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bahrom7893 Group TitleBest ResponseYou've already chosen the best response.0
same base, just set the powers equal to each other and solve: 2x+3=x^2 solve for x
 2 years ago

eon.burkett Group TitleBest ResponseYou've already chosen the best response.0
I am having trouble remembering how to do that. I know that it should be two numbers, but can't remember how to get the answer.
 2 years ago

kumar2006 Group TitleBest ResponseYou've already chosen the best response.1
You just need to set the powers up and solve it step by step.
 2 years ago

eon.burkett Group TitleBest ResponseYou've already chosen the best response.0
\[\sqrt{ x ^{2}}=\sqrt{2x+3}\]
 2 years ago

eon.burkett Group TitleBest ResponseYou've already chosen the best response.0
so \[x=\sqrt{2x+3}\]
 2 years ago

eon.burkett Group TitleBest ResponseYou've already chosen the best response.0
what's the next step?
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
Starting with 2x+3=x^2 rearrange: subtract 2x+3 from both sides 0= x^22x3 or x^22x3 = 0 This is a quadratic equation. You can solve it by factoring, or using the quadratic formula (Sometimes factoring is easy, especially if you are in practice. But the quadratic formula always works) In this case, where the coefficient of x^2 is 1, focus on the constant term 3 the minus sign means the two factors will have opposite signs. the factors of 3 are 1,3 (nice and simple) now look at the coefficient on the x term: 2x we need two factors that add to get 2 from our list of 1,3 we don't have much of a choice: 3 and +1 = 2 (the other possibility 1+3 = +2 does not match) so the factorization is (x3)(x+1)=0 (multiply out to check that it is correct) this means either (x3)= 0 or (x+1)=0 solve each: x3= 0, x=3 x+1=0 , x= 1
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
Here's a video on using the quadratic formula http://www.khanacademy.org/math/algebra/quadtratics/v/quadraticformula1 It is easier to watch than read a boring description.
 2 years ago
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