SOLUTION: Ok I need some help square 2x^2+5x+6=x If I subtract x I get 2x^2+4x+6=0 If I do that then it can't be factored right?

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Question 133080: Ok I need some help square 2x^2+5x+6=x If I subtract x I get 2x^2+4x+6=0 If I do that then it can't be factored right?
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
Does your equation look like this: sqrt%282x%5E2%2B5x%2B6%29=x ???


sqrt%282x%5E2%2B5x%2B6%29=x Start with the given equation


2x%5E2%2B5x%2B6=x%5E2 Square both sides


2x%5E2%2B5x%2B6-x%5E2=0 Subtract x%5E2 from both sides


x%5E2%2B5x%2B6=0 Combine like terms



%28x%2B3%29%28x%2B2%29=0 Factor the left side (note: if you need help with factoring, check out this solver)



Now set each factor equal to zero:
x%2B3=0 or x%2B2=0

x=-3 or x=-2 Now solve for x in each case


So our possible solutions are

x=-3 or x=-2



However, we must check our answers:

Let's check the first possible solution x=-3

sqrt%282x%5E2%2B5x%2B6%29=x Start with the given equation



sqrt%282%28-3%29%5E2%2B5%28-3%29%2B6%29=-3 Plug in x=-3


sqrt%2818-15%2B6%29=-3 Square and multiply


sqrt%289%29=-3 Combine like terms


3=-3 Take the square root of 9 to get 3


Since the two sides are clearly not equal, this means that x=-3 is an extraneous solution. In other words, x=-3 is not a real solution.


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Let's check the first possible solution x=-2

sqrt%282x%5E2%2B5x%2B6%29=x Start with the given equation



sqrt%282%28-2%29%5E2%2B5%28-2%29%2B6%29=-2 Plug in x=-2


sqrt%288-10%2B6%29=-2 Square and multiply


sqrt%284%29=-2 Combine like terms


2=-3 Take the square root of 4 to get 2


Since the two sides are clearly not equal, this means that x=-2 is an extraneous solution. In other words, x=-2 is not a real solution.



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Answer:


So this shows us that sqrt%282x%5E2%2B5x%2B6%29=x does not have any solutions.

Answer by solver91311(12126) About Me  (Show Source):
You can put this solution on YOUR website!
Well, that is not precisely correct. ALL quadratic trinomials can be factored, just not necessarily to integers, rationals, or even real numbers. In other words, for any second degree polynomial ax%5E2%2Bbx%2Bc there exist two numbers p and q such that %28x-p%29%28x-q%29=ax%5E2%2Bbx%2Bc. It is just that in the case of your equation, 2x%5E2%2B4x%2B6=0, p and q are a conjugate pair of complex numbers, namely -1%2B-i%2Asqrt%282%29, where i%5E2=-1.

That means your factors would be %28x-%28-1%2Bi%2Asqrt%282%29%29%29%28x-%28-1-i%2Asqrt%282%29%29%29

Having said all of that, I suspect you meant that the polynomial cannot be factored over the integers. That is an absolutely correct statement.