -1 by the 2.
6. You will have . . . (x + 3)(x + -1/2)
7. Since you should have all whole numbers and not fractions, take the 2 from the denominator and put it in front of the 2, and thus you will have the factored from of 2x^2+5x-3:
(x + 3)(2x + -1)
Now to solve:
1. Set each of the binomials to equal zero, and solve for x.
x+3 = 0 x = -3
2x + -1 = 0
2x = 1
x = 1/2
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