SOLUTION: I need help with sloving quadractic equations by factoring: {{{2x^2-3x-5=0}}}

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Question 214983: I need help with sloving quadractic equations by factoring:
2x%5E2-3x-5=0

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
First, let's factor 2x%5E2-3x-5


Looking at the expression 2x%5E2-3x-5, we can see that the first coefficient is 2, the second coefficient is -3, and the last term is -5.


Now multiply the first coefficient 2 by the last term -5 to get %282%29%28-5%29=-10.


Now the question is: what two whole numbers multiply to -10 (the previous product) and add to the second coefficient -3?


To find these two numbers, we need to list all of the factors of -10 (the previous product).


Factors of -10:
1,2,5,10
-1,-2,-5,-10


Note: list the negative of each factor. This will allow us to find all possible combinations.


These factors pair up and multiply to -10.
1*(-10) = -10
2*(-5) = -10
(-1)*(10) = -10
(-2)*(5) = -10

Now let's add up each pair of factors to see if one pair adds to the middle coefficient -3:


First NumberSecond NumberSum
1-101+(-10)=-9
2-52+(-5)=-3
-110-1+10=9
-25-2+5=3



From the table, we can see that the two numbers 2 and -5 add to -3 (the middle coefficient).


So the two numbers 2 and -5 both multiply to -10 and add to -3


Now replace the middle term -3x with 2x-5x. Remember, 2 and -5 add to -3. So this shows us that 2x-5x=-3x.


2x%5E2%2Bhighlight%282x-5x%29-5 Replace the second term -3x with 2x-5x.


%282x%5E2%2B2x%29%2B%28-5x-5%29 Group the terms into two pairs.


2x%28x%2B1%29%2B%28-5x-5%29 Factor out the GCF 2x from the first group.


2x%28x%2B1%29-5%28x%2B1%29 Factor out 5 from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.


%282x-5%29%28x%2B1%29 Combine like terms. Or factor out the common term x%2B1


So 2x%5E2-3x-5 factors to %282x-5%29%28x%2B1%29.


In other words, 2x%5E2-3x-5=%282x-5%29%28x%2B1%29.


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Now let's use this factorization to solve 2x%5E2-3x-5=0


2x%5E2-3x-5=0 Start with the given equation


%282x-5%29%28x%2B1%29=0 Factor the left side (using the factorization from above)



Now set each factor equal to zero:
2x-5=0 or x%2B1=0


x=5%2F2 or x=-1 Now solve for x in each case


So the solutions are x=5%2F2 or x=-1